Class 10

Haryana State Board HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations :
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (- 2) (3 – x)
(iii (x – 2)(x + 1) = (x – 1) (x + 3)
(iv) (x – 3)(2x + 1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
(vi) x² + 3x + 1 = (x – 2)²
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³.
Solution :
(i) The given equation is (x + 1)² = 2(x – 3)
⇒ x² + 2x + 1 = 2x – 6
⇒ x² + 2x + 1 – 2x + 6 = 0
⇒ x² + 7 = 0
⇒ 1.x² + 0.x + 7 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(ii) The given equation is :
x² – 2x = (- 2)(3 – x)
⇒ x² – 2x = – 6 + 2x
⇒ x² – 2x – 2x + 6 = 0
⇒ x² – 4x + 6 = 0
⇒ 1.x² + (- 4).x + 6 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(iii) The given equation is :
(x – 2)(x + 1) = (x – 1)(x + 3)
⇒ x² + x – 2x – 2 = x² + 3x – x – 3
⇒ x² – x – 2 = x² + 2x – 3
⇒ x² – x – 2 – x² – 2x + 3 = 0
⇒ – 3x + 1 = 0
⇒ 0.x² + (- 3) x + 1 = 0
∵ a = 0
∴ It is not of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is not a quadratic equation.

(iv) The given equation is :
(x – 3) (2x + 1) = x (x + 5)
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 = x² + 5x
⇒ 2x² – 5x – 3 = x² – 5x = 0
⇒ x² – 10x – 3 = 0
⇒ 1.x² + (- 10).x + (- 3) = 0
It is of the form ax² + bx + c = 0 where a ≠ 0.
Hence, the given equation is a quadratic equation.

(v) The given equation is :
(2x – 1)(x – 3) = (x + 5)(x- 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² – 7x + 3 = x² + 4x – 5
⇒ 2x² – 7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
⇒ 1.x² + (- 11)x + 8 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

(vi) The given equation is :
x² + 3x + 1= (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ x² + 3x + 1 – x² + 4x – 4 = 0
⇒ 7x – 3 = 0
⇒ 0.x² + 7.x + (- 3) = 0
∵ a = 0
∴ It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(vii) The given equation is :
(x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x
⇒ x³+ 6x² + 12x + 8 – 2x³ + 2x = 0
⇒ – x³ + 6x² + 14x + 8=0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

(viii) The given equation is :
x³ – 4x² – x +1= (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ x³– 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
⇒ 2x² – 13x + 9 = 0
⇒ 2.x² + (- 13).x + 9 = 0
It is of the form ax² + bx + c = 0, where a ≠ 0.
Hence, the given equation is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution :
(i) Let the breadth of plot he x m.
Then length = (2x + 1) m. According to question,
x(2x + 1) = 528
[∵ Area of rectangle = Length x Breadth]
⇒ 2x² + x = 528
⇒ 2x² + x – 528 = 0
Hence, the quadratic equation is 2x² + x – 528 = 0.

(ii) Let the two consecutive integers be x and (x + 1)
According to question, x(x+ 1) = 306
⇒ x² + x = 306
⇒ x² + x – 306 = 0
Hence, the quadratic equation is x² + x – 306 = 0.

(iii) Let the Rohan’s present age be x years.
Then, his mother’s present age be (x + 26) years.
After three years, age of Rohan = (x + 3) years
After three years, age of his mother = x + 26 + 3
= (x + 29) years
According to question,
(x + 3)(x + 29) = 360
⇒ x² + 29x +3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
Hence, the quadratic equation is x² + 32x – 273 = 0.

(iv) Let the speed of the train be x km/h
Time taken to travel 480 km = \(\frac{480}{x}\) hours
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
If speed is 8 km/h less then new speed = (x – 8) km/h
Time taken to travel 480 km = \(\frac{480}{x-8}\) hours
According to question,
\(\frac{480}{x-8}-\frac{480}{x}\) = 3
⇒ \(\frac{480 x-480(x-8)}{x(x-8)}\) = 3
⇒ 480x – 480x + 480 x 8 = 3x (x – 8)
⇒ 3840 = 3x² – 24x
⇒ 3x² – 24x – 3840 = 0
⇒ x² – 8x – 1280 = 0
Hence, the quadratic equation is x² – 8x – 1280 = 0.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two Mends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution :
Let the present age of Ani be x years.
and present age of Biju be y years.
Therefore, Dharam’s age = 2x years
Cathy’s age = \(\frac{y}{2}\) years
According to question,
x – y = 3 …………………(1)
– x + y = 3 ………………….(2)
also 2x – \(\frac{y}{2}\) = 30
On multiplying both sides by 2
⇒ 4x – y = 60 …………….(3)
On subtracting equation (1) from (3)

On substituting the value of x in equation (1), we get
19 – y = 3
⇒ y = 19 – 3 = 16
So, x = 19, y = 16 is the required solution.
On adding equation (2) and (3)

On substituting the value of x in (2), we get
– 21 + y = 3 .
y = 3 + 21 = 24
So, another required solution is x = 21, y = 24
Hence, present age of Ani is 19 years and present age of Biju is 16 years or present age of Ani is 21 years and present age of Biju is 24 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital?
Solution:
Let the capital one has be ₹ x
and the capital other has be ₹ y
According to question,
x + 100 = 2(y – 100)
⇒ x + 100 = – 200
x – 2y = – 200 – 100
x – 2y = – 300 …………….(1)
and y + 10 = 6(x – 10)
y + 10 = 6x – 60
6x – y = 10 + 60
6x – y = 70 ……………..(2)
Multiplying equation (2) by 2, we get
12x – 2y = 140 ………………..(3)
Subtracting equation (3) from (1), we get

Substituting the value of x in equation (1), we get
40 – 2y = – 300
⇒ – 2y = – 300 – 40
⇒ – 2y = – 340
⇒ y = \(\frac{-340}{-2}\) = 170
So, x = 40, y = 170 is the required solution.
Hence, the capital of one be ₹ 40.
and the capital of other be ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/hr faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 kmlhr, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the speed of a train be x km/hr
and time taken by a train be y hours.
Then distance covered by a train = xy km
Since, distance covered by train in each case is same.
Therefore, according to question,
(x + 10) (y – 2) = xy
[∵ Distance = Speed × Time]
⇒ xy – 2x + 10y – 20 = xy
⇒ – 2x + 10y – 20 = 0
⇒ – 2x + 10y = 20
and (x – 10) (y + 3) = xy
⇒ xy + 3x – 10y – 30 = xy
⇒ 3x – 10y – 30 = 0
⇒ 3x – 10y = 30
Adding equation (1) and equation (2), we get

Substituting the value of x in equation (2), we get
3 × 50 – 10y = 30
150 – 10y = 30
– 10y = 30 – 150 = – 120.
y = \(\frac{-120}{-10}\) = 12
So, x = 50, y = 12 is the required solution.
Hence, distance covered by a train = xy = 50 × 12 = 600 km.

Question 4.
The students of a class are made to stands in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less min a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of rows be x and
number of students in each row be y.
Then total students in the class = xy
According to question,
(y + 3) (x – 1) = xy
⇒ xy – y + 3x – 3 = xy
⇒ 3x – y – 3 = 0
⇒ 3x – y = 3 ……………….(1)
and (y – 3) (x + 2) = xy
⇒ xy + 2y – 3x – 6 = xy
⇒ – 3x + 2y – 6 = 0
⇒ – 3x + 2y = 6 ………………..(2)
Adding equation (1) and equation (2), we get

Substituting the value of y in equation (1), we get
3x – 9 = 3
3x = 3 + 9= 12
x = \(\frac{12}{3}\) = 4
So, x = 4, y = 9 is the required solution.
Hence, total students in the class = xy = 4 × 9 = 36.

Question 5.
In a ∆ABC, ∠A = 3 ∠B = 2 (∠A + ∠B). Find the three angles.
Solution:
Let ∠A = x°, ∠B = y°
∵ ∠C = 3∠B (given)
∴ ∠C = 3y°
also 3∠B = 2(∠A + ∠B) (given)
⇒ 3y° = 2(x° + y°)
⇒ 3y = 2x° + 2y°
⇒ y° = 2x°
⇒ 2x°- y° = 0° …………………(1)
Since, the sum of ∠s of a triangle is 180°.
∠A + ∠B + ∠C = 180°
⇒ x° + y° + 3z° = 180°
⇒ x° + 4y° = 180° ………………(2)
Multiplying equation (1) by 4 and adding it to equation (2), we get

Substituting the value of x in equation (1), we get
2 × 20° – y° = 0°
⇒ 40° – y° = 0°
⇒ y° = 40°
So, x = 20°, y = 40° is the required solution.
Hence, ∠A = 20°,
∠B = 40°,
∠C = 3 × 40° = 120°.

Question 6.
Draw the graphs of the equations 5x -y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
We draw the graphs of given equa-tions as follows :
5x – y =5
⇒ y = 5x – 5
Putting the different values of x in above equation.
We get different values of y and we prepare the table of x, y.

and 3x – y =3
y = 3x – 3
Putting the different values of x in above equation.
We get different values of y and we prepare the table of x, y.

Plotting the points (0, – 5), (2, 5) and (3, 10) on a graph paper and draw a graph which passes through these points.
So, we obtain the graph of equation 5x – y = 5.
Again plotting the points (0, – 3), (2, 3) and (3, 6) on the same graph paper and draw a graph which passes through these points.
So, we obtain the graph of equation 3x – y = 3.

We observe that graphs of the given equations intersects each other at point A and at y-axis at points B and C respectively. So, the coordinates of the triangle so formed are A (1, 0), B (0, – 3) and C (0, – 5).

Question 7.
Solve the following pair of linear equations :
(i) px + qy = p – q
qx – py = p + q

(ii) ax + by = c
bx + ay = 1 + c

(iii) \(\frac{x}{a}-\frac{y}{b}\) = 0
ax + by = a² + b²

(iv) (a – b)x + (a + b)y = a² – 2ab – b²
(a + b)(x + y) = a² + b²

(v) 152x – 378y = – 74
– 378x + 152y = – 604
Solution :
(i) The given equations are :
px + qy = p – q …………………..(1)
qx – py = p + q ………………….(2)
Multiplying equation (1) by b and equation (2) by a and adding them, we get

Substituting the value of x in equation (1), we get

p × 1 + qy = p – q
qy = p – q – p
qy = – q
y = – \(\frac{q}{q}\) = – 1
Hence, x = 1, y = – 1 is the required solution.

(ii) The given equations are:
ax + by = c ………………..(1)
bx + ay = 1 + c
Multiplying equation (1) by b and equation (2) by a and subtracting them, we get

Substituting the value of y in equation (1), we get
ax + b × \(\left[\frac{c(a-b)+a}{a^2-b^2}\right]\) = c
ax + \(\frac{a b c-b^2 c+a b}{a^2-b^2}\) = c
⇒ a³x – ab²x + abc – b²c + ab = a²c – b²c
⇒ a(a² – b²)x = a²c – b²c – abc + b²c – ab
⇒ a(a² – b²)x = a²c – abc – ab
⇒ a(a² – b²)x = a(ac – bc – b)
⇒ a(a² – b²)x = a[c(a – b)- bi
⇒ x = \(\frac{a[c(a-b)-b]}{a\left(a^2-b^2\right)}\)
⇒ x = \(\frac{c(a-b)-b}{a^2-b^2}\)
Hence, x = \(\frac{c(a-b)-b}{a^2-b^2}\), y = \(\frac{c(a-b)+a}{a^2-b^2}\) the required solution.

(iii) The given equations are:
\(\frac{x}{a}-\frac{y}{b}\) = 0
bx – ay = 0 ……………….(1)
and ax + by = a² + b² …………………..(2)
Multiplying equation (1) by b and equation (2) by a and adding them, we get
b²x – aby = 0
a²x + aby = a(a² + b²)
(a² + b²) x = a (a² + b²)
⇒ x = \(\frac{a\left(a^2+b^2\right)}{\left(a^2+b^2\right)}\)
⇒ x = a
Substituting the value of x in equation (1), we get
b × a – ay = 0
– ay = – ab
y = \(\frac{-a b}{-a}\) = b
Hence, x = a, y = b is the required solution.

(iv) The given equations are
(a – b) x + (a + b) y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
(a + b) x + (a + b) y = a² + b²
Multiplying equation (1) by (a + b) and equation (2) by (a – b) and subtracting them, we get

(a² + b² + 2ab – a² + b²) y = a³ – 2a²b – ab² + a²b – 2ab² – b³ – (a³ + ab² – ab – b³)
(2b² + 2ab) y = a³ – a²b – 3ab² – b³ – a³ – ab² + a²b + b³
2b(a + b)y = – 4ab²
y = \(\frac{-4 a b^2}{2 b(a+b)}=\frac{-2 a b}{a+b}\)
Substituting the value of y in equation (2), we get
(a + b)x + (a + b)\(\left(-\frac{2 a b}{a+b}\right)\) = a² + b²
(a + b)x – 2ab = a² + b²
(a + b)x = a² + b² + 2ab
= (a + b)x = (a + b)²
x = \(\frac{(a+b)^2}{(a+b)}\) = a + b
Hence, x = a + b, y = \(-\frac{2 a b}{a+b}\) is the required solution.

(v) The given equations are :
152x – 378y = – 74 ……………(1)
– 378x + 152y = – 604 …………..(2)
Adding equation (1) and equation (2), we get

⇒ x + y = 3 …………….(3)
Subtracting equation (2) from equation (1) we get

Substituting the value of x in equation (3), we get
2 + y = 3
y = 3 – 2 = 1
Hence, x = 2, y = 1 is the required solution.

Question 8.
ABCD is a cyclic quadrilateral (See Figure). Find the angles of the cyclic quadri lateral.

Solution:
Since, sum of opposite angles of cyclic quadrilateral is 180°.
∴ ∠A + ∠C = 180°
⇒ 4y° + 20 – 4x° = 180°
⇒ 4x° + 4y° = 180° – 20°
⇒ – 4x° + 4y° = 160°
⇒ – x° + y° = 40°
and ∠B +∠D = 180°
3y° – 5 – 7x° + 5 = 180°
– 7x° + 3y° = 180°
Multiplying equation (1) by 7 and subtracting equation (2) from it, we get

4y° = 100°
y° = \(\frac{100}{4}\) = 25°
Substituting the value ofy in equation (1), we get
– x° + 25 = 40°
– x° = 40° – 25° = 15°
x° = 15°
So, x° = – 15, y° = 25 is the required solution.
Hence,
∠A = 4° × 25° + 20° = 120°,
∠B = 3° × 25° – 5° = 70°,
∠C = – 4 × (- 15°) = 60°,
∠D = – 7 × (- 15°) + 5 = 110°.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

Question 1.
Solve the following pairs of linear equations by reducing them to a pair of linear equations:

(i) \(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = – 1

(iii) \(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23

(iv) \(\frac{5}{x-1}+\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1

(v) \(\frac{7 x-2 y}{x y}\) = 5
\(\frac{8 x+7 y}{x y}\) = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = – 2

(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)

Solution:
(i) The given equations are:
\(\frac{1}{2 x}+\frac{1}{3 y}\) = 2 …………….(1)
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\) ………………(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
∴ \(\frac{1}{2}\) a + \(\frac{1}{3}\) b = 2 ……………(3)
and \(\frac{1}{3}\) a + \(\frac{1}{2}\) b = \(\frac{13}{6}\) ……………(4)
Multiplying equation (3) by 4 and equation (4) by 6, we get
2a + \(\frac{4}{3}\)b = 8 ………….(5)
2a + 3b = 13 …………..(6)
Subtracting equation (6) from equation (5), we get
2a+ \(\frac{4}{3}\) b = 8

Substituting the value of b in equation (6), we get
2a + 3 × 3 = 13
⇒ 2a + 9 = 13
⇒ 2a = 13 – 9 = 4
⇒ a = \(\frac{4}{2}\) = 2
⇒ \(\frac{1}{x}\) = 2 [∵ a = \(\frac{1}{x}\)]
⇒ x = \(\frac{1}{2}\)
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{3}\) is the required solution.

(ii) The given equations are:
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2 …………….(1)
\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}\) = – 1 ……………..(2)
Let \(\frac{1}{\sqrt{x}}\) = a, \(\frac{1}{\sqrt{y}}\) = b;
∴ 2a + 3b = 2 ……………..(3)
and 4a – 9b = – 1 ………………..(4)
Multiplying equation (3) by 3, we get
6a + 9b = 6 ……………(5)
Adding equation (5) and equation (4), we get

⇒ x = 4 [Squaring on both sides]
Substituting the value of a in equation (3), we get
2 × \(\frac{1}{2}\) + 3b = 2
⇒ 1 + 3b = 2
⇒ 3b = 2 – 1 = 1
⇒ b = \(\frac{1}{3}\)
⇒ \(\frac{1}{\sqrt{y}}=\frac{1}{3}\) [∵ b = \(\frac{1}{\sqrt{y}}\)]
⇒ √y = 3
⇒ y = 9 [Squaring on both sides]
Hence, x = 4, y= 9 is the required solution.

(iii) The given equations are :
\(\frac{4}{x}\) + 3y = 14 ……………….(1)
\(\frac{3}{x}\) – 4y = 23 ………………..(2)
Let \(\frac{1}{x}\) = a,
4a + 3y = 14 ………………(3)
3a – 4y = 23 ……………….(4)
Multiplying equation (3) by 4 and equation (4) by 3, we get
16a + 12y = 56 ……………….(5)
9a – 12y = 69 ……………….(6)
Adding equation (5) and equation (6), we get

⇒ a = 5
⇒ \(\frac{1}{x}\) = 5
⇒ x = \(\frac{1}{5}\)
Substituting the value of a in equation (3), we get
4 × 5 + 3y = 14
⇒ 20 + 3y = 14
⇒ 3y = 14 – 20 = – 6
⇒ y = \(-\frac{6}{3}\) = – 2
Hence, x = \(\frac{1}{5}\), y = – 2 is the required solution.

(iv) The given equations are:
\(\frac{5}{x-1}+\frac{1}{y-2}\) = 2 ……………..(1)
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1 ……………..(2)
Let \(\frac{1}{x-1}\) = p, \(\frac{1}{y-2}\) = q
∴ 5p + q = 2 ……………….(3)
and 6p – 3q = 1 ……………..(4)
Multiplying equation (3) by 3, we get
15p + 3q = 6 ………………..(5)
Adding equation (5) and equation (4), we get

x – 1 = 3
x = 3 + 1 = 4
Substituting the value of p in equation (3), we get
5 × \(\frac{1}{3}\) + q = 2
\(\frac{5}{3}\) + q = 2
q = 2 – \(\frac{5}{3}\)
q = \(\frac{4}{4}\)
\(\frac{1}{y-2}=\frac{1}{3}\) [∵ q = \(\frac{1}{y-2}\)]
y – 2 = 3
y = 3 + 2 = 5
Hence, x = 4, y = 5 is the required solution.

(v) The given equations are:

– 2p + 7q = 5 ………………….(3)
7p + 8q = 15 …………………..(4)
Multiplying equation (3) by 7 and equation (4) by 2, we get
– 14p + 49q = 35 ………………(5)
14p + 16q = 30 ………………….(6)
Adding equation (5) and equation (6), we get

⇒ y = 1
Substituting the value of q in equation (4), we get
7p + 8 × 1 = 15
⇒ 7p = 15 – 8 = 7
⇒ p = \(\frac{7}{7}\) = 1
⇒ \(\frac{1}{x}\) = 1 [∵ p = \(\frac{1}{x}\)]
⇒ x = 1
Hence, x = 1, y = 1 is the required solution.

(vi) The given equations are :
6x + 3y = 6xy
⇒ \(\frac{6 x}{x y}+\frac{3 y}{x y}=\frac{6 x y}{x y}\)
[Dividing both sides of the equation by xy]
\(\frac{6}{y}+\frac{3}{x}\) = 6
\(\frac{3}{x}+\frac{6}{y}\) = 6 ………………(1)
and 2x + 4y = 5xy
⇒ \(\frac{2 x}{x y}+\frac{4 y}{x y}=\frac{5 x y}{x y}\)
[Dividing both sides of the equation by xy]
⇒ \(\frac{2}{y}+\frac{4}{x}\) = 5
⇒ \(\frac{4}{x}+\frac{2}{y}\) = 5 …………..(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b
∴ 3a + 6b = 6
⇒ a + 2b = 2 ……………(3)
and 4a + 2b = 5 ……………….(4)
Subtracting equation (4) from (3), we get

⇒ x = 1
Substituting the value of a in equation (3), we get
⇒ 1 + 2b = 2
2b = 2 – 1 = 1
⇒ b = \(\frac{1}{2}\)
⇒ \(\frac{1}{y}=\frac{1}{2}\) [∵ b = \(\frac{1}{y}\)]
⇒ y = 2
Hence, x = 1, y = 2 is the required solution.

(vii) The given equations are :
\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4 ……………….(1)
and \(\frac{15}{x+y}-\frac{5}{x-y}\) = – 2 ……………(2)
Let \(\frac{1}{x+y}\) = a, \(\frac{1}{x-y}\) = b
10a + 2b = 4
⇒ 5a + b = 2 ……………….(3)
and 15a – 5b = – 2 ……………..(4)
Adding equation (5) and equation (4), we get

⇒ x + y = 5
Substituting the value of a in equation (4), we get
15 × \(\frac{1}{5}\) – 5b = – 2
⇒ 3 – 5b = – 2
⇒ – 5b = – 2 – 3 = – 5
⇒ b = \(\frac{-5}{-5}\) = 1
⇒ \(\frac{1}{x-y}\) = 1
⇒ x – y = 1 …………………..(7)
Adding equation (6) and equation (7), we get

Substituting the value of x in equation (6), we get
3 + y = 5
⇒ y = 5 – 3 = 2
Hence, x = 3, y = 2 is the required solution.

(viii) The given equations are:
\(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ………………..(1)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) …………………(2)
Let \(\frac{1}{3 x+y}\) = a, \(\frac{1}{3 x-y}\) = b
∴ a + b = \(\frac{3}{4}\) ……………(3)
and \(\frac{1}{2} a-\frac{1}{2} b=-\frac{1}{8}\) …………….(4)
Multiplying equation (4) by 2, we get
a – b = – \(\frac{2}{8}\)
or a – b = – \(\frac{1}{4}\) ………………(5)
Adding equation (3) and equation (5), we get

⇒ 3x + y = 4 ………………(6)
Substituting the value of a in equation (3), we get
⇒ \(\frac{1}{4}\) + b = \(\frac{3}{4}\)
⇒ b = \(\frac{3}{4}-\frac{1}{4}=\frac{2}{4}\)
⇒ b = \(\frac{1}{2}\)
⇒ \(\frac{1}{3 x-y}=\frac{1}{2}\) [∵ b = \(\frac{1}{3 x-y}\)]
⇒ 3x – y = 2 ……………….(7)
Adding equation (6) and equation (7), we get

Substituting the value of x in equation (6), we get
3 × 1 + y = 4
y = 4 – 3 = 1
y = 1
Hence, x = 1, y = 1 is the required solution.

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
(i) Let speed of her rowing in still water be x km/hr.
and speed of current be y km/hr.
Speed of rowing downstream= (x + y) km/hr.
Speed of rowing upstream = (x – y) km/hr.
According to question,
\(\frac{20}{x+y}\) = 2 [∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
⇒ 2x + 2y = 20
⇒ x + y = 10 ……………..(1)
and \(\frac{4}{x-y}\) = 2
2x – 2y = 4
⇒ x – y = 2 ……………..(2)
Adding equation (1) and equation (2), we get

Substituting the value of x in equation (1), we
6 + y = 10
⇒ y = 10 – 6 = 4
Hence, the equations are : x + y = 10, x – y = 2 and x = 6, y = 4 is the required solution.
Therefore,speed of her rowing in still water = 6 km/hr and speed of current = 4 km/hr.

(ii) Let time taken by 1 woman to finish the embroidery work = x days
and time taken by 1 man to finish the embroidery work = y days
1 woman’s 1 day’s work = \(\frac{1}{x}\)
2 women’s 1 day’s work = \(\frac{2}{x}\)
and 1 man’s 1 day’s work = \(\frac{1}{y}\)
5 men’s 1 day’s work = \(\frac{5}{y}\)
Since 2 women and 5 men finish the work in 4 days.
∴ \(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\)
⇒ \(\frac{8}{x}+\frac{20}{y}\) = 1 ……………..(1)
Again 3 women and 6 men finish the work in 3 days.
∴ \(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\)
⇒ \(\frac{9}{x}+\frac{18}{y}\) = 1 ……………..(2)
Let \(\frac{1}{x}\) = a, \(\frac{1}{y}\) = b,
∴ 8a + 20b = 1 ……………….(3)
and 9a + 186 = 1 ……………..(4)
Multiplying equation (3) by 9 and equation (4) by 8, we get
72a + 180b = 9 ………………..(5)
72 a + 144b = 8 ………………..(6)
Subtracting equation (6) from equation (5), we get

⇒ y = 36
Substituting the value of b in equation (3), we get
⇒ 8a + 20 × \(\frac{1}{36}\) = 1
⇒ 8a = 1 – \(\frac{20}{36}\)
⇒ 8a = \(\frac{36-20}{36}=\frac{16}{36}\)
⇒ a = \(\frac{16}{36 \times 8}=\frac{2}{36}\)
⇒ a = \(\frac{1}{18}\)
⇒ \(\frac{1}{x}=\frac{1}{18}\) [∵ a = \(\frac{1}{x}\)]
⇒ x = 18
Hence, the equations are :\(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\), \(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) and x = 18, y = 36 is the required solution.
Therefore, 1 woman can finish the embroidery work in 18 days and 1 man can finish the embroidery work in 36 days.

(iii) Let the speed of train be x km/hr and speed of bus be y km/hr.
According to question,
\(\frac{60}{x}+\frac{(300-60)}{y}\) = 4
[∵ Time = \(\frac{\text { Distance }}{\text { Speed }}\)]
⇒ \(\frac{60}{x}+\frac{240}{y}\) = 4 …………….(1)
and \(\frac{100}{x}+\frac{(300-100)}{y}=4+\frac{10}{60}\)
\(\frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}\)
\(\frac{100}{x}+\frac{200}{y}=\frac{\varepsilon}{6}\) ………………..(2)
Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q
∴ 60p + 240q = 4 ……………..(3)
and 100p + 200q = \(\frac{25}{6}\) ………………(4)
Multiplying equation (3) by 5 and equation (4) by 3, we get
300p + 1200q = 20 ……………..(5)
300p + 600q = \(\frac{25}{2}\) …………….(6)
Subtracting equation (6) from equation (5), we get

y = 80
Substituting the value of q in equation (3), we get
60p + 240 × \(\frac{1}{80}\) = 4
⇒ 60p + 3 = 4
⇒ 60p = 4 – 3 = 1
⇒ p = \(\frac{1}{60}\)
⇒ \(\frac{1}{x}=\frac{1}{60}\) [∵ p = \(\frac{1}{x}\)]
⇒ x = 60

Hence, the equations are \(\frac{10}{x}+\frac{200}{y}=\frac{25}{6}\), \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\)
and x = 60, y = 80 is the required solution.
Therefore,speed of the train = 60 km/hr
and speed of the bus = 80 km/hr

Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0

(ii) 2x + y = 5
3x + 2y = 8

(iii) 3x – 5y = 20
6x – 10y = 40

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0.
Solution :
(i) The given equations are;
x – 3y – 3 = 0
3x – 9y – 2 =0
Comparing the given equations with the stan-dard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 1, b₁ = – 3, c₁ = – 3, a₂ = 3, b₂ = – 9, c₂ = – 2
\(\frac{a_1}{a_2}=\frac{1}{3}\)
\(\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}\)
and \(\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

So, pair of linear equations has no solution.

(ii) We can write the given equations in general form :
2x + y – 5 = 0
3x + 2y – 8 = 0
Comparing the given equations with the standard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 2, b₁ = 1, c₁ = – 5, a₂ = 3, b₂ = 2, c₂ = – 8
\(\frac{a_1}{a_2}=\frac{2}{3}\)
\(\frac{b_1}{b_2}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-8}=\frac{5}{8}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

So, the pair of linear equations has a unique solution.

By Cross Multiplication Method:

x = 2 and y = 1.
Hence, x = 2, y = 1 is the required solution.

(iii) We can write the given equations in general form :
3x – 5y – 20 = 0
6x – 10y – 40 = 0
Comparing the given equations with the standard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 3, b₁ = – 5, c₁ = – 20, a₂ = 6, b₂ = 10,
c₂ = – 40
\(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\)
\(\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

So, the pair of linear equations has infinitely many solutions.

(iv) The given equations are:
x – 3y – 7 = 0
3x – 3y – 15 = 0
Comparing the given equations with the standard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 1, b₁ = – 3, c₁ = – 7, a₂ = – 3, b₂ = – 3,
c₂ = – 15
\(\frac{a_1}{a_2}=\frac{1}{3}\)

\(\frac{b_1}{b_2}=\frac{-3}{-3}\) = 1

and \(\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}\)

Since, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
So, the pair of linear equations has a unique solution.

By Cross Multiplication Method:

x = 4, and y = – 1
Hence, x = 4, y = – 1 is the required solution.

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7
(a – b) x + (a + b) y =3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) We can write the given equations in general form :
2x + 3y – 7 = 0
(a – b) x + (a + b) y – 3a – b + 2 = 0
Comparing the given equations with the standard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 2, b₁ = 3, c₁ = – 7, a₂ = a – b, b₂ = a + b,
c₂ = – 3a – b + 2
Condition for infinite number of solutions is:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
\(\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-3 a-b+2}\)
\(\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3 a+b-2}\)
\(\frac{2}{a-b}=\frac{3}{a+b}\) and
\(\frac{3}{a+b}=\frac{7}{3 a+b-2}\)
⇒ 2a + 2b = 3a – 3b
and 3(3a + b – 2) = 7 (a + b)
⇒ 2a – 3a + 2b + 3b = 0
and 9a + 3b – 6 = 7a + 7b
⇒ – a + 5b = 0
and 9a – 7a + 3b – 7b – 6 = 0
⇒ – a + 5b = 0
and 2a – 4b – 6 = 0
We will solve these equations by elimination method. The equations are :
– a + 56 = 0 ……………..(1)
2a – 46 = 6 ………………(2)
Multiplying equation (1) by 2 and adding it to equation (2), we get

Substituting the value of 6 in equation (1), we get
– a + 5 × 1 = 0
⇒ – a = – 5
⇒ a = 5
Hence, a = 5, b = 1.

(ii) We can write the given equations in gen¬eral form :
3x + y – 1 = 0
(2k – 1) x + (k – 1) y – 2k – 1 = 0
Comparing the given equations with the stan-dard form of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 3, b₁ = 1, c₁ = – 1, a₂ = 2k – 1, b₂ = k – 1, c₂ = – (2k + 1)
The condition for no solution is :
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
\(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{-1}{-(2 k+1)}\)
\(\frac{3}{2 k-1}=\frac{1}{k-1}\)
and \(\frac{1}{k-1} \neq \frac{1}{2 k+1}\)
3k – 3 = 2k – 1
k – 1 ≠ 2k + 1
3k – 2k = – 1 + 3
3k – 2k = – 1 + 3
and k – 2k ≠ 1 + 1
k = 2 and – k ≠ 2
k = 2 and k ≠ – 2
Hence, k = 2.

Question 3.
Solve the following pair of linear equations by substitution method and cross multiplication method :
8x + 5y = 9
3x + 2y = 4
Solution :
By Substitution Method :
The given equations are ;
8a + 5y = 9 ………………(1)
3x + 2y = 4 ………………(2)
From equation (2), we get
3a = 4 – 2y
x = \(\left(\frac{4-2 y}{3}\right)\)
Substituting the value of x in equation (1), we get
8(\(\left(\frac{4-2 y}{3}\right)\)) + 5y = 9
⇒ \(\frac{32-16 y}{3}\) + 5y = 9
⇒ \(\frac{32-16 y+15 y}{3}\) = 9
⇒ \(\frac{-y+32}{3}\) = 9
⇒ – y + 32 = 27
⇒ – y = 27 – 32 = – 5
⇒ y = 5
Putting the v1ue ofy in equation (1), we get
8x + 5 × 5 = 9
⇒ 8x + 25 = 9
⇒ 8x = 9 – 25 = – 16
⇒ x = \(-\frac{16}{8}\) = – 2
Hence, x = – 2; y = 5 is the required solution.

By Cross Multiplication Method:
The given equations are
8x + 5y = 9
3x + 2y = 4
By cross multiplication, we get

Hence, x = – 2; y = 5 is the required solution.

Question 4.
Form the pair of linear equations in the following problems and find their solutions (If they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes when \(\frac{1}{2}\) is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its denominator. Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded foi each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are speeds of the two cars?

(v) The area of rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:
(i) Let the fixed charge of hostel be ₹ x and cost of food per day be ₹ y.
According to question,
x + 20y = 1000
and x + 26y = 1180
By Cross Multiplication Method, we get

x = 400 and y = 30
Hence, equations are x + 20y = 1000, x + 26y = 1180 and x = 400, y = 30 is the required solution.
Therefore, Fixed charge of hostel = 400
and cost of food per day = 30

(ii) Let the numerator be x and denomin cor be y,
then fraction = \(\frac{x}{y}\)
According to question,
\(\frac{x-1}{y}=\frac{1}{3}\)
3x – 3 = y
3x – y – 3 = 0 ……………….(1)
and \(\frac{x}{y+8}=\frac{1}{4}\)
4x = y + 8
4x – y – 8 = 0 ………………(2)
By Cross Multiplication Method, we get

x = 5 and y = 12
Hence, the equations are 3x – y – 3 = 0, 4x – y – 8 = 0 and x = 5, y = 12 is the required solution.
Therefore, the fraction = \(\frac{5}{12}\).

(iii) Let the number of right answer be = x and
number of wrong answers be = y
Then total questions = x + y
According to question
3x – y = 40 ………………..(1)
and 4x – 2y = 50 ……………..(2)
By Cross Multiplication Method, we get

Hence, the equations are; 3x – y = 40, 4x – 2y = 50 and x = 15, y = 5 is the required solution.
Therefore, total questions = 15 + 5 = 20.

(iv) Let the speed of first car be x km/hr and speed of second car be y km/hr.
Case I :
When they travel in same direction, they meet at point C.

Distance covered by first car in 5 hours = 5 x km.
[∵ Distance = Speed × Time]
Distance covered by second car in 5 hours = 5y km.
Since AC – BC = AB
Therefore, 5x – 5y = 100
⇒ x – y = 20 ……………(1)

Case II:
When they travel in opposite direc-tions. They meet at point C.

Distance covered by first car in 1 hour = x km.
Distance covered by second car in 1 hour = y km.
Since AC + BC = AB
Therefore, x + y = 100 ……………..(2)
So, the equations are
x – y = 20
x + y = 100
By Cross Multiplication Method, we get

x = \(\frac{120}{2}\) and y = \(\frac{80}{2}\)
⇒ x = 60 and y = 40
Hence, the equations are; x – y = 20, x + y = 100.
And x = 60, y = 40 is the required solution.
Therefore, speed of first car = 60 km/hr
and speed of second car = 40 km/hr.

(v) Let the length of rectangle be x units and breadth of rectangle be y units
∴ Area of rectangle = xy square units
According to question,
xy – (x – 5)(y + 3) = 9
⇒ xy – [xy + 3x – 5y – 15] =9
⇒ xy – xy – 3 + 5y + 15 = 9
⇒ – 3x + 5y + 15 – 9= 0
⇒ – 3x + 5y + 6 = 0
⇒ 3x – 5y – 6 = 0 ………….(1)
and (x + 3)(y + 2) = xy + 67
⇒ 3 – xy + 2x + 3y + 6 = xy + 67
⇒ xy + 2x + 3y – xy + 6 – 67 = 0
⇒ 2x + 3y – 61 = 0 …………….(2)
So, the equations are :
3x – 5y – 6 = 0
2x + 3y – 61 = 0
By Cross Multiplication Method, we get

x = 17 and y = 9
Hence, equations are 3x – 5y – 6 = 0, 2x + 3y – 61 = 0.
And x = 17 and y = 9 is the required solution.
Therefore, length = 17 units, breadth = 9 units.

Haryana State Board HBSE 10th Class Social Science Notes History Chapter 1 यूरोप में राष्ट्रवाद का उदय Notes.

Haryana Board 10th Class Social Science Notes History Chapter 1 यूरोप में राष्ट्रवाद का उदय

यूरोप में राष्ट्रवाद का उदय Class 10 Notes In Hindi

1. फ्रांसीसी क्रांति और राष्ट्र का विचार
→ राष्ट्रवाद की पहली स्पष्ट अभिव्यक्ति 1789 में फ्रांसीसी क्रांति के साथ हुई। जैसा कि आपको याद होगा. 1789 में फ़्रांस एक ऐसा राज्य था जिसके संपूर्ण भू-भाग पर एक निरंकुश राजा का आधिपत्य था।

→ फ्रांसीसी क्रांति सेजो राजनीतिक और संवैधानिक बदलाव हुए उनसे प्रभुसत्ता राजतंत्र से निकल कर फ्रांसीसी नागरिकों के समूह में हस्तांतरित हो गई। क्रांति ने घोषणा की कि अब लोगों द्वारा राष्ट्र का गठन होगा और वे ही उसकी नियति तय करेंगे।

→ प्रारंभ से ही फ्रांसीसी क्रांतिकारियों ने ऐसे अनेक कदम उठाए जिनसे फ्रांसीसी लोगों में एक सामूहिक पहचान की भावना पैदा हो सकती थी। पितृभूमि (la patrie) और नागरिक (le citoyen) जैसे विचारों ने एक संयुक्त समुदाय के विचार पर बल दिया जिसे एक संविधान के अंतर्गत समान अधिकार प्राप्त थे।

→ अतः एक नया फ्रांसीसी झंडा-तिरंगा (the tricolour) चुना गया जिसने पहले के राजध्वज की जगह ले ली। इस्टेट जेनरल का चुनाव सक्रिय नागरिकों के समूह द्वारा किया जाने लगा और उसका नाम बदल कर नेशनल एसेंबली कर दिया गया।

→ नयी स्तुतियाँ रची गईं, शपथें ली गई, शहीदों का गुणगान हुआ-और यह सब राष्ट्र के नाम पर हुआ। एक केंद्रीय प्रशासनिक व्यवस्था लागू की गई जिसने अपने भू-भाग में रहने वाले सभी नागरिकों के लिए समान कानून बनाए।

→ आंतरिक आयात-निर्यात शुल्क समाप्त कर दिए गए और भार तथा नापने की एकसमान व्यवस्था लागू की गई। क्षेत्रीय बोलियों को हतोत्साहित किया गया और पेरिस में फ्रेंच जैसी बोली और लिखी जाती थी, वही राष्ट्र की साझा भाषा बन गई।

→ क्रांतिकारियों ने यह भी घोषणा की कि फ़्रांसीसी राष्ट्र का यह भाग्य और लक्ष्य था कि वह यूरोप के लोगों को निरंकुश शासकों से मुक्त कराए। दूसरे शब्दों में, फ़्रांस यूरोप के अन्य लोगों को राष्ट्रों में गठित होने में मदद देगा।

→ जब फ्रांस की घटनाओं की ख़बर यूरोप के विभिन्न शहरों में पहुँची तो छात्र तथा शिक्षित मध्य-वर्गों के अन्य सदस्य जैकोबिन क्लबों की स्थापना करने लगे। उनकी गतिविधियों और अभियानों ने उन फ्रेंच सेनाओं के लिए रास्ता तैयार किया जो 1790 के दशक में हॉलैंड, बेल्जियम, स्विट्ज़रलैंड और इटली के बड़े इलाके में घुसी।

→ क्रांति कारी युद्धों के शुरू होने के साथ ही फ़्रांसीसी सेनाएँ राष्ट्रवाद के विचार को विदेशों में ले जाने लगीं। नेपोलियन के नियंत्रण में जो विशाल क्षेत्र आया वहाँ उसने ऐसे अनेक सुधारों की शुरुआत की जिन्हें फ्रांस में पहले ही आरंभ किया जा चुका था। फ्रांस में राजतंत्र वापस लाकर नेपोलियन ने नि:संदेह वहाँ प्रजातंत्र को नष्ट किया था।

→ मगर प्रशासनिक क्षेत्र में उसने क्रांतिकारी सिद्धांतों का समावेश किया था ताकि पूरी व्यवस्था अधिक तर्कसंगत और कुशल बन सके। 1804 की नागरिक संहिता जिसे आमतौर पर नेपोलियन की संहिता के नाम से जाना जाता है, ने जन्म पर आधारित विशेषाधिकार समाप्त कर दिए थे।

→ उसने कानून के समक्ष बराबरी और संपत्ति के अधिकार को सुरक्षित बनाया। इस संहिता को फ्रांसीसी नियंत्रण के अधीन क्षेत्रों में भी लागू किया गया। डच गणतंत्र, स्विट्जरलैंड, इटली और जर्मनी में नेपोलियन ने प्रशासनिक विभाजनों को सरल बनाया, सामंती व्यवस्था को खत्म किया और किसानों को भू-दासत्व और जागीरदारी शुल्कों से मुक्ति दिलाई।

→ शहरों में भी कारीगरों के श्रेणी-संघों के नियंत्रणों को हटा दिया गया। यातायात और संचार-व्यवस्थाओं को सुधारा गया। किसानों, कारीगरों, मज़दूरों और नए उद्योगपतियों ने नयी-नयी मिली आज़ादी चखी।

→ उद्योगपतियों और खासतौर पर समान बनाने वाले लघु उत्पादक यह समझने लगे कि एकसमान क़ानून, मानक भार तथा नाप और एक राष्ट्रीय मुद्रा से एक इलाके से दूसरे इलाके में वस्तुओं और पूँजी के आवागमन में सहूलियत होगी।

→ लेकिन जीते हुए इलाकों में स्थानीय लोगों की फ्रांसीसी शासन के प्रति मिली-जुली प्रतिक्रियाएँ थीं। शुरुआत में अनेक स्थानों जैसे हॉलैंड और स्विट्ज़रलैंड और साथ ही कई शहरों जैसे-ब्रसेल्स, मेंज, मिलान और वॉरसा में फ्रांसीसी सेनाओं का स्वतंत्रता का तोहफा देने वालों की तरह स्वागत किया गया।

→ मगर यह शुरुआती उत्साह शीघ्र ही दुश्मनी में बदल गया जब यह साफ़ होने लगा कि नयी प्रशासनिक व्यवस्थाएँ राजनीतिक स्वतंत्रता के अनुरूप नहीं थीं।

→ बढ़े हुए कर, सेंसरशिप और बाक़ी यूरोप को जीतने के लिए फ्रेंच सेना में जबरन भर्ती से हो रहे नुकसान प्रशासनिक परिवर्तनों से मिले फ़ायदों से कहीं ज्यादा नज़र आने लगे।

HBSE 10th Class History Chapter 1 यूरोप में राष्ट्रवाद का उदय

2. यूरोप में राष्ट्रवाद का निर्माण
→ अगर आप मध्य अठारहवीं सदी के यूरोप के नक्शे को देखें तो उसमें वैसे ‘राष्ट्र-राज्य’ नहीं मिलेंगे जैसे कि आज हैं। जिन्हें आज हम जर्मनी, इटली और स्विट्ज़रलैंड के रूप में जानते हैं वे तब राजशाहियों, डचियों (duchies) और कैंटनों (cantons) में बँटे हुए थे, जिनके शासकों के स्वायत्त क्षेत्र थे।

→ पूर्वी और मध्य यूरोप निरंकुश राजतंत्रों के अधीन थे और इन इलाकों में तरह-तरह के लोग रहते थे। वे अपने आप को एकसामूहिक पहचान या किसी समान संस्कृति का भागीदार नहीं मानते थे। अकसर वे अलग-अलग भाषाएँ बोलते थे और विभिन्न जातीय समूहों के सदस्य थे।

→ उदाहरण के तौर पर, ऑस्ट्रिया-हंगरी पर शासन करने वाला हैब्सबर्ग साम्राज्य कई अलग-अलग क्षेत्रों और जनसमूहों को जोड़ कर बना था। इसमें ऐल्प्स के टिरॉल, ऑस्ट्रिया और सुडेटेनलैंड जैसे इलाकों के साथ-साथ बोहेमिया भी शामिल था जहाँ के कुलीन वर्ग में जर्मन भाषा बोलने वाले ज़्यादा थे। हैब्सबर्ग साम्राज्य में लॉम्बार्डी और वेनेशिया जैसे इतालवी-भाषी – प्रांत भी शामिल थे।

→ सामाजिक और राजनीतिक रूप से ज़मीन का मालिक कुलीन वर्ग यूरोपीय महाद्वीप का सबसे प्रभुत्वशाली वर्ग था। इस वर्ग के सदस्य एक साझा जीवन शैली से बँधे हुए थे जो क्षेत्रीय विभाजनों के आर-पार व्याप्त थी। वे ग्रामीण इलाकों में जायदाद और शहरी-हवेलियों के मालिक थे।

→ राजनीतिक कार्यों के लिए तथा उच्च वर्गों के बीच वे फ्रेंच भाषा का प्रयोग करते थे। उनके परिवार अकसर वैवाहिक बंधनों से आपस में जुड़े होते थे। मगर यह शक्तिशाली कुलीन वर्ग संख्या के लिहाज़ से एक छोटा समूहथा। जनसंख्या के अधिकांश लोग कृषक थे।

→ पश्चिम में ज्यादातर ज़मीन पर किराएदार और छोटे काश्तकार खेती करते थे जबकि पूर्वी और मध्य यूरोप में भूमि विशाल जागीरों में बँटी थी जिस पर भूदास खेती करते थे।

→ पश्चिमी और मध्य यूरोप के हिस्सों में औद्योगिक उत्पादन और व्यापार में वृद्धि से शहरों का विकास और वाणिज्यिक वर्गों का उदय हुआ जिनका अस्तित्व बाज़ार के लिए उत्पादन पर टिका था।

→ इंग्लैंड में औद्योगीकरण अठारहवीं सदी के दूसरे भाग में आरंभ हुआ लेकिन फ़्रांस और जर्मनी के राज्यों के कुछ हिस्सों में यह उन्नीसवीं शताब्दी के दौरान ही हुआ।

→ यूरोप में उन्नीसवीं सदी के शुरुआती दशकों में राष्ट्रीय एकता से संबंधित विचार उदारवाद से करीब से जुड़े थे। उदारवाद यानी liberalism शब्द लातिन भाषा के मूल liber पर आधारित है जिसका अर्थ है ‘आज़ाद’।

→ नए मध्य वर्गों के लिए उदारवाद का मतलब था व्यक्ति के लिए आजादी और कानून के समक्ष सबकी बराबरी। राजनीतिक रूप से उदारवाद एक ऐसी सरकार पर जोर देता था जो सहमति से बनी हो।

→ फ्रांसीसी क्रांति के बाद से उदारवाद निरंकुश शासक और पादरीवर्ग के विशेषाधिकारों की समाप्ति, संविधान तथा संसदीय प्रतिनिधि सरकार का पक्षधर था। उन्नीसवीं सदी केउदारवादी निजी संपत्ति के स्वामित्व की अनिवार्यता पर भी बल देते थे।

Chapter 1 यूरोप में राष्ट्रवाद का उदय History 10th Class

3. क्रांतियों का युग : 1830-1848
→ जैसे-जैसे रूढ़िवादी व्यवस्थाओं ने अपनी ताकत को और मज़बूत बनाने की कोशिश की, यूरोप के अनेक क्षेत्रों में उदारवाद और राष्ट्रवाद को क्रांति से जोड़ कर देखा जाने लगा।

→ इटली और जर्मनी के राज्य, ऑटोमन साम्राज्य के सूबे, आयरलैंड और पोलैंड ऐसे ही कुछ क्षेत्र थे। इन क्रांतियों का नेतृत्व उदारवादी-राष्ट्रवादियों ने किया जो शिक्षित मध्यवर्गीय विशिष्ट लोग थे। इनमें प्रोफेसर, स्कूली-अध्यापक, क्लर्क और वाणिज्य व्यापार में लगे मध्यवर्गों के लोग शामिल थे।

→ प्रथम विद्रोह फ़्रांस में जुलाई 1830 में हुआ। बूबों राजा, जिन्हें 1815 के बाद हुई रूढ़िवादी प्रतिक्रिया के दौरान सत्ता में बहाल किया गया था, उन्हें अब उदारवादी क्रांतिकारियों ने उखाड़ फेंका। उनकी जगह एक संवैधानिक राजतंत्र स्थापित किया गया जिसका अध्यक्ष लुई फिलिप था।

→ मैटरनिख ने एक बार यह टिप्पणी की थी कि ‘जब फ्रांस छींकता है तो बाकी यूरोप की सर्दी-जुकाम हो जाता है।’ जुलाई क्रांति से ब्रसेल्स में भी विद्रोह भड़क गया जिसके फलस्वरूप यूनाइटेड किंगडम ऑफ़ द नीदरलैंड्स से अलग हो गया।

→ एक घटना जिसने पूरे यूरोप के शिक्षित अभिजात वर्ग में राष्ट्रीय भावनाओं का संचार किया, वह थी, यूनान का स्वंतत्रता संग्राम। पंद्रहवीं सदी से यूनान ऑटोमन साम्राज्य का हिस्सा था। यूरोप में क्रांतिकारी राष्ट्रवाद की प्रगति से यूनानियों का आजादी के लिए संघर्ष 1821 में आरंभ हो गया।

→ यूनान में राष्ट्रवादियों को निर्वासन में रह रहे यूनानियों के साथं पश्चिमी यूरोप के अनेक लोगों का भी समर्थन मिला जो प्राचीन यूनानी संस्कृति (Hellenism) के प्रति सहानुभूति रखते थे।

→ कवियों और कलाकारों ने यूनान को यूरोपीय सभ्यता का पालना बता कर प्रशंसा की और एक मुस्लिम साम्राज्य के विरुद्ध यूनान के संघर्ष के लिए जनमत जुटाया।

→ अंग्रेज़ कवि लॉर्ड बायरन ने धन इकट्ठा किया और बाद में युद्ध में लड़ने भी गए जहाँ 1824 में बुखार से उनकी मृत्यु हो गई। अंततः 1832 की कुस्तुनतुनिया की संधि ने यूनान को एक स्वतंत्र राष्ट्र की मान्यता दी।

4. जर्मनी और इटली का निर्माण
जर्मनी-क्या सेना राष्टं की निर्माता हो सकती है?
→ 1848 के बाद यूरोप में राष्ट्रवाद का जनतंत्र और वंति से – अलगाव होने लगा। राज्य की सना को बढ़ाने और पूरे यूरोप पर राजनीतिक प्रभुत्व हासिल करने के लिए रूढ़िवादियों ने अकसर राष्ट्रवादी भावनाओं का इस्तेमाल किया।

→ इसे उस प्रक्रिया में देखा जा सकता है जिससे जर्मनी और इटली एकीछत होकर राष्ट्र-राज्य बने। जैसा आपने देखा है, राष्ट्रवादी भावनाए!

→ मध्यवर्गीय जर्मन लोगों में काफी व्याप्त थीं और उन्होंने 1848 में जर्मन महासंघ के विभिन्न इलाकों को जोड़ कर एक निर्वाचित संसद द्वारा शासित राष्ट्र-राज्य बनाने का प्रयास किया था।

→ मगर राष्ट्र निर्माण की यह उदारवादी पहल राजशाही और फ़ौज की ताक़त ने मिलकर दबा दी। उसके पश्चात प्रशासन ने राष्ट्रीय एकीकरण के आंदोलन का नेतृत्व सँभाल लिया।

→ उसका प्रमुख मंत्री, ऑटो वॉन बिस्मार्क इस प्रक्रिया का जनक था जिसने प्रशासन की सेना और नौकरशाही की मदद ली। सात वर्ष के दौरान ऑस्ट्रिया, डेन्मार्क और फ़्रांस से तीन युणे में प्रशासन को जीत हुई और एकीकरण की प्रक्रिया पूरी हुई।

→ जनवरी 1871 में, वर्साय में हुए एक समारोह में प्रशासन के राजा विलियम प्रथम को जर्मनी का सम्राट घोषित किया गया।

→ 18 जनवरी 1871 की सुबह बेहद ठंडी थी। जर्मन राज्यों के राजकुमारों, सेना के प्रतिनिमिायों और प्रमुखमंत्री ऑटो वॉन बिस्मार्क समेत प्रशा के महन्वपूर्ण मंत्रियों की एक बैठक वर्साय के महल के बेहद ठंडे शीशमहल (हॉल ऑफ़ मिरर्स) में हुई।

→ सभा ने प्रशा के काइज़र विलियम प्रथम के नेतृत्व में नए जर्मन साम्राज्य की घोषणा की।

→ जर्मनी में राष्ट्र निर्माण प्रक्रिया ने प्रशा राज्य की शक्ति के प्रभुत्व को दर्शाता था।

इटली
→ जर्मनी की तरह इटली में भी राजनीतिक विखंडन का एक लंबा इतिहास था। इटली अनेक वंशानुगत राज्यों तथा बहु-राष्ट्रीय हैब्सबर्ग साम्राज्य में बिखरा हुआ था।

→ उन्नीसवीं सदी के समय में इटली सात राज्यों में बँटा हुआ था जिनमें से केवल एक-सार्डिनिया पीडमॉण्ट में एक इतालवी राजघराने का शासन था।

→ उनरी भाग ऑस्ट्रियाई हैब्सबर्गो के अमीन था, इस समय इलाकों पर पोप का शासन था और दक्षिणी क्षेत्र स्पेन के बूढे राजाओं के अमीन थे। इतालवी भाषा ने भी साझा रूप हासिल नहीं किया था और अभी तक उसके विविमा क्षेत्रीय और स्थानीय रूप मौजूद थे।

→ 1830 के दशक में ज्युसेपे मेत्सिनी ने एकीकृत इतालवी गणराज्य के लिए एक सुविचारित कार्यक्रम प्रस्तुत करने की कोशिश की थी। उसने अपने उद्देश्यों के प्रसार के लिए यंग इटली नामक एक गुप्त संगठन भी बनाया था।

→ 1831 और 1848 में क्रांतिकारी विद्रोहों की असफलता से युद्ध के ज़रिये इतालवी राज्यों को जोड़ने की ज़िम्मेदारी सार्डिनिया-पीडमॉण्ट के शासक विक्टर इमेनुएल द्वितीय पर आ गई।

→ इस क्षेत्र के शासक अभिजातवर्ग की नज़रों में एकीकृत इटली उनके लिए आर्थिक विकास और राजनीतिक प्रभुत्व की संभावनाएँ उत्पन्न करता था।

→ मंत्री प्रमुख कावूर, जिसने इटली के प्रदेशों को एकीकृत करने वाले आंदोलन का नेतृत्व किया, न तो एक क्रांतिकारी था और न ही जनतंत्र में विश्वास रखने वाला। इतालवी अभिजात वर्ग के तमाम अमीर और शिक्षित सदस्यों की तरह वह इतालवी भाषा से कहीं बेहतर फ्रेंच बोलता था।

5. राष्ट्र की दृश्य-कल्पना
→ किसी शासक को एक चित्र या मूर्ति के रूप में अभिव्यक्त करना आसान है किंतु एक राष्ट्र को चेहरा कैसे दिया जा सकता है?

→ अठारहवीं और उन्नीसवीं सदी में कलाकारों ने राष्ट्र का मानवीकरण करके इस प्रश्न को हल किया। दूसरे शब्दों में, उन्होंने एक देश को कुछ यूँ चित्रित किया जैसे वह कोई व्यक्ति हो। उस समय राष्ट्रों को नारी भेष में प्रस्तुत किया जाता था।

→ राष्ट्र को व्यक्ति का जामा पहनाते हुए जिस नारी रूप को चुना गया वह असल जीवन में कोई खास महिला नहीं थी। यह तो राष्ट्र के अमूर्त विचार को ठोस रूप प्रदान करने का प्रयास था। यानी नारी की छवि राष्ट्र का रूपक बन गई।

→ आपको याद होगा कि फ्रांसीसी क्रांति के दौरान कलाकारों ने स्वतंत्रता, न्याय और गणतंत्र जैसे विचारों को व्यक्त करने के लिए नारी रूपक का प्रयोग किया। इन आदर्शों को विशेष वस्तुओं या प्रतीकों से व्यक्त किया गया था।

6. राष्ट्रवाद और साम्राज्यवाद
→ उन्नीसवीं सदी की अंतिम चौथाई तक राष्ट्रवाद का वह आदर्शवादी उदारवादी-जनतांत्रिक स्वभाव नहीं रहा जो सदी के प्रथम भाग में था। अब राष्ट्रवाद सीमित लक्ष्यों वाला संकीर्ण सिद्धांत बन गया।

→ इस बीच के दौर में राष्ट्रवादी समूह एक-दूसरे के प्रति अनुदार होते चले गए और लड़ने के लिए हमेशा तैयार रहते थे। साथ ही प्रमुख यूरोपीय शक्तियों ने भी अपने साम्राज्यवादी उद्देश्यों की प्राप्ति के लिए अधीन लोगों की राष्ट्रवादी आकांक्षाओं का इस्तेमाल किया।

→ 1871 के बाद यूरोप में गंभीर राष्ट्रवादी तनाव का स्रोत बाल्कन क्षेत्र था। इस क्षेत्र में भौगोलिक और जातीय भिन्नता थी। इसमें आधुनिक रोमानिया, बुल्गेरिया, अल्बेनिया, यूनान, मेसिडोनिया, क्रोएशिया, बोस्निया-हर्जेगोविना, स्लोवेनिया, सर्बिया और मॉन्टिनिग्रो शामिल थे।

→ क्षेत्र के निवासियों को आमतौर पर स्लाव पुकारा जाता था। बाल्कन क्षेत्र का एक बड़ा हिस्सा ऑटोमन साम्राज्य के नियंत्रण में था।

→ बाल्कन क्षेत्र में रूमानी राष्ट्रवाद के विचारों के फैलने और ऑटोमन साम्राज्य के विघटन से स्थिति काफ़ी विस्फोटक हो गई। उन्नीसवीं सदी में

→ ऑटोमन साम्राज्य ने आधुनिकीकरण और आंतरिक सुधारों के जरिए मज़बूत बनना चाहा था किंतु इसमें इसे बहुत कम सफलता मिली। एक के बाद एक उसके अधीन यूरोपीय राष्ट्रीयताएँ उसके चंगुल से निकल कर स्वतंत्रता की घोषणा करने लगीं।

→ बाल्कन लोगों ने आज़ादी या राजनीतिक अधिकारों के अपने दावों को राष्ट्रीयता का आधार दिया। उन्होंने इतिहास का इस्तेमाल यह साबित करने के लिए किया कि वे कभी स्वतंत्र थे किंतु तत्पश्चात विदेशी शक्तियों ने उन्हें अधीन कर लिया।

→ अतः बाल्कन क्षेत्र के विद्रोही राष्ट्रीय समूहों ने अपने संघर्षों को लंबे समय से खोई आजादी को वापस पाने के प्रयासों के रूप में देखा।

Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\)= – 1 and x – \(\frac{y}{3}\) = 3.
Solution :
(i) By Elimination Method :
The given equations are :
x + y = 5 ………………..(1)
2x – 3y = 4 …………………(2)
Multiplying equation (1) by 3, we get
3x + 3y = 15 ……………..(3)
Adding equation (3) and equation (2), we get

Substituting the value of x in equation (2), we get
2 × \(\frac{19}{5}\) – 3y = 4
\(\frac{38}{5}\) – 3y = 4
– 3y = \(\frac{4}{1}-\frac{38}{5}\)

– 3y = \(\frac{20-38}{5}=-\frac{18}{5}\)

y = \(\frac{-18}{-3 \times 5}=\frac{6}{5}\)

Hence, x = \(\frac{19}{5}\), y = \(\frac{6}{5}\) is the required solution.

By Substitution Method:
The given equations are:
x + y = 5 ……………..(1)
2x – 3y = 4 ……………….(2)
From equation (1), we get
y = 5 – x
Substituting the value of y in equation (2), we get
2x – 3(5 – x) = 4
⇒ 2x – 15 + 3x = 4
⇒ 5x = 4 + 15
⇒ 5x = 19
⇒ x = \(\frac{19}{5}\)

Putting the value of x in equation (1), we get
\(\frac{19}{5}\) + y = 5
y = \(\frac{5}{1}-\frac{19}{5}\)
y = \(\frac{25-19}{5}\)
y = \(\frac{6}{5}\)
Hence, x = \(\frac{19}{5}\), y = \(\frac{6}{5}\) is the required solution.

(ii) By Elimination Method :
The given equations are :
3x + 4y = 10 …………..(1)
2x – 2y = 2
or x – y = 1 …………….(2)
Multiplying the equation (2) by 3, we get
3x – 3y = 3 ……………….(3)
Subtracting equation (3) from equation (1), we get

Substituting the value of y in equation (2), we get
x – 1 = 1
x = 1 + 1 = 2
Hence, x = 2, y = 1 is the required solution.

By Substitution Method: .
The given equations are: –
3x + 4y = 10 …………………(1)
2x – 2y = 2 ………………….(2)
or x – y = 1
From equation (2), we get
x = 1 + y
Substituting the value of x in equation (1), we get
3 (1 + y) + 4y = 10
3 + 3y + 4y = 10
7y = 10 – 3
7y = 7
y = \(\frac{7}{7}\) = 1
Putting the value of y in equation (2), we get
x – 1 = 1
⇒ x = 1 + 1 = 2
Hence, x = 2, y = 1 is the required solution.

(iii) By Elimination Method :
The given equations are :
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ………………(1)
and 9x = 2y + 7
⇒ 9x – 2y = 7 ……………….(2)
Multiplying equation (1) by 3, we get
9x – 15y = 12 ………………(3)
Subtracting equation (2) from equation (3), we get

By Substitution Method :
The given equations are :
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ………………(1)
and 9x = 2y + 7
⇒ 9x – 2y = 7 ……………….(2)
From equation (1), we get
3x = 4 + 5y
⇒ x = \(\frac{4+5 y}{3}\)
Substituting the value of x in equation (2), we get
\(\frac{9(4+5 y)}{3}\) – 2y = 7
3 (4 + 5y) – 2y = 7
12 + 15y – 2y = 7
13y = 7 – 12
13y = – 5
y = – \(\frac{5}{13}\)
Putting the value of y in equation (1), we get
3x – 5 × (- \(\frac{5}{13}\)) = 4
3x + \(\frac{25}{13}\) = 4
3x = \(\frac{4}{1}-\frac{25}{13}\)
3x = \(\frac{52-25}{13}\)
3x = \(\frac{27}{13}\)
x = \(\frac{27}{3 \times 13}=\frac{9}{13}\)
Hence, x = \(\frac{9}{13}\), y = – \(\frac{5}{13}\) is the required.

(iv) By Elimination Method:
The given equations are :

Substituting the value of y in equation (2), we get
3x – (- 3) = 9
⇒ 3x + 3 = 9
⇒ 3x = 9 – 3 = 6
⇒ x = \(\frac{6}{3}\) = 2
Hence, x = 2, y = 3 is the required solution.

By Substitution Method :
The given equations are :
\(\frac{x}{2}+\frac{2 y}{3}\) = – 1

\(\frac{3 x+4 y}{6}\) = – 1

3x + 4y = – 6 ………………(1)
and x – \(\frac{y}{3}\) = 3
\(\frac{3 x-y}{3}\) = 3
3x – y = 9 ……………..(2)
From equation (2), we get
3x = 9 + y
x = \(\frac{9+y}{3}\)
Substituting the value of x in equation(1), we get
3 × \(\left(\frac{9+y}{3}\right)\) + 4y = – 6
9 + y + 4y = – 6
5y = – 6 – 9 = – 15
y = – \(\frac{15}{5}\) = – 3
Putting the value of y in equation (2), we get
3x – (-3) = 9
3x + 3 = 9
⇒ 3x = 9 – 3 = 6
x = \(\frac{6}{3}\) = 2
Hence, x = 2, y = – 3 is the required solution.

Question 2.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method.
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nun was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nun and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digit. Find the number.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of 50 and 100 she received?

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter, Santa paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:
(i) Let the numerator be x and denominator be y, then
The fraction = \(\frac{x}{y}\)
According to question,
\(\frac{x+1}{y-1}\) = 1
⇒ x + 1 = y – 1
⇒ x – y = – 1 – 1
⇒ x – y = – 2 ……………..(1)
and \(\frac{x}{y+1}=\frac{1}{2}\)
⇒ 2x = y + 1
⇒ 2x – y = 1
Subtracting equation (2) from equation (1), we get

Substituting the value of x in equation (1), we
3 – y = – 2
⇒ – y = – 2 – 3 = – 5
y = 5
Hence, the equations are x – y = – 2, 2x – y = 1 and x = 3, y = 5 is the required solution.
Therefore, fraction = \(\frac{3}{5}\).

(ii) Let the present age of Nun be x years and present age of Sonu be y years.
Five years ago age of Nuri = (x – 5) years
Five years ago age of Sonu = (y – 5) years
According to question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 15 + 5
⇒ x – 3y = – 10 ………………(1)
Ten years later age of Nuri = (x + 10) years
Ten years later age of Sonu= (y + 10) years
According to question,
x + 10 = 2 (y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 20 – 10
⇒ x – 2y = 10 ……………..(2)
Subtracting equation (2) from equation (1), we get

⇒ y = 20
Substituting the value of y in equation (2), we get
x – 2 × 20 = 10
⇒ x – 40 = 10
⇒ x = 10 + 40 = 50
Hence, the equations are :
x – 3y = – 10,
x – 2y = 10 and
x = 50, y = 20 is the required solution.
Therefore,present age of Nuri = 50 years and
present age of Sonu = 20 years.

(iii) Let the digit of unit’s place be x and digit of ten’s place be y
Then required number = 10y + x
After reversing the digits, digit of unit’s place be y and ten’s place be x.
The required number after interchange of digits = 10x + y
According to question, x + y = 9
and 9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 20x – 9x + 2y – 90y = 0
⇒ 11x – 88y = 0
⇒ x – 8y = 0
Subtracting equation (2) from equation (1), we get

Substituting the value of y in equation (1), we get
x + 1 = 9
⇒ x = 9 – 1
⇒ x = 8.
Hence, the equations are x + y = 9, x – 8y = 0 and x = 8, y = 1 is the required solution.
Therefore, the required number = 10 × 1 + 8 = 18.

(iv) Let the number of ₹ 50 notes be x
and number of ₹ 100 notes be y.
Total Amount of ₹ 50 notes = 50x and
total amount of ₹ 100 notes = 100y
According to question,
50x + 100y = 2000
⇒ x + 2y = 40 …………….(1)
and x + y = 25 ……………(2)
Subtracting equation (2) from equation (1), we get

Substituting the value of y in equation (2), we get
x + 15 = 25
⇒ x = 25 – 15
⇒ x = 10
Hence, the equations are x + 2y = 40, x + y = 25 and x = 10, y = 15 is the required solution.
Therefore, ₹ 50 notes = 10
and ₹ 100 notes = 15.

(v) Let the fixed charge be ₹ x
and charge for each extra day be ₹ y
According to question,
x + 4y = 27 ………………(1)
and x + 2y = 21 ……………(2)
Subtracting equation (2) from equation (1), we get

Substituting the value of y in equation (1), we get
x + 4 × 3 = 27
⇒ x + 12 = 27
⇒ x = 27 – 12
⇒ x = 15
Hence, the equations are x + 4y = 27, x + 2y = 21 and x = 15, y = 3 is the required solution.
Therefore, fixed charge = ₹ 15
and charge for each extra day = ₹ 3

Haryana State Board HBSE 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 5 Arithmetic Progressions Ex 5.1

Question 1.
In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹ 15 for the first km and ₹ 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹ 150 for the first metre and rises by ₹ 50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8% per annum.
Solution :
(i) We have,
Taxi fare for 1 km = ₹ 15
Taxi fare for each additional km = ₹ 8
Let a₁, a₂, a₃, a₄ …………… be the fares for 1 km, 2 km, 3 km, 4 km ………….. respectively.
∴ a₁ = ₹ 15
a₂ = 15 + 8 = ₹ 23
a₃ = 15 + 2 × 8 = ₹ 31
a₄ = 15 + 3 × 8 = ₹ 39 and so on.
The sequence is 15, 23, 31, 39, …………..
Now, a₂ – a₁ = 23 – 15 = 8
a₃ – a₂ = 31 – 23 = 8
a₄ – a₃ = 39 – 31 = 8
a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The sequence 15, 23, 31, 39, …………….. forms an AP.

(ii) Let the amount of air present in the cylinder be x units
and let the amount of air remained after each pump be a₂, a₃, a₄, … respectively.
∴ a₁ = x units.

∵ a₃ – a₂ ≠ a₂ – a₁
Hence, the sequence x, \(\frac{3 x}{4}, \frac{9 x}{16}, \frac{27 x}{64}, \ldots\) does not form an AP.

(iii) Let the cost of 1m, 2m, 3m, …….. be ₹ a₁, ₹ a₂, ₹ a₃, … respectively.
a₁ = ₹ 150
a₂ = 150 + 50 = ₹ 200
a₃ = 150 + 2 × 50 = ₹ 250 and so on.
The sequence is 150, 200, 250, ………………
Now, a₂ – a₁ = 200 – 150 = ₹ 50
a₃ – a₂ = 250 – 200 = ₹ 50
∵ a₂ – a₁ = a₃ – a₂
Hence, the sequence 150, 200, 250, ………….. forms an A.P.

(iv) Amount deposited in the bank = ₹ 10000
Rate of interest = 8% per annum.
Let the amounts in bank after 1 year, 2 year, 3 years, … are ₹ a₁, ₹ a₂, ₹ a₃, ………….. respectively.

= ₹ 12597.12 and so on.
The sequence is 10800, 11664, 12597.12, ………………
Now, a₂ – a₁ = 11664 – 10800 = 864
a₃ – a₂ = 12597.12 – 11664 = 933.12
∵ a₂ – a₁ ≠ a₃ – a₂
Hence, the sequence 10800, 11664, 12597.12, …………… does not form an AP.

Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows :
(i) a = 10, d = 10
(ii) a = – 2, d = 0
(iii) a = 4, d = – 3
(iv a = – 1, d = \(\frac{1}{2}\)
(v) a = – 1.25, d = – 0.25.
Solution :
(i) We have, a = 10, d = 10
First four terms of required AP are
a₁ = a = 10
a₂= a + d = 10 + 10 = 20
a₃ = a₂ + d
= 20 + 10 = 30
a₄ = a₃ + d
= 30 + 10 = 40
Hence, the first four terms of AP are 10, 20, 30, 40.

(ii) We have, a = – 2, d = 0
First four terms of required AP are
a₁ = a = – 2
a₂ = a₁ + d
= – 2 + 0 = – 2
a₃ = a₂ + d
= – 2 + 0 = – 2
a₄ = a₃ + d
= – 2 + 0 = – 2
Hence, the first four terms of AP are – 2,- 2, – 2, – 2.

(iii) We have, a = 4, d = – 3
First four terms of required AP are
a₁ = a = 4
a₂ = a₁ + d
= 4 – 3 = 1
a₃ = a₂ + d
= 1 – 3 = – 2
a₄ = a₃ + d
= – 2 – 3 = – 5
Hence, the first four terms of AP are 4, 1, – 2, – 5.

(iv) We have, a = – 1, d = \(\frac{1}{2}\)
First four terms of required AP are
a₁ = a = – 1
a₂ = a₁ + d
= – 1 + \(\frac{1}{2}\) = – \(\frac{1}{2}\)
a₃ = a₂ + d
= – \(\frac{1}{2}\) + \(\frac{1}{2}\) = 0
a₄ = a₃ + d
= 0 + \(\frac{1}{2}\) = \(\frac{1}{2}\)
Hence, the first four terms of AP are – 1, – \(\frac{1}{2}\), 0, \(\frac{1}{2}\).

(v) We have, a = – 1.25, d = – 0.25
First four terms of required AP are
a₁ = a = – 1.25
a₂ = a₁ + d
= – 1.25 – 0.25 = – 1.50
a₃ = a₂ + d
= – 1.50 – 0.25 = – 1.75
a₄ = a₃ + d
= – 1.75 – 0.25 = – 2.00
Hence, the first four terms of AP are – 1.25, – 1.50, – 1.75, – 2.00.

Question 3.
For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, ……………
(ii) – 5, – 1, 3, 7, ……………..
(iii) \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
(iv) 0.6, 1.7, 2.8, 3.9, ……………
Solution :
(i) The given sequence is 3, 1, – 1, – 3, ……………..
Here, a = 3, d = 1 – 3 = – 2
Hence, a = 3, d = – 2.

(ii) The given equation is – 5,- 1, 3, 7, …………..
Here, a = – 5
d = – 1 – (- 5) = – 1 + 5 = 4
Hence a = – 5, d = 4.

(iii) The given sequence is \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
Here, a = \(\frac{1}{3}\)
d = \(\frac{5}{3}-\frac{1}{3}=\frac{4}{3}\)
Hence, a = \(\frac{1}{3}\), d = \(\frac{4}{3}\)

(iv) The given sequence is 0.6, 1.7, 2.8, 3.9, …………….
Here, a = 0.6
d = 1.7 – 0.6 = 1.1
Hence, a = 0.6, d = 1.1.

Question 4.
Which of the following are APs? If they form an AP, find the common difference d and write next three more terms:
(i) 2, 4, 8, 16, …………….
(ii) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), ……………….
(iii) – 1.2, – 3.2, – 5.2, – 7.2, ……………
(iv) – 10, – 6, – 2, 2, …………….
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …………..
(vi) 0.2, 0.22, 0.222, 0.2222, …………….
(vii) 0, – 4, – 8, – 12, …
(viii) \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\)
(ix) 1, 3, 9, 27, …………….
(x) a, 2a, 3a, 4a, …………….
(xi) a, a², a³, a⁴, …………….
(xii) √2, √8, √18, √32, ………………
(xiii) √3, √6, √9, √12, ……………….
(xiv) 1², 3², 5², 7², …………….
(xv) 1², 5², 7², 73, …………….
Solution:
(i) The given sequence is 2, 4, 8, 16, ……………
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ = 8 – 4 = 4
∵ a₂ – a₁ ≠ a₃ – a₂
The given sequence is not in AP.

(ii) The given sequence is 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\), …………………
a₂ – a₁ = \(\frac{5}{2}\) – 2
= \(\frac{5-4}{2}=\frac{1}{2}\)

a₃ – a₂ = 3 – \(\frac{5}{2}\)
= \(\frac{6-5}{2}=\frac{1}{2}\)

a₄ – a₃ = \(\frac{7}{2}\) – 3
= \(\frac{7-6}{2}=\frac{1}{2}\)

∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The given sequence is an AP and
d = \(\frac{5}{2}\) – 2
= \(\frac{5-4}{2}=\frac{1}{2}\)

a₅ = a₄ + d
= \(\frac{7}{2}+\frac{1}{2}=\frac{8}{2}\) = 4

and a₆ = a₅ + d
= 4 + \(\frac{1}{2}\)
= \(\frac{8+1}{2}=\frac{9}{2}\)
and a₇ = a₆ + d
= \(\frac{9}{2}+\frac{1}{2}=\frac{10}{2}\) = 5
Hence, d = \(\frac{1}{2}\) the next three tes are 4, \(\frac{9}{2}\), 5.

(iii) The given sequence is – 1.2, – 3.2, – 5.2, – 7.2, …………..
a₂ – a₁ = – 3.2 – (- 1.2)
= – 3.2 + 1.2 = – 2
a₃ – a₂ = – 5.2 – (- 3.2)
= – 5.2 + 3.2 = – 2
a₄ – a₃ = – 7.2 – (- 5.2)
= – 7.2 + 5.2 = – 2
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
∴ The sequence is an AP and d = a₂ – a₁
= – 3.2 – (- 1.2)
= – 3.2 + 1.2 = – 2
a₅ = a₄ + d
a₅ = – 7.2 – 2 = – 9.2 and
a₆ = a₅ + d
= – 9.2 – 2 = – 11.2 and
a₇ = a₆ + d
= – 11.2 – 2 = – 13.2
Hence, d = – 2, the next three terms are – 9.2, – 11.2, – 13.2.

(iv) The given sequence is – 10, – 6, – 2, 2, …………..
a₂ – a₄ = – 6 – (- 10)
= – 6 + 10 = 4
a₃ – a₂ = – 2 – (- 6)
= – 2 + 6 = 4
a₄ – a₃ = 2 – (- 2)
= 2 + 2 = 4
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
∴ The given sequence is an AP.
and d = a₂ – a₁
= – 6 – (- 10)
= – 6 + 10 = 4
∴ a₅ = a₄ + d
= 2 + 4 = 6
and a₆ = a₅ + d
= 6 + 4 = 10
∴ a₇ = a₆ + d
= 10 + 4 = 14
Hence, d = 4, the next three terms are 6, 10, 14.

(v) The given sequence is
3, 3 + √2, 3 + 2√2, 3 + 3√2, ……………..
a₂ – a₁ = 3 + V2 – 3 = V2
a₃ – a₂ = 3 + 2V2 – 3 – V2 = V2
a₄ – a₃ = 3 + 3V2 – 3 – 2V2 = V2
a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The given sequence is an AP and
d = a₂ – a₁
= 3 + √2 – 3 = √2
a₅ = a₄ + d
= 3 + 3√2 + √2 = 3 + 4√2 and
a₆ = a₅ + d
= 3 + 4√2 + √2 = 3 + 5√2 and
a₇ = a₆ + d
= 3 + 5√2 + √2 = 3 + 6√2
Hence, d = √2 , the next three terms are 3 + 4√2, 3 + 5√2, 3 + 6√2 .

(vi) The given sequence is 0.2, 0.22, 0.222, 0.2222, …
a₂ – a₁ = 0.22 – 0.2 = 0.02
a₃ – a₂ = 0.222 – 0.22 = 0.002
∵ a₂ – a₁ ≠ a₃ – a₂
∴ The given sequence is not an AP.

(vii) The given sequence is 0, – 4, – 8, – 12, ………….
a₂ – a₁ = – 4 – 0 = – 4
a₃ – a₂ = – 8 – (- 4)
= – 8 + 4 = – 4
a₄ – a₃ = – 12 – (- 8)
= – 12 + 8 = – 4
a₂ – a₁ = a₃ – a₂ = a₄ – a₃
∴ The given sequence is an AP and
d = a₂ – a₁
= – 4 – 0 = – 4
a₅ = a₄ + d
= – 12 – 4 = – 16 and
a₆ = a₅ + d
= – 16 – 4 = – 20 and
a₇ = a₆ + d
= – 20 – 4 = – 24
Hence, d = – 4, the next three terms are – 16, – 20, – 24.

(viii) The given sequence is \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\)
a₂ – a₁ = \(-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
a₃ – a₂ = \(-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
a₄ – a₃ = \(-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The given sequence is an AP and
d = a₂ – a₁
= \(-\frac{1}{2}-\left(-\frac{1}{2}\right)\)
= \(-\frac{1}{2}+\frac{1}{2}\) = 0
a₅ = a₄ + d
= \(-\frac{1}{2}\) + 0 = \(-\frac{1}{2}\)
and a₆ = a₅ + d
= \(-\frac{1}{2}\) + 0 = \(\frac{1}{2}\)
and a₇ = a₆ + d
= \(-\frac{1}{2}\) + 0 = \(-\frac{1}{2}\)
Hence, d = 0, the next three terms are \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\).

(ix) The given sequence is 1, 3, 9, 27, ……………
a₂ – a₁ = 3 – 1 = 2
a₃ – a₂ = 9 – 3 = 6
∵ a₂ – a₁ ≠ a₃ – a₂
∴ The given sequence is not an AP.

(x) The given sequence is a, 2a, 3a, 4a, …………..
a₂ – a₁ = 2a – a = a
a₃ – a₂ = 3a – 2a = a
a₄ – a₃= 4a – 3a = a
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
∴ The given sequence is an AP and d = a₂ – a₁
= 2a – a = a
∴ a₅ = a₄ + d
= 4a + a = 5a
and a₆ = a₅ + d
= 5a + a = 6a and
a₇ = a₆ + d
= 6a + a = 7a
Hence, d = a, the next three terms are 5a, 6a, 7a.

(xi) The given sequence is a, a², a³, a⁴, ……………
a₂ – a₁ = a² – a
a₃ – a₂ = a³ – a²
∵ a₂ – a₁ ≠ a₃ – a₂
∴ The given sequence is not an AP.

(xii) The given sequence is √2, √8, √18, √32, …………
a₂ – a₁ = √8 – √2
= 2√2 – √2 = √2
a₃ – a₂ = √18 – √8
= 3√2 – 2√2 = √2
a₄ – a₃ = √32 – √18
= 4√2 – 3√2 = √2
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The given sequence is an AP and d = a₂ – a₁
= √8 – √2
= 2√2 – √2 = √2
∴ a₅ = a₄ + d
= √32 + √2
= 4√2 + √2
= 5√2 = √50 and
a₆ = a₅ + d
= √50 + √2
= 5√2 + √2 = 6√2 = √72
and a₇ = a₆ + d
= √72 + √2
= 6√2 + √2
= 7√2 = √98
Hence, d = √2, the next three terms are √50, √72, √98.

(xiii) The given sequence is √3, √9, √12, ………………..
a₂ – a₁ = √6 – √3
a₃ – a₂ = √9 – √6
= 3 – √6
∵ a₂ – a₁ ≠ a₃ – a₂
The given sequence is not an AP.

(xiv) The given sequence is 1², 3², 5², 7², ……………..
a₂ – a₁ = 3² – 1²
= 9 – 1 = 8
a₃ – a₂ = 5² – 3²
= 25 – 9 = 16
∵ a₂ – a₁ ≠ a₃ – a₂
∴ The given sequence is not an AP.

(xv) The given sequence is 1², 5², 7², 73, ……………..
a₂ – a₁ = 5² – 1²
a₃ – a₂ = 7² – 5²
= 49 – 25 = 24
a₄ – a₃ = 73 – 7²
= 73 – 49 = 24
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃
The given sequence is an AP and
d = a₂ – a₁
= 5² – 1²
= 25 – 1 = 24
∴ a₅ = a₄ + d
= 73 + 24 = 97
and a₆ = a₅ + d
= 97 + 24 = 121
a₇ = a₆ + d
= 121 + 24 = 145
Hence, d = 24, the next three terms are 97, 121, 145.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x².
Solution :
(i) p(x) = x³ – 3x² + 5x – 3
g(x) = x² – 2
Now, we divide p(x) by g(x) as follows :

Hence, quotient q(x) = x – 3 and remainder r(x) = 7x – 9.

(ii) p(x) = x⁴ – 3x² + 4x + 5
and g(x) = x² + 1 – x
We write g(x) in standard form of decreasing power of x.
So, g(x) = x² – x + 1
Now, we divide p(x) by g(x) as follows :

Hence, quotient q(x) = x² + x – 3 and remainder r(x) = 8.

(iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
We write g(x) in standard form of decreasing power of x.
So, g(x) = – x² + 2
Now, we divide p(x) by g(x) as follows :

Hence, the quotient q(x) = – x² – 2 and remainder r(x) = – 5x + 10.

Question 2.
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
(ii) x^x + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution :
(i) We have,
f(t) = 2t⁴ + 3t³ – 2t² – 9t – 12
g(t) = t² – 3
Now, we divide f(t) by g(t) as follows :

Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial.

(ii) We have,
f(x) = 3x⁴ + 5x³ – 7x² + 2x + 2 and g(x) = x² + 3x + 1
Now, we divide f(x) by g(x) as follows :

Since, the remainder is zero therefore, the second polynomial is divisible by first polynomial.
Hence, the first polynomial is a factor of the second polynomial.

(iii) We have,
f(x) = x⁵ – 4x³ + x² + 3x + 1
g(x) = x³ – 3x + 1
Now, we divide f(x) by g(x) as follows :

Since remainder = 2 ≠ 0 therefore, second polynomial is not divisible by first polynomial.
Hence, first polynomial is not a factor of second polynomial.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)
Solution :
We know that if a is zero of the polynomial then (x – α) is a factor of given polynomial.
Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of the polynomial f(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Hence, (x – \(\sqrt{\frac{5}{3}}\)) (x + \(\sqrt{\frac{5}{3}}\)) = x² – \(\frac{5}{3}\) is a factor of f(x).
Now, we divide f(x) by x² – \(\frac{5}{3}\) to obtain the other zeroes as follows :

According to division algorithm of polynomial,
∴ 3x⁴ + 6x³ – 2x² – 10x – 5 = (x² – \(\frac{5}{3}\)) (3x² + 6x + 3)
= (x² – \(\frac{5}{3}\)) × 3(x² + 2x + 1)
= 3(x² – \(\frac{5}{3}\)) (x + 1)²
For zeroes of the polynomial f(x) = 0
∴ 3 (x² – \(\frac{5}{3}\)) (x + 1)² = 0
⇒ (x + \(\sqrt{\frac{5}{3}}\)) (x – \(\sqrt{\frac{5}{3}}\)) (x + 1) (x + 1) = 0
⇒ x = – \(\sqrt{\frac{5}{3}}\), \(\sqrt{\frac{5}{3}}\) , – 1, – 1
Hence, the other zeroes of polynomial are – 1, – 1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder are x – 2 and – 2x + 4, respectively. Find g(x).
Solution :
We have,
f(x) = x³ – 3x² + x + 2
q(x) = x – 2
r(x) = – 2x + 4; g(x) = ?
According to division algorithm of polynomials
f(x) = g(x) × q(x) + r(x)
⇒ f(x) – r(x) = g(x) × q(x)

Hence, g(x) = x² – x + 1.

Question 5.
Give examples of polynomial p(x), g(x) and Kx), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg Kx)
(iii) deg Kx) = 0.
Solution :
(i) Let p(x) = 4x² + 4x + 12 and g(x) = 4.
Now, we divide p(x) by g(x) and we get
q(x) = \(\frac{4 x^2+4 x+12}{4}\)
= x² + x + 3
∴ r(x) = 0.
Hence, p(x) = 4x² + 4x + 12, g(x) = 4, q(x) = x² + x + 3, r(x) = 0
∴ deg p(x) = deg q(x)

(ii) Let p(x) = x³ + x² + x + 1
and g(x) = x² – 1
Now, we get q(x) and r(x) on dividing p(x) by g(x) as follows :

Hence, p(x) = x³ + x² + x + 1, g(x) = x² – 1, r(x) = 2x + 2, q(x) = x + 1
deg q(x) = deg r(x).

(iii) Let p(x) = x³ + 4x² – x + 7
and g(x) = x² – 1
Now, we get q(x), r(x) on dividing p(x) by g(x) as follows :

Hence, p(x) = x³ + 4x² – x + 7, g(x) = x² – 1, g(x) = x + 4, r(x) = 11.
∴ deg r(x) = 0.

Haryana State Board HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution :
(i) The given equation is :
2x² – 3x + 5 = 0
Comparing the given equation with the general
form of quadratic equation, ax² + bx + c = 0, we get
a = 2, b = – 3, c = 5
∴ D = b² – 4ac
⇒ D = (- 3)² – 4 × 2 × 5
⇒ D = 9 – 40
⇒ D = – 31.
∵ D < 0
The real roots do not exist.

(ii) The given equation is :
3x² – 4√3x + 4 = 0
Comparing the given equation with the general form of quadratic equation, ax² + bx + c = 0, we get
a = 3, b = – 4√3 , c = 4
∴ D = b² – 4ac
⇒ D = (- 4√3)² – 4 × 3 × 4
⇒ D = 48 – 48
⇒ D = 0
The roots of the equation are real and equal.
x = \(-\frac{b}{2 a} \text { and }-\frac{b}{2 a}\)
x = \(\frac{4 \sqrt{3}}{2 \times 3} \text { and } \frac{4 \sqrt{3}}{2 \times 3}\)
x = \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)
Hence, the roots of the equation are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

(iii) The given equation is :
2x² – 6x + 3 = 0
Comparing the given equation with the general form of quadratic equation ax2 + bx + c = 0, we get a = 2, b = – 6, c = ∴ D = b² – 4ac
⇒ D = (- 6)² – 4 × 2 × 3
⇒ D = 36 – 24
⇒ D = 12
∵ D > 0
The roots of the equation are real and distinct.

Hence, the roots of the equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\).

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots :
(i) 2×2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0.
Solution:
(i) The given equation is :
2x² + kx + 3 = 0
Comparing the given equation with general form of the quadratic equation ax2 + bx + c = 0, we get
a = 2, b = k, c = 3
D = b² – 4ac
⇒ D = k² – 4 × 2 × 3
⇒ D = k² – 24
Since, the roots of the given equation are equal.
Therefore, D = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = ± √24
⇒ k = ± 2√6
Hence, k = ± 2√6

(ii) The given equation is : kx (x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Comparing the given equation with the general form of quadratic equation ax² + bx + c = 0, we get
a = k, b = – 2k, c = 6
D = b² – 4ac
⇒ D = (- 2k)² – 4 × k × 6
⇒ D= 4k² – 24k
Since the roots of the given equation are equal.
Therefore, D = 0
⇒ 4k² – 24k = 0
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or & k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 [Not possible]
Hence, k = 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Solution:
Let the breadth of the rectangular grove be x m.
Then length of rectangular grove = 2x m
According to question,
2x × x = 800
[∵ Area of rectangle = Length × Breadth]
⇒ 2x² = 800
⇒ x² = \(\frac{800}{2}\)
⇒ x² = 400
⇒ x = √400
⇒ x = ± 20
Reject x = – 20
[∵ Breadth cannot be negative.]
∴ x = 20
Hence, the rectangular grove is possible and its breadth = 20 m
and length 2 × 20 = 40 m.

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution :
Let the present age of first friend be x years.
Then present age of second friend = (20 – x) years
Four years ago age of first friend = (x – 4) years
Four years ago age of second friend = (20 – x – 4) years = (16 – x) years
According to question,
(x – 4)(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ – x² + 20x – 64 – 48 = 0
⇒ – x² + 20x – 112 = 0
⇒ x² – 20x + 112 = 0
Here, a = 1, b = – 20, c = 112
∴ D = b² – 4ac
⇒ D = (- 20)² – 4 × 1 × 112
D = 400 – 448
⇒ D = – 48
∵ D < 0
∴ The equation has no real value of x. Hence, the situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Solution:
Let the breadth of rectangular park be x m.
Perimeter of rectangular park= 2 (length + breadth)
⇒ 80 = 2(length + x)
⇒ \(\frac{80}{2}\) = length + x
⇒ 40 = length + x
⇒ length = (40 – x) m
Now, according to question,
Area of rectangular park = Length × Breadth
400 = (40 – x) × x
⇒ 400 = 40x – x²
⇒ x² – 40x + 400 = 0
Here, a = 1, b = – 40, c = 400
D = b² – 4ac
⇒ D = (- 40)² – 4 × 1 × 400
⇒ D = 1600 – 1600
⇒ D = 0
The equation has real and equal roots.
⇒ x = 20 and 20
Hence, the rectangular park is possible and its length = 20 m and breadth = 20 m.

Haryana State Board HBSE 10th Class Science Notes Chapter 13 विद्युत धारा का चुम्बकीय प्रभाव Notes.

Haryana Board 10th Class Science Notes Chapter 13 विद्युत धारा का चुम्बकीय प्रभाव

→ दिक्सूचक एक छोटा चुम्बक होता है जो सदा उत्तर व दक्षिण दिशाओं की ओर संकेत करता है।

→ उत्तरी ध्रुव व दक्षिणी ध्रुव (North pole and South pole)–दिक्सूचक का जो सिरा उत्तर की ओर संकेत करता है . उसे उत्तरी ध्रुव तथा जो सिरा दक्षिण की ओर संकेत करता है उसे दक्षिणी ध्रुव कहते हैं।

→ ध्रुव (Pole)-चुम्बकों के सजातीय ध्रुवों में परस्पर प्रतिकर्षण एवं विजातीय ध्रुवों में परस्पर आकर्षण होता है।

→ चुम्बकीय बल (Magnetic force)—किसी चुम्बक के चारों ओर एक चुम्बकीय क्षेत्र होता है जिसमें उस चुम्बक के बल का संसूचन किया जा सकता है। |

→ चुम्बकीय क्षेत्र रेखायें (Magnetic field lines) किसी चुम्बकीय क्षेत्र के निरूपण के लिए चुम्बकीय क्षेत्र रेखाओं का उपयोग किया जाता है। चुम्बकीय क्षेत्र के किसी बिन्दु पर क्षेत्र की दिशा उस बिन्दु पर रखे उत्तर ध्रुव । की गति की दिशा द्वारा दिखाई जाती है। चुम्बकीय क्षेत्र के प्रबल होने पर ये रेखाएँ निकट होती हैं।

→ चुम्बकीय क्षेत्र (Magnetic field)-चुम्बकीय क्षेत्र एक ऐसी राशि है जिसमें परिमाण और दिशा दोनों होते हैं।

→ चुम्बकीय क्षेत्र रेखाएँ बंद वक्र होती हैं तथा एक-दूसरे को कभी भी प्रतिच्छेदित नहीं करती।

→ किसी धात्विक चालक में विद्युत धारा प्रवाहित करने पर उसके चारों ओर एक चुम्बकीय क्षेत्र उत्पन्न हो जाता है।

→ किसी विद्युत धारावाही धातु के तार से एक चुम्बकीय क्षेत्र संबद्ध होता है। तार के चारों ओर क्षेत्र रेखाएँ अनेक संकेन्द्री वृत्तों के रूप में होती हैं, जिनकी दिशा दक्षिण हस्त अंगुष्ठ नियम द्वारा ज्ञात की जाती है।

→ परिनालिका (Solenoid)-पास-पास लिपटे विद्युतरोधी ताँबे के तार की बेलन की आकृति की अनेक फेरों वाली कुंडली को परिनालिका कहते हैं।

→ परिनालिका के भीतर चुम्बकीय क्षेत्र रेखाएँ समान्तर सरल रेखाओं की तरह होती हैं। किसी परिनालिका के भीतर सभी बिन्दुओं पर चुम्बकीय क्षेत्र समान होता है।

→ विद्युत चुम्बक (Electromagnet) विद्युत चुम्बक में नर्म लौह क्रोड होता है जिसके चारों ओर विद्युतरोधी ताँबे के तार की कुंडली लिपटी रहती है।

→ कोई विद्युत धारावाही चालक चुम्बकीय क्षेत्र में रखे जाने पर बल का अनुभव करता है। यदि चुम्बकीय क्षेत्र तथा विद्युत धारा की दिशाएँ परस्पर लम्बवत् हैं तो इस दशा में चालक पर आरोपित बल की दिशा इन दोनों के लम्बवत् होती है, जिसे फ्लेमिंग के वामहस्त नियम से प्राप्त किया जाता है।

→ विद्युत मोटर, विद्युत जनित्र, विद्युत मापक यंत्र का सम्बन्ध विद्युत धारावाही तथा चुंबकीय क्षेत्र से है।

→ विद्युत मोटर एक ऐसी युक्ति है जो विद्युत ऊर्जा को यांत्रिक ऊर्जा में रूपान्तरित करती है।

→ विद्युत मोटरों का उपयोग विद्युत पंखों, रेफ्रिजरेटरों, वाशिंग मशीनों, कम्प्यूटर आदि में किया जाता है।

→ दिक्परिवर्तक (Commutator)-विद्युत मोटर में विद्युतरोधी तार की एक आयताकार कुंडली किसी चुम्बकीय क्षेत्र के दो ध्रुवों के बीच रखी जाती है। वह युक्ति जो परिपथ में विद्युत धारा के प्रवाह को उत्क्रमित कर देती है, उसे दिक्परिवर्तक कहते हैं।

→ विद्युत चुम्बकीय प्रेरण (Electromagnetic Induction)-वैद्युत चुम्बकीय प्रेरण एक ऐसी परिघटना है जिसमें किसी कुंडली में जो किसी ऐसे क्षेत्र में स्थित है जहाँ समय के साथ चुम्बकीय क्षेत्र परिवर्तित होता है, एक प्रेरित विद्युत धारा उत्पन्न होती है।

→ चुम्बकीय क्षेत्र में परिवर्तन किसी चुम्बक तथा उसके पास स्थित किसी कुंडली के बीच आपेक्षिक गति के कारण हो सकता है। यदि कुंडली किसी विद्युत धारावाही चालक के निकट रखी है तब कुंडली से संबद्ध चुम्बकीय क्षेत्र या तो चालक से प्रवाहित विद्युत धारा में अन्तर के कारण हो सकता है अथवा चालक तथा कुंडली के बीच आपेक्षिक गति के कारण हो सकता है। प्रेरित विद्युत धारा की दिशा फ्लेमिंग के दक्षिण-हस्त नियम द्वारा प्राप्त की जाती है।

→ जब कुंडली की गति की दिशा चुम्बकीय क्षेत्र की दिशा के लम्बवत् होती है तब कुंडली में उत्पन्न प्रेरित विद्युत धारा अधिकतम होती है।

→ विद्युत जनरेटर (Electric generator)-विद्युत उत्पन्न करने की युक्ति को विद्युत धारा जनित्र (ac जनित्र) कहते हैं।। 1

→ दिष्ट धारा प्राप्त करने के लिए विभक्त वलय प्रकार के दिक्परिवर्तक का उपयोग किया जाता है।

→ दिष्ट धारा (dc) सदा एक ही दिशा में प्रवाहित होती है लेकिन प्रत्यावर्ती धारा (ac) एक निश्चित काल अंतराल के बाद अपनी दिशा उत्क्रमित करती रहती है।

→ हमारे देश में प्रत्यावर्ती धारा की आवृत्ति 50 ह है।

→ हम अपने घरों में प्रत्यावर्ती विद्युत शक्ति 220 V पर प्राप्त करते हैं। आपूर्ति का एक तार लाल विद्युत रोधन युक्त होता है जिसे विद्युन्मय तार (धनात्मक तार) कहते हैं। दूसरे पर काला विद्युत रोधन होता है जिसे उदासीन तार (ऋणात्मक तार) कहते हैं। इन दोनों तारों के बीच 220 V का विभवांतर होता है। तीसरा तार भूसंपर्क तार (अर्थ वॉयर) होता है जिस पर हरा विद्युत रोधन होता है।

→ अतिभारण (Over loading)-जब विद्युन्मय और उदासीन तार सीधे सम्पर्क में आते हैं तो अतिभारण हो सकता है।।

→ भूसंपर्कण एक सुरक्षा उपाय है जो यह सुनिश्चित करता है कि साधित्र के धात्विक आवरण में यदि कोई विद्युत धारा का कोई भी क्षरण होता है तो उस साधित्र का उपयोग करते समय व्यक्ति को झटका न लगे।

→ विद्युत फ्यूज (Electric fuse) विद्युत फ्यूज सभी घरेलू परिपथों का महत्वपूर्ण हिस्सा होता है। यह अतिभारण के कारण होने वाली हानि से बचाता है। !

→ फ्यूजों में होने वाला तापन फ्यूज को पिघला देता है जिससे विद्युत परिपथ टूट जाता है।

Haryana State Board HBSE 10th Class Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4.
Solution :
(i) We have, f(x) = x² – 2x – 8
= x² – (4 – 2)x – 8
= x² – 4x + 2x – 8
= (x² – 4x) + (2x – 8)
= x(x – 4) + 2(x – 4)
= (x – 4) (x + 2)
To find the zeroes of the polynomial f(x), put f(x) = 0
⇒ (x – 4) (x + 2) = 0
⇒ x = 4, – 2
Therefore, the zeroes of polynomial f(x) are : α = 4, β = – 2.
Now, Sum of the zeroes = α + β = (4 – 2)
= 2 = \(\frac{-(-2)}{1}\)
= \(\frac{\text { – Coefficient of } x}{\text { Coefficient of } x^2}\)

Product of zeroes = α × β
= 4 × (- 2) = \(-\frac{8}{1}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)
Hence, zeroes of the polynomial f(x) are 4, – 2.

(ii) We have,
f(s) = 4s² – 4s + 1
= 4s² – (2 + 2)s + 1
= 4s² – 2s – 2s + 1
= (4s² – 2s) – (2s – 1)
= 2s (2s – 1) – 1 (2s – 1)
= (2s – 1) (2s – 1)
To find the zeroes of the polynomial f(s), put f(s) = 0.
⇒ (2s – 1) (2s – 1) = 0
⇒ s = \(\frac{1}{2}\), \(\frac{1}{2}\)
Therefore, zeroes of the polynomial f(s) are α = \(\frac{1}{2}\), β = \(\frac{1}{2}\).
∴ It has equal roots.
Now, Sum of zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
= \(\frac{-(-4)}{4}\)
= \(\frac{\text { – Coefficient of } s}{\text { Coefficient of } s^2}\)

Product of zeroes = α × β
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } s^2}\)
Hence, zeroes of polynomial f(s) = \(\frac{1}{2}\), \(\frac{1}{2}\).

(iii) We have,
p(x) = 6x² – 3 – 7x
= 6x² – 7x – 3
= 6x² – (9 – 2) x – 3
= 6x² – 9x + 2x – 3
= 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
To find the zeroes of the polynomial p(x)
put p(x) = 0
⇒ (2x – 3) (3x + 1) = 0
⇒ x = \(\frac{3}{2}\), – \(\frac{1}{3}\)
Therefore, zeroes of the polynomial p(x) are : α = \(\frac{3}{2}\), β = – \(\frac{1}{3}\)
Now, Sum of zeroes = α + β
= \(\frac{3}{2}\) – \(\frac{1}{3}\)
= \(\frac{9-2}{6}=\frac{7}{6}=\frac{-(-7)}{6}\)
= \(\frac{\text { – Coefficient of } x}{\text { Coefficient of } x^2}\)

Product of zeroes = α × β
= \(\frac{3}{2} \times\left(-\frac{1}{3}\right)=-\frac{3}{6}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)
Hence, zeroes of the polynomial p(x) = \(\frac{3}{2}\), – \(\frac{1}{3}\)

(iv) We have,
f(u) = 4u² + 8u
= 4 u(u + 2)
To find the zeroes of the polynomial f(u), put f(u) = 0
⇒ 4 u(u + 2) = 0
⇒ u = 0, – 2.
Therefore, zeroes of the polynomial f(u) are : α = 0, and β = – 2.
Now, Sum of zeroes = α + β
= 0 + (- 2) = – 2 = \(\frac{-8}{4}\)
= \(\frac{\text { – Coefficient of } u}{\text { Coefficient of } u^2}\)

Product of zeroes = α × β
= 0 × (- 2)
= \(\frac{0 \times(-8)}{4}=\frac{0}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } u^2}\)
Hence, zeroes of the polynomial f(u) = 0, – 2.

(v) We have,
f(t) = t² – 15
= (t)² – (√15)²
= (t + √15) (t – √15)
To find the zeroes of polynomial f(t), put f(t) = 0.
⇒ (t + √15) (t – √15) = 0.
⇒ t = – √15, √15
Therefore, zeroes of polynomial f(t) are :
α = – √15 and β = √15
Now, Sum of zeroes = α + β
= – √15 + √15 = 0 = – \(\frac{0}{1}\)
= \(\frac{\text { – Coefficient of } t}{\text { Coefficient of } t^2}\)

Product of zeroes = α × β
= – √15 × √15
= \(\frac{\text { Constant term }}{\text { Coefficient of } t^2}\)
Hence, zeroes of the polynomial f(t) = – √15 , √15

(vi) We have,
f(x) = 3x² – x – 4
= 3x² – (4 – 3)x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1 (3x – 4)
= (3x – 4) (x + 1)
To find the zeroes of the polynomial f(x),
put f(x) = 0
⇒ (3x – 4) (x + 1) = 0
⇒ x = \(\frac{4}{3}\), – 1
Therefore, zeroes of the polynomial f(x) are : α = \(\frac{4}{3}\) and β= – 1
Now, Sum of zeroes = α + β
= \(\frac{4}{3}\) – 1
= \(\frac{4-3}{3}=\frac{1}{3}=\frac{-(-1)}{3}\)
= \(\frac{\text { – Coefficient of } x}{\text { Coefficient of } x^2}\)

Product of zeroes = α × β
= \(\frac{4}{3}\) × (- 1)
= – \(\frac{4}{3}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)
Hence, zeroes of polynomial f(x) = \(\frac{4}{3}\) , – 1.

Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\frac{1}{4}\), – 1
(ii) √2, \(\frac{1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) – \(\frac{1}{4}\), \(\frac{1}{4}\)
(vi) 4, 1.
Solution :
(i) We have,
Sum of zeroes = (α + β) = \(\frac{1}{4}\)
Product of zeroes = α × β = – 1
So, the required quadratic polynomial is f(x) = x² – (α + β)x + α × β
= x² – \(\frac{1}{4}\) x – 1
= \(\frac{1}{4}\) (4x² – x – 4)
∴ Quadratic polynomial = 4x² – x – 4.

(ii) We have,
Sum of zeroes = (α + β) = √2
Product of zeroes = α × β = \(\frac{1}{3}\)
So, the required quadratic polynomial is :
p(x) = x² – (α + β)x + α × β
= x² – √2x + \(\frac{1}{3}\)
= \(\frac{1}{3}\) (3x² – 3√2x + 1)
∴ Quadratic polynomial = 3x² – 3√2x + 1

(iii) We have,
Sum of zeroes = (α + β) = 0.
Product of zeroes = α × β = √5
So, the required quadratic polynomial is
f(x) = x² – (α + β)x + α × β
= x² – 0 × x + √5
= x² +√5

(iv) We have,
Sum of zeroes = α + β = 1
Product of zeroes = α × β = 1
So, the required quadratic polynomial is f(x) = x² – (α + β)x + α × β
= x² – 1.x + 1
= x² – x + 1

(v) We have,
Sum of zeroes = α + β = – \(\frac{1}{4}\)
Product of zeroes = α × β = \(\frac{1}{4}\)
So, the required quadratic polynomial is f(x) = x² – (α + β)x + α × β
= x² – (- \(\frac{1}{4}\)) x + \(\frac{1}{4}\)
= x² + \(\frac{1}{4}\) x + \(\frac{1}{4}\)
= \(\frac{1}{4}\) (4x² + x + 1)
∴ Quadratic polynomial = 4x² + x + 1.

(vi) We have,
Sum of zeroes = α + β = 4
Product of zeroes = α × β = 1
So, the required polynomial is
f(x) = x² – (α + β)x + α × β
= x² – 4x + 1