Haryana State Board HBSE 10th Class Maths Important Questions Chapter 4 Quadratic Equations Important Questions and Answers.

## Haryana Board 10th Class Maths Important Questions Chapter 4 Quadratic Equations

Short/Long Answer Type Questions

Question 1.

If x = 3 is one root of quadratic equation x^{2} – 2kx – 6 = 0, then find the value of k.

Solution :

Since, x = 3 is one root of the quadratic equation x^{2} – 2kx – 6 = 0. So, we put x = 3 in the given equation, we get

(3)^{2} – 2k × 3 – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ – 6k + 3 = 0

⇒ k = \(\frac {3}{6}\)

⇒ k = \(\frac {1}{2}\)

Question 2.

If one root of the quadratic equation 6x^{2} – x – k = 0 is \(\frac {2}{3}\) then find the value of k.

Solution :

Since, \(\frac {2}{3}\) is one root of the quadratic equation 6x^{2} – x – k = 0.

So, we put x = \(\frac {2}{3}\) in the given equation, we get

6 × (\(\frac {2}{3}\))^{2} – \(\frac {2}{3}\) – k = 0

⇒ 6 × \(\frac{4}{9}-\frac{2}{3}\) – k = 0

⇒ \(\frac{8}{3}-\frac{2}{3}\) – k = 0

⇒ \(\frac {6}{3}\) – k = 0

⇒ 2 – k = 0

⇒ k = 2

Question 3.

Find the possible root of \(\sqrt{3 x^2+6}\) = 9.

Solution :

The given equation is

\(\sqrt{3 x^2+6}\) = 9

Squaring on both sides, we get

3x^{2} + 6 = 81

⇒ 3x^{2} = 81 – 6

⇒ 3x^{2} = 75

⇒ x^{2} = \(\frac {75}{3}\)

⇒ x^{2} = 25

Taking square root both sides, we get

\(\sqrt{x^2}\) = \(\sqrt{25}\)

x = ± 5

Question 4.

Solve : \(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}=\frac{1}{x}\) where a + b ≠ 0.

Solution :

The given equation is :

⇒ – (a + b) × ab = (a + b)x (a + b + x)

⇒ (a + b)x (a + b + x) + ab (a + b) = 0

⇒ (a + b) [x(a + b + x) + ab] = 0

⇒ x (a + b + x) + ab = 0

⇒ ax + bx + x^{2} + ab = 0

⇒ x^{2} + ax + bx + ab = 0

⇒ x(x + a) + b (x + a) = 0

⇒ (x + a) (x + b) = 0

⇒ (x + a) = 0 and (x + b) = 0

⇒ x = – a and x = – b

Question 5.

A train travels 300 km at a uniform speed. If the speed had been 10 km/hr more, it would have 1 hr less for same journey. Find the speed of train.

Solution :

Let the speed of train be x km/hr

Distance = 300 km (given)

Time taken by train = \(\frac {300}{x}\)hrs.

[∵ Time = Distance / Speed]

If speed had been 10 km/hr more then new speed = (x + 10) km/hr.

Time taken = \(\frac{360}{x+10}\) hrs.

According to question

Question 6.

The diagonal of a rectangular field is 25 metres more than the shorter side. If longer side is 23 metres more than the shorter side, find the sides of the field.

Solution :

Let the shorter side be x m. Then diagonal = (x + 25)m and longer side = (x + 23)m.

We known that each angle of a rectangle is 90°. In a right ΔABC, we have (x + 25)^{2} = (x + 23)^{2} + x^{2}.

⇒ x^{2} + 50x + 625 = x^{2} + 46x + 529 + x^{2}

⇒ 2x^{2} + 46x + 529 – x^{2} – 50x – 625 = 0

⇒ x^{2} – 4x – 96 = 0

⇒ x^{2} – (12 – 8)x – 96 = 0

⇒ x^{2} – 12x + 8x – 96 = 0

⇒ (x^{2} – 12x) + (8x – 96) = 0

⇒ x(x – 12) + 8 (x – 12) = 0

⇒ (x – 12) (x + 8) = 0

⇒ x – 12 = 0 and x + 8 = 0

⇒ x = 12 and x = – 8 (Reject)

Hence, longer side 12 + 23 i.e., 35 m and shorter side = 12 m.

Question 7.

In a flight of 600 km, an discraft was slowed down due to bad weather. The average speed of the strip was reduced by 200 km/hr and time of flight increased by 30 minutes. Find the duration of flight.

Solution :

Distance = 600 km.

Let the speed of discraft be x km/hr. Time taken by discraft (T_{1}) = \(\frac {600}{x}\) hrs.

It speed of is craft was slow down by 200 km/ hr then new speed = (x – 200) km/hr.

Time taken by discraft (T_{2}) = \(\frac{600}{x-200}\) hrs

According to question,

⇒ 2,40,000 = x^{2} – 200x

⇒ x^{2} – 200x – 2,40,000 = 0

⇒ x^{2} – (600 – 400)x – 2,40,000 = 0

⇒ x^{2} – 600x + 400x – 2,40,000 = 0

⇒ x(x – 600) + 400(x – 600) = 0

⇒ (x – 600) (x + 400) = 0

⇒ x – 600 = 0 and x + 400 = 0

⇒ x = 600 and x = – 400 (Reject)

So, speed of discraft = 600 km/hr and duration of flight = \(\frac{600}{600-200}\) hrs

= \(\frac{600}{400}=\frac{3}{2}\) = 1\(\frac {1}{2}\)hrs.

Question 8.

The speed of a boat in still water is 18 km/hr. It takes \(\frac {1}{2}\) an hours extra in going 12 km upstream instread of going the same distance down stream. Find the speed of the stream.

Solution :

Let the speed of stream be x km/hr. Speed of boat in still water = 18 km/hr (given)

Speed of boat in down stream = (18 + x) km/hr

Time taken in going down stream (T_{1}) = \(\frac{12}{(18+x)}\) hrs

Speed of boat in upstream (18 – x) km/hr.

Time taken in going upstream (T_{2}) = \(\frac{12}{(18-x)}\) hrs

According to question,

⇒ 48x = – x^{2} + 324

⇒ x^{2} + 48x – 324 = 0

⇒ x^{2} + (54 – 6)x – 324 = 0

⇒ x^{2} + 54x – 6x – 324 = 0

⇒ x(x + 54) – 6(x + 54) = 0

⇒ (x + 54) (x – 6) = 0

⇒ x + 54 = 0 and x – 6 = 0

⇒ x = – 54 (Reject) and x = 6

Hence, speed of stream = 6 km/hr.

Question 9.

I can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately ?

Solution :

Let the time taken by larger pipe be x hours and time taken by smaller pipe be y hours

The portion of tank filled by larger pipe in 1 hour = \(\frac {1}{x}\)part

The portion of tank filled by smaller pipe in 1 hour = \(\frac {1}{y}\)part

Two pipe can fill the tank in 12 hours.

∴ The portion of tank filled by both pipe in 1 hr = \(\frac {1}{12}\)part

According to condition 1st

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\) ……………(1)

If larger pipe used for 4 hours, then the portion of tank filled by larger pipe in 4 hours = \(\frac {4}{x}\)part

And smaller pipe used for 9 hours, then. Portion of tank filled by smaller pipe in 9 hours = \(\frac {9}{y}\)part

According to condition 2nd

\(\frac{4}{x}+\frac{9}{y}=\frac{1}{2}\) ………..(2)

Let \(\frac{1}{x}\) be a and \(\frac{1}{y}\) be b, from equations (1) and (2) we get

a + b = \(\frac {1}{12}\) ………(3)

4a + 9b = \(\frac {1}{2}\) ………….(4)

By cross-multiplication, we get

Putting the values of a and b, we get

\(\frac{1}{x}=\frac{1}{20}\) and \(\frac{1}{y}=\frac{1}{30}\)

⇒ x = 20 and y = 30

Hence, time taken by larger pipe is 20 hours and smaller pipe is 30 hours.

Question 10.

Find the value of k for which are roots of the equation 3x^{2} – 10x + k = 0 are reciprocal of each other.

Solution :

The given equation is :

3x^{2} – 10x + k = 0

Let one root of the equation be a since roots are reciprocal each other so, other root is \(\frac {1}{α}\)

Product of two roots = \(\frac {c}{a}\)

⇒ α × \(\frac {1}{α}\) = \(\frac {k}{3}\)

⇒ 1 = \(\frac {k}{3}\)

⇒ k = 3

Question 11.

For what value of k, the given quadratic equation kx^{2} – 6x – 1 = 0 has no real roots ?

Solution :

The given equation is :

kx^{2} – 6x – 1 = 0

The condition for no real roots is :

D < 0

⇒ b^{2} – 4ac < 0

⇒ (-6)^{2} – 4 × k × -1 < 0

⇒ 36 + 4k < 0

⇒ 4k < – 36

⇒ k < –\(\frac {36}{4}\)

⇒ k < – 9

Hence, k should be less than – 9 i.e., (-10, – 11, …..).

Question 12.

For what values of k, the roots of the equation kx^{2} + 4x + k = 0 are real ?

Solution :

The given equation is :

x^{2} + 4x + k = 0

∵ The given equation has real roots

∴ D ≥ 0

⇒ b^{2} – 4ac ≥ 0

⇒ (4)^{2} – 4 × 1 × k ≥ 0

⇒ 16 – 4k ≥ 0

⇒ – 4k ≥ – 16

⇒ k ≤ \(\frac {-16}{-4}\)

⇒ k ≤ + 4

Therefore, k should be ≤ 4 i.e., (4, 3, 2, 1, ….)

Question 13.

Find the k so that the quadratie equation (k + 1)x^{2} – 2 (k + 1)x + 1 = 0 has equal roots.

Solution :

The given equation is :

(k + 1)x^{2} – 2(k + 1) x + 1 = 0

Since, equation has equal roots

∴ D = 0

⇒ b^{2} – 4ac = 0

⇒ [-2(k + 1)]^{2} – 4 × (k + 1) × 1 = 0

⇒ 4(k^{2} + 2k + 1) – 4k – 4 = 0

⇒ 4k^{2} + 8k + 4 – 4k – 4 = 0

⇒ 4k^{2} + 4k = 0

⇒ 4k(k + 1) = 0 and k + 1 = 9

⇒ k = – 1

k = – 1 (Reject since if we put the k = – 1 in the equation (-1 + 1)x will be zero

k = 0

Question 14.

If – 3 is a root of the quadratic equation 2x^{2} + Px – 15 = 0 while the quadratic equation x^{2} – 4Px + k = 0 has equal root. Find the value of k.

Solution :

Since, – 3 is a root of the equation

2x^{2} + Px – 15 = 0.

So, we put x = – 3 in this equation, we get

2(-3)^{2} + P(-3) – 15 = 0

⇒ 18 – 3P – 15 = 0

⇒ 3 – 3P = 0

⇒ P = \(\frac {-3}{-3}\) = 1

Putting the value of P in the equation, we get

x^{2} – 4 × 1 + k = 0

⇒ x^{2} – 4x + k = 0

The equation x^{2} – 4x + k = 0 has equal roots.

So, D = 0

⇒ b^{2} – 4ac = 0

⇒ (-4)^{2} – 4 × 1 × k = 0

⇒ 16 – 4k = 0

⇒ k = \(\frac {-16}{-4}\)

⇒ k = 4

Question 15.

If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0 are equal, then show that a = b = c.

Solution :

The given equation is :

(x – a) (x – b) + (x – b) (x – c) + (x – c)(x – a) = 0

x^{2} – ax – bx + ab + x^{2} – bx – xc + bc + x^{2} – cx – ax + ac = 0

⇒ 3x^{2} – 2ax – 2x – 2x + ab + bc + ac = 0

⇒ 3x^{2} – 2 (a + b + c) x + (ab + bc + ac) = 0

Since, roots are equal, then

D = 0

⇒ b^{2} – 4ac = 0

⇒ [- 2 (a + b + c)]^{2} – 4 × 3 × (ab + bc + ac) = 0

⇒ 4(a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca) – 12 (ab + bc + ac) = 0

⇒ 4[a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca – 3ab – 3bc – 3ac] = 0

⇒ a^{2} + b^{2} + c^{2} – ab – bc – ac = 0

⇒ \(\frac {1}{2}\)[2a^{2} + 2b^{2} + 2c^{2} – 2ab – 2bc – 2ac] = 0

⇒ \(\frac {1}{2}\)[(a^{2} + b^{2} – 2ab) + (b^{2} + c^{2} – 2bc) + (a^{2} + c^{2} – 2ac)] = 0

⇒ (a – b)^{2} + (b – c)^{2} + (a – c)^{2} = 0

⇒ (a – b)^{2} = 0, (b – c)^{2} = 0, (a – c)^{2} = 0

⇒ a = b, b = c, a = c

∴ a = b = c

Hence Proved.

Question 16.

If the root of the quadratic equation (c^{2} – ab)x^{2} – 2 (a^{2} – bc)x + b^{2} – ac = 0 in x are equal, then show that either a = 0 or a^{3} + b^{3} + c^{3} = 3abc.

Solution :

The given equation is:

(c^{2} – ab)x^{2} – 2(a^{2} – bc)x + (b^{2} – ac) = 0

Since, roots are equal, then

D = 0

⇒ b^{2} – 4ac = 0

⇒ [- 2(a^{2} – bc)]^{2} – 4 × (c^{2} – ab) (b^{2} – ac) = 0

⇒ 4(a^{4} + b^{2}c^{2} – 2abc) – 4(b^{2}c^{2} – ac^{3} – ab^{3} + a^{2}bc) = 0

⇒ 4[a^{4} + b^{2}c^{2} – 2a^{2}bc – b^{2}c^{2} + ac^{3} + ab^{3} – a^{2}bc) = 0

⇒ [a^{4} – 3a^{2}bc + ac^{3} + ab^{3}] = 0

⇒ a[a^{3} + b^{3} + c^{3} – 3abc] = 0

⇒ a = 0 or a^{3} + b^{3} + c^{3} – 3abc = 0

⇒ a = 0 or a^{3} + b^{3} + c^{3} = 3abc

Hence proved.

Question 17.

A pole has to erected at a point on the boundary of a circular park of diameter 17 metres in such a way that the differences of its distances from two dimetrically opposite fixed gates A and B on the boundary is 7 metres. It is possible to do so? If yes, at what distances from the two gates should the pole be erected ?

Solution :

Let us first draw the diagram. Let P be location of the pole. Let the distance of pole from the gate B be x m ie., BP = x m. Now distances of the pole from the two gates = AP – BP (or BP – AP) = 7 m.

Therefore AP = (x + 7) m.

Now,

AB = 17 m.

∠APB = 90° (angle in a pamicircle is 90°) In a right triangle APB, we have

AB^{2} = AP^{2} + BP^{2}

[By Pythagoaes theorem]

⇒ 17^{2} = (x + 7)^{2} + x^{2}

⇒ 289 = x^{2} + 49 + 14x + x^{2}

⇒ 289 = 2x^{2} + 14x + 49

⇒ 2x^{2} + 14x + 49 – 289 = 0

⇒ x^{2} + 7x – 120 = 0

⇒ x^{2} + (15 – 8) x – 120 = 0

⇒ x^{2} + 15x – 8x – 120 = 0

⇒ x(x + 15) – 8(x + 15) = 0

⇒ (x – 8) (x + 15) = 0

⇒ x – 8 = 0 or x + 15 = 0

⇒ x = 8 or x = – 15 (Reject)

Hence, the pole has to be erected on the boundary of the park at a distance of 8 m from gate Band 8 + 7 i.e., 15 m from gate A.

Fill in the Blanks

Question 1.

A quardratic equation ax^{2} + bx + c = 0 has two………..real roots, if b^{2} – 4ac > 0.

Solution :

Distinct

Question 2.

A quadratic equation ax^{2} + bx + c = 0 has no ………… roots, if b^{2} – 4ac < 0.

Solution :

Real

Question 3.

If we can factorise ax^{2} + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation ax^{2} + bx + c = 0 can be found by equating each ………. to zero.

Solution :

Factor

Question 4.

A quadratic equation can also be solved by the method of completing the …………..

Solution :

Square

Question 5.

ar^{2} + bx + c = 0, a ≠ 0 is called the standard form in a quadratic ………..

Solution :

Equation

Question 6.

A quadratic equation has atmost ……….. roots.

Solution :

Two.

Multiple Choice Questions

Question 1.

The root of the quadratic equation x^{2} – 0.04 = 0 are:

(a) ± 0.2

(b) ± 0.02

(c) 0.4

(d) 2

Solution :

(a) ± 0.2

x^{2} – 0.04 = 0

⇒ x^{2} – (0.2)^{2} = 0

⇒ (x + 0.2) (x – 0.2) = 0

⇒ x + 0.2 = 0

and x – 0.2 = 0

⇒ x = – 0.2

and x = 0.2

⇒ x = ± 0.2

Question 2.

If x^{2} + 2kx + 4 = 0 has a root x = 2, then value of k is:

(a) – 1

(b) – 2

(c) 2

(d) – 4.

Solution :

(b) – 2

x^{2} + 2kx + 4 = 0

one root x = 2 then

(2)^{2} + 2k(2) + 4 = 0

4k + 8 = 0

⇒ k = – 2

So correct choice is (b).

Question 3.

(x^{2} + 1)^{2} – x^{2} = 0 has

[NCERT Exemplar Problems]

(a) four real roots

(b) two real roots

(c) no real roots

(d) one real roots.

Solution :

(c) no real roots

(x^{2} + 1)^{2} – x^{2} = 0 get x^{2} = y, then

(y + 1)^{2} – y = 0

y^{2} + 1 + 2y – y = 0

y^{2} + y + 1 = 0

∵ b^{2} < 4ac

So, no real root.

Hence correct choice is (c).

Question 4.

Which of the following equations has two distinct real roots ?

[NCERT Exemplar Problems]

(a) 2x^{2} – 3\(\sqrt{2}\)x + \(\frac {9}{4}\)

(b) x^{2} + x – 5 = 0

(c) x^{2} + 3x + 2\(\sqrt{2}\) = 0

(d) 5x^{2} – 3x + 1 = 0.

Solution :

(b) x^{2} + x – 5 = 0

(a) 2x^{2} – 3\(\sqrt{2}\)x + \(\frac {9}{4}\) = 0

D = (-3\(\sqrt{2}\))^{2} – 4 × 2 × \(\frac {9}{4}\) = 0

(b) x^{2} + x – 5 = 0

D = (1)^{2} – (4) (1) (-5) = 21

D > 1

Hence correct choice is (b).

Question 5.

The roots of the equation 3x^{2} – 4x + 3 = 0 are :

(a) real and unequal

(b) real and equal

(c) imaginary

(d) none of these.

Solution :

(c) imaginary

3x^{2} – 4x + 3 = 0

a = 3, b = – 4, c = 3

b^{2} = (-4)^{2} = 16

4ac = 4 × 3 × 3 = 36

∵ b^{2} < 4ac

So roots are imaginary Hence correct choice (c).