HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Haryana State Board HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation :
(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2x² – x + \(\frac{1}{8}\) = 0
(v) 100x² – 20x + 1 = 0.
Solution :
(i) The given equation is :
x² – 3x – 10 = 0
x² – (5 – 2)x – 10 = 0
[∵ 1 × – 10 = – 10
5 × – 2 = – 10
5 – 2 = – 3]
⇒ x² – 3x + 2x – 10 = 0
⇒ x(x – 5) + 2(x – 5) = 0
⇒ (x – 5)(x + 2) = 0
⇒ x – 5 = 0 or x + 2 = 0
⇒ x = 5 or x = – 2
Hence, x = 5 and x = – 2 are the roots of the given quadratic equation.

(ii) The given equation is :
2x² + x – 6 = 0
⇒ 2x² + (4 – 3)x – 6 = 0
[∵ – 6 × 2 = – 12
– 3 × 4 = – 12
4 – 3 = 1]
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x + 2) = 0
⇒ (x + 2)(2x – 3) = 0
⇒ x + 2 = 0 or 2x – 3 = 0
⇒ x = – 2 or x = – \(\frac{3}{2}\)
Hence, x = – 2 and x = – \(\frac{3}{2}\) are the roots of the given quadratic equation.

(iii) The given equation is :
√2x² + 7x + 5√2 = 0
√2x² + (2 + 5)x + 5√2 = 0
[∵ 5√2 × √2 = 10
2 × 5 = 10
2 + 5 = 7]
√2x² + 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2)(√2x + 5) = 0
x + √2 = 0 or √2x + 5 = 0
x = – √2or x = – \(\frac{5}{\sqrt{2}}\)
Hence, x = – √2 and x = – \(\frac{5}{\sqrt{2}}\) are the roots of the given quadratic equation.

(iv) The given equation is:
2x² – x + \(\frac{1}{8}\) = 0
16x² – 8x + 1 = 0
[∵ 1 × 16 = 16
4 × 4 = 16
4 + 4 = 8]
16x² – 8x + 1 = 0
16x² – (4 + 4)x + 1 = 0
16x² – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x – 1) = 0
(4x – 1)(4x – 1) = 0
4x – 1 = 0 or 4x – 1 = 0
x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)
Hence, x = \(\frac{1}{4}\) and x = \(\frac{1}{4}\) are the roots of given quadratic equation.

(v) The given equation is :
100x² – 20x + 1 = 0
⇒ 100x² – (10 + 10)x +1 = 0
[∵ 100 × 1 = 100
10 × 10 = 100.
10 + 10 = 20]
⇒ 100x² – 1ox – 10x + 1 = 0
⇒ 10x (10x – 1) – 1 (10x – 1) = 0
⇒ (10x – 1) (10x – 1) = 0
⇒ 10x – 1 = 0 or 10x – 1 = 0
⇒ x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)
Hence, x = \(\frac{1}{10}\) and x = \(\frac{1}{10}\) are the roots of the given quadratic equation.

Question 2.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in ₹) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John has be x.
Then, the number of marbles Jivanti has = 45 – x.
The number of marbles left with John, when he lost 5 marbles = x – 5
and the number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
According to question,
(x – 5) (40 – x) = 124
⇒ 40x – x² – 200 + 5x = 124
⇒ – x² + 45x – 200 – 124 = 0
⇒ – x² + 45x – 324 = 0
⇒ x² – 45x + 324 = 0
⇒ x² – (36 + 9)x + 324 = 0
⇒ x² – 36x – 9x + 324 = 0
[∵ 324 × 1 = 324
36 × 9 = 324
36 + 9 = 45]
⇒ x(x – 36) – 9(x – 36) = 0
⇒ (x – 36)(x – 9) = 0
⇒ x – 36 = 0 or x – 9 = 0
x = 36 or x = 9
Hence, either John had 36 marbles and Jivanti had 9 marbles or vice-versa.

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production of each toy = ₹ (55 – x)
So, the total cost of production on that day = ₹ x(55 – x)
According to question,
x(55 – x) = 750
⇒ 55x – x² = 750
⇒ – x² + 55x – 750 = 0
⇒ x² – 55x + 750 = 0
⇒ x² – (25 + 30)x + 750 = 0
⇒ x² – 25x – 30x + 750 = 0
[∵ 750 × 1 = 750
25 × 30 = 750
25 + 30 = 55]
⇒ x(x – 25) – 30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
⇒ x – 25 = 0 or x – 30 = 0
⇒ x = 25 or x = 30
Hence, the number of toys produced on that day = 25 or 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.‘
Solution :
Let 1st number be x, then 2nd number = 27 – x
According to question,
x(27 – x) = 182
⇒ 27x – x² = 182
⇒ – x² + 27x – 182 = 0
⇒ x² – 27x + 182 = 0
⇒ x² – (14 + 13)x + 182= 0
[∵ 182 × 1 = – 182
14 × – 13 = – 182
14 – 13 = 1]
⇒ x² – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 14)(x – 13) = 0
⇒ x – 14 = 0 or x – 13
⇒ x = 14 or x = 13
Hence, the numbers are 14, 13 or 13, 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the consecutive positive integers be x and x + 1.
According to question,
x² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ x² + x – 182 = 0
[∵ 182 × 1 = 182
14 × 13 = 182
14 + 13 = 27]
⇒ x² + x (14 – 13) – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
⇒ x + 14 = 0 or x – 13 = 0
⇒ x = – 14 or x = 13
Since, x is a positive integer, it cannot be negative.
Therefore, x = 13.
Hence, the consecutive positive integers are 13, 14.

Question 5.
The Edtitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of a right trismgle be x cm.
then altitude of the right triangle = (x – 7) cm
and hypotenuse = 13 cm (given)
In a right ∆ABC,

By Pythagoras theorem, we get
AC² = BC² + AB²
⇒ 13² = x² + (x – 7)²
⇒ 169 = x² + x² – 14x + 49
2x² – 14x + 49 – 169
⇒ 2x² – 14x – 120 = 0
[∵ – 60 × 1 = – 60
12 × – 5 = – 60
12 – 5 = 7]
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
⇒ x – 12 = 0 or x + 5 = 0
⇒ x = 12 or x = – 5
Since, x is length of base, it cannot be negative.
Therefore, x = 12.
Hence, the base of triangle = 12 cm
and altitude of triangle = 12 – 7 = 5 cm.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in Rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90. Find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day be x,
then cost of one Eirticle = ₹ (2x + 3)
According to question,
x(2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
[∵ 2 × (- 90) = – 180
15 × (- 12) = – 180
15 – 12 = 3]
⇒ 2x² + (15 – 12)x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒ x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
⇒ 2x + 15 = 0
or x – 6=0
⇒ x = – \(\frac{15}{2}\) or x = 6
Since, x is number of articles. It cannot be negative.
Therefore, x = 6
Hence,the number of articles = 6
Cost of each article = ₹ 2 × 6 + 3 = ₹ 15