Well-designed AP Board Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Quadratic Equations Class 10 Exercise 4.2 Solutions – 10th Class Maths 4.2 Exercise Solutions

Question 1.

Find the roots of the following quadratic equations by factorisation :

i) x^{2} – 3x – 10 = 0

ii) 2x^{2} + x – 6 = 0

iii) \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

iv) 2x^{2} – x + \(\frac{1}{8}\) =0

v) 100x^{2} – 20x + 1 = 0

Solution:

i) x^{2} – 3x – 10 = 0

Given quadratic equation is

x^{2} – 3x – 10 = 0

x^{2} – 5x + 2x- 10 = 0

x(x – 5) + 2(x – 5) = 0

(x + 2) (x – 5) = 0

The roots of x^{2} – 3x – 10 = 0 are the values of x for which (x + 2) (x – 5) = 0

Therefore, x + 2 = 0 or x – 5 = 0

i.e., x = -2 (or) x = 5

Therefore, the roots of x^{2} – 3x – 10 = 0 are -2 and 5.

ii) 2x^{2} + x – 6 = 0

Given quadratic equation is

2x^{2} + x – 6 = 0

2x^{2} + 4x – 3x – 6 = 0

2x (x + 2) – 3 (x + 2) = 0

(2x – 3) (x + 2) = 0

The roots of 2x^{2} + x – 6 = 0 are the values of x for which (2x – 3) (x + 2) = 0

Therefore, 2x – 3 = 0 (or) x + 2 = 0 3

i.e. x = \(\frac{3}{2}\) (or) x = -2

Therefore, the roots of 2x^{2} + x – 6 = 0 are \(\frac{3}{2}\) and -2.

iii) \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

Given quadratic equation is

\(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

\(\sqrt{2}\)x^{2} + 5x + 2x + 5\(\sqrt{2}\) = 0

x(\(\sqrt{2}\)x + 5) + \(\sqrt{2}\)(\(\sqrt{2}\)x + 5) = 0

(\(\sqrt{2}\)x + 5)(x + \(\sqrt{2}\)) = 0

The roots of \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0 are the values of x for which (\(\sqrt{2}\)x + 5)(x + \(\sqrt{2}\)) = 0

Therefore, \(\sqrt{2}\)x + 5 = 0 (or) x + \(\sqrt{2}\) = 0

i.e. x = \(\frac{-5}{\sqrt{2}}\) (or) x = – \(\sqrt{2}\)

Therefore, the roots of \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0 are \(\frac{-5}{\sqrt{2}}\) and –\(\sqrt{2}\).

iv) 2x^{2} – x + \(\frac{1}{8}\) = 0

Given quadratic equation is

The roots of 2x^{2} – x + \(\frac{1}{8}\) = 0 are the values of x for which (2x – \(\frac{1}{2}\))(x – \(\frac{1}{4}\)) = 0

Therefore, (2x – \(\frac{1}{2}\)) = 0 (or) (x – \(\frac{1}{4}\)) = 0

i.e., x = \(\frac{1}{4}\) (or) x = \(\frac{1}{4}\)

Therefore, the roots of 2x^{2} – x + \(\frac{1}{8}\) = 0 are \(\frac{1}{4}\), \(\frac{1}{4}\).

v) 100x^{2} – 20x + 1 = 0

Given quadratic equation is

100x^{2} – 20x + 1 = 0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) – 1 (10x – 1) = 0

(10x – 1) (10x – 1) = 0

The roots of 100x^{2} – 20x + 1 = 0 are the values of x for which

(10x – 1) (10x – 1) = 0

Therefore, 10x – 1 = 0 (or) 10x – 1 = 0

i.e., x = \(\frac{1}{10}\) (or) x = \(\frac{1}{10}\)

Therefore, the roots of

100x^{2} – 20x + 1 = 0 are \(\frac{1}{10}\), \(\frac{1}{10}\).

Question 2.

Solve the problems given in example 1.

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 mar-bles they had to start with.

Solution:

Let the number of marbles John had = x

Number of marbles Jivanti had = 45 – x

When he lost 5 marbles,

The number of marbles left with John = x – 5

The number of marbles left with Jivanti = 45 – x – 5 = 40 – x

Therefore, their product = 124

(x – 5) (40 – x) = 124

⇒ 40x – x^{2} – 200 + 5x = 124

⇒ -x^{2} + 45x – 200 = 124

⇒ x^{2} – 45x + 200 + 124 = 0

⇒ x^{2} – 45x + 324 = 0

It is a quadratic equation.

x^{2} – 36x – 9x + 324 = 0

x (x – 36) – 9 (x – 36) = 0

(x – 9) (x – 36) = 0

Therefore, x = 9 (or) x = 36

Marbles John had x = 9 (or) 36

Marbles Jivanti had = 45 – x (or) 45 – x

= 45 – 9

= 45 – 36

= 36 = 9

ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) 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:

Let the number of toys produced on that day be x.

The cost of production of each toy on that day = (55 – x)/-

So, the total cost of production that day

∴ x (55 – x) = 750 ⇒ 55x – x^{2} = 750

∴ x (55 – x) = 750 ⇒ 55x – x^{2} = 750

⇒ x^{2} – 55x + 750 = 0

⇒ x^{2} – 30x – 25x + 750 = 0

⇒ x (x – 30) – 25 (x – 30) = 0

⇒ (x – 25) (x – 30) = 0

Therefore, x = 25 (or) 30

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 the two numbers are x and y.

Sum of two numbers = 27

x + y = 27 ⇒ y = 27 – x ……….. (1)

Product of two numbers = 182

x∙y = 182 ……… (2)

Put (1) in (2) ⇒ x (27 – x) = 182

⇒ 27x – x^{2} = 182

⇒ x^{2} – 27x + 182 = 0

⇒ x^{2} – 14x – 13x + 182 = 0

⇒ x(x – 14) – 13(x – 14) = 0

⇒ (x – 13) (x – 14) = 0

Therefore, x = 13 (or) 14 Numbers are 13 and 14.

Question 4.

Find two consecutive positive integers, sum of whose squares is 365.

Solution:

Let two consecutive positive integers are x and (x + 1).

Sum of their squares = 365

⇒ x^{2} + (x + 1)^{2} = 365

⇒ x^{2} + x^{2} + 2x + 1 – 365 = 0

⇒ 2x^{2} + 2x – 364 = 0

⇒ x^{2} + x – 182 = 0

⇒ x^{2} + 14x – 13x – 182 = 0

⇒ x (x + 14) – 13 (x + 14) = 0

⇒ (x – 13)(x + 14) = 0

Therefore, x = 13 (or) -14.

Numbers are 13, 13 + 1, i.e. 13 and 14.

Question 5.

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Let the base of the right triangle be x cm.

Then altitude = base – 7 cm = (x – 7) cm.

By Pythagorous theorem,

x^{2} + (x – 7)^{2} = 13^{2}

x^{2} + x^{2} – 14x + 49 = 169

2x^{2} – 14x + 49 – 169 = 0

2x^{2} – 14x – 120 = 0

x^{2} – 7x – 60 = 0

x^{2} – 12x + 5x – 60 = 0

x (x – 12) + 5 (x – 12)

(x + 5) (x – 12) = 0

Therefore, x = -5 (or) 12

∴ Base of the triangle =12 cm

Altitude = (x – 7) cm

∴ Altitude = 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 num-ber 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 = x

Cost of each article = ₹ (2x + 3)/-

Total cost of production = x (2x + 3) = 90

Therefore, x (2x + 3) = 90

⇒ 2x^{2} + 3x – 90 = 0

⇒ 2x^{2} + 15x – 12x – 90 = 0

⇒ x(2x + 15) – 6(2x + 15) = 0

⇒ (x – 6) (2x + 15) = 0

Therefore, x = +6 (or) \(\frac{-15}{2}\)

Number of articles produced = 6.

Cost of each article = 2x + 3

= 2(6) + 3 = ₹ 15/-