# AP 10th Class Maths 4th Chapter Quadratic Equations Exercise 4.2 Solutions

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) x2 – 3x – 10 = 0
ii) 2x2 + x – 6 = 0
iii) $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
iv) 2x2 – x + $$\frac{1}{8}$$ =0
v) 100x2 – 20x + 1 = 0
Solution:
i) x2 – 3x – 10 = 0
x2 – 3x – 10 = 0
x2 – 5x + 2x- 10 = 0
x(x – 5) + 2(x – 5) = 0
(x + 2) (x – 5) = 0
The roots of x2 – 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 x2 – 3x – 10 = 0 are -2 and 5.

ii) 2x2 + x – 6 = 0
2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
2x (x + 2) – 3 (x + 2) = 0
(2x – 3) (x + 2) = 0
The roots of 2x2 + 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 2x2 + x – 6 = 0 are $$\frac{3}{2}$$ and -2.

iii) $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
$$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
$$\sqrt{2}$$x2 + 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 $$\sqrt{2}$$x2 + 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 $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0 are $$\frac{-5}{\sqrt{2}}$$ and –$$\sqrt{2}$$.

iv) 2x2 – x + $$\frac{1}{8}$$ = 0

The roots of 2x2 – 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 2x2 – x + $$\frac{1}{8}$$ = 0 are $$\frac{1}{4}$$, $$\frac{1}{4}$$.

v) 100x2 – 20x + 1 = 0
100x2 – 20x + 1 = 0
100x2 – 10x – 10x + 1 = 0
10x(10x – 1) – 1 (10x – 1) = 0
(10x – 1) (10x – 1) = 0
The roots of 100x2 – 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
100x2 – 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 – x2 – 200 + 5x = 124
⇒ -x2 + 45x – 200 = 124
⇒ x2 – 45x + 200 + 124 = 0
⇒ x2 – 45x + 324 = 0
x2 – 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 – x2 = 750
∴ x (55 – x) = 750 ⇒ 55x – x2 = 750
⇒ x2 – 55x + 750 = 0
⇒ x2 – 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 – x2 = 182
⇒ x2 – 27x + 182 = 0
⇒ x2 – 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
⇒ x2 + (x + 1)2 = 365
⇒ x2 + x2 + 2x + 1 – 365 = 0
⇒ 2x2 + 2x – 364 = 0
⇒ x2 + x – 182 = 0
⇒ x2 + 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,
x2 + (x – 7)2 = 132
x2 + x2 – 14x + 49 = 169
2x2 – 14x + 49 – 169 = 0
2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0
x2 – 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
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 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/-