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

Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x3 – 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) We have:
x2 – 3x – 10 = 0
or x2 – 5x + 2x – 10 = 0
or x(x – 5) + 2(x – 5) = 0
or (x – 5)(x + 2) = 0
i.e; x – 5 = 0 or x + 2 = 0
i.e; x = 5 or x = – 2
Thus, x = 5 and x = – 2 are two solutions of the equation x2 – 3x – 10 = 0. That is, roots of the equation are 5 and – 2.

(ii) We have:
2x2 + x – 6 = 0
or 2x2 + 4x – 3x – 6 = 0
or 2x(x + 2) – 3(x + 2) = 0
or (x + 2)(2x – 3) = 0
i.e; x + 2 = 0 or 2x – 3 = 0
i.e; x = – 2 or x = $$\frac{3}{2}$$
Thus, – 2 and $$\frac{3}{2}$$ are two roots of the equation 2x2 + x – 6 = 0.

(iii) We have:
$$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0
or $$\sqrt{2}$$x(x + $$\sqrt{2}$$) + 5(x + $$\sqrt{2}$$) = 0
or (x + $$\sqrt{2}$$)($$\sqrt{2}$$x + 5) = 0
i.e; x + $$\sqrt{2}$$ = 0 or $$\sqrt{2}$$x + 5 = 0
i.e; x = – $$\sqrt{2}$$
or x = $$\frac{-5}{\sqrt{2}}=\frac{-5 \sqrt{2}}{2}$$
Thus, – $$\sqrt{2}$$ and $$\frac{-5 \sqrt{2}}{2}$$ are two roots of the equation $$\sqrt{2}$$x2 + 7x + 5$$\sqrt{2}$$ = 0.

(iv) We have:
2x2 – x + $$\frac{1}{8}$$ = 0
or 16x2 – 8x + 1 = 0
or 16x2 – 4x – 4x + 1 = 0
or 4x(4x – 1) – 1(4x – 1) = 0
or 4x – 1 = 0 or 4x – 1 = 0
i.e; x = $$\frac{1}{4}$$ or x = $$\frac{1}{4}$$
Thus, $$\frac{1}{4}$$ and $$\frac{1}{4}$$ are two equal roots of the equation 2x2 – x + $$\frac{1}{8}$$ = 0

(v) We have:
100x2 – 20x + 1 = 0
or 100x2 – 10x – 10x + 1 = 0
or 10x(10x – 1) – 1(10x – 1) = 0
or (10x – 1)(10x – 1) = 0
i.e; 10x – 1 = 0 or 10x – 1 = 0
i.e; x = $$\frac{1}{10}$$ or x = $$\frac{1}{10}$$
Thus, $$\frac{1}{10}$$ and $$\frac{1}{10}$$ are two equal roots of the equation 100x2 – 20x = 1 = 0.

Question 2.
Solve the problems given in Example 1, i.e; to solve
(i) x2 – 45x + 324 = 0
(ii) x2 – 55x + 750 = 0
using factorisation method.
Solution:
(i) We have: x2 – 45x + 324 = 0
or x2 – 9x – 36x + 324 = 0
or x(x – 9) – 36(x – 9) = 0
or (x – 9)(x – 36) = 0
i.e; x – 9 = 0 or x – 36 = 0
i.e; x = 9 or x = 36
Thus, 9 and 36 are two roots of the equation x2 – 45x + 324 = 0.

(ii) We have:
x2 – 55x + 750 = 0
or x2 – 30x – 25x + 750 = 0
or x(x – 30) – 25(x – 30) = 0
or (x – 30)(x – 25) = 0
i.e; x – 30 = 0 or x – 25 = 0
i.e; x = 30 or x = 25
Thus, 30 and 25 are two roots of the equation.
x2 – 55x + 750 = 0.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the required numbers be x and 27 – x. Then,
x(27 – x) = 182
or 27x – x2 = 182
or x2 – 27x + 182 = 0
or x2 – 13x – 14x + 182 = 0
or x(x – 13) – 14(x – 13) = 0
or (x – 13)(x – 14) = 0
i.e; x – 13 = 0 or x – 14 = 0
i.e; x = 13 or x = 14
Hence, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let the two consecutive positive integers be x and x + 1.
Then, x2 + (x + 1)2 = 365
or x2 + x2 + 2x + 1 = 365
or 2x2 + 2x – 364 = 0
or x2 + x – 182 = 0
or x(x + 14) – 13(x + 14) = 0
or (x + 14)(x – 13) = 0
i.e; x + 14 = 0 or x – 13 = 0
i.e; x = – 14 or x = 13
Since x, being a positive integer, cannot be negative. Therefore, x = 13.
Hence, the two consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right traiangle 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, its altitude is (x – 7) cm.
It is given that the hypotenuse = 13 cm
So, $$\sqrt{x^{2}+(x-7)^{2}}$$ = 13
or x2 + (x2 – 14x + 49) = 169
or 2x2 – 14x – 120 = 0
or x2 – 7x – 60 = 0
or x2 – 12x + 5x – 60 = 0
or x(x – 12) + 5(x – 12) = 0
or (x – 12)(x + 5) = 0
i.e., x – 12 = 0 or x + 5 = 0
i.e., x = 12 or x = – 5
So, x = 12
[Since side of a triangle can never be negative]
∴ Length of the base = 12 cm
and length of the altitude
= (12 – 7) cm
= 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 Rs 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced in a day = x
Then, cost of production of each article = Rs (2x + 3)
It is given that the total cost of production = Rs 90
∴ x × (2x + 3) = 90
or 2x2 + 3x – 90 = 0
or 2x2 – 12x + 15x – 90 = 0
or 2x(x – 6) + 15(x – 6) = 0
or (x – 6)(2x + 15) = 0
i.e; x – 6 = 0 or 2x + 15 = 0
i.e; x = 6 or x = $$\frac{-15}{2}$$
So, x = 6 [Since the number of articles produced cannot be negative]
Cost of each article = Rs (2 × 6 + 3) = Rs 15
Hence, the number of articles produced are 6 and the cost of each article is Rs 15.