# Class10 NCRT Quadratic Equations Exercise – 4.2 pdf || UP Board

1. Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0

x2 – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x  + 2) = 0

x – 5 = 0, x + 2 = 0

x = 5, x = -2

Hence, the roots of the quadratic equation are 5 and -2.

(ii) 2x2 + x – 6 = 0

2x2 – 4x + 3x – 6 = 0

2x(x – 4) + 3(x – 6) = 0

(x – 4) (2x + 3) = 0

x – 4 = 0, 2x + 3 = 0

x = 4, x = -3/2

Hence, the roots of the quadratic equation are 2 and -3/2.

(iii) √2x2 + 7x + 5√2 = 0

√2x2 + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2(√2x + 5) = 0

(√2x + 5)(x + √2) = 0

√2x + 5 = 0, x + √2 = 0

x = -5/√2, x = -√2

Hence, the roots of the quadratic equation are -5/√2 and -√2.

(iv) 2x2 – x + 1/8 = 0

16x2 – 8x + 1 = 0

16x2 – 4x – 4x + 1 = 0

4x(4x – 1) – 1(4x – 1) = 0

(4x – 1) (4x – 1) =0

4x – 1 = 0, 4x – 1 = 0

x = ¼, x = ¼

Hence, the roots of the quadratic equation are ¼ and ¼.

(v) 100x2 – 20x + 1 = 0

100x2 – 10x – 10x + 1 = 0

10x(10x – 1) -1(10x – 1) = 0

(10x – 1)( 10x – 1) = 0

10x – 1 = 0, 10x – 1 = 0

x = 1/10, x = 1/10

Hence, the roots of the quadratic equation are 1/10 and 1/10.

2. Solve the problems given in Example 1.

[The problems given in the example x2 – 45x + 324 = 0 and x2 – 55x + 750 = 0.]

x2 – 45x + 324 = 0

x2 – 36x – 9x + 324 = 0

x(x – 36) – 9(x – 36) = 0

(x – 36)(x – 9) = 0

x – 36 = 0, x – 9 = 0

x = 36, x = 9

So, John and Jivanti initially had 36 and 9 marbles.

x2 – 55x + 750 = 0

x2 – 30x – 25x + 750 = 0

x(x – 30) – 25(x – 30) = 0

(x – 30)(x – 25) = 0

x = 30, x = 25

Hence, the number of toys manufactured on that day is 30 or 25.

3. Find two numbers whose sum is 27 and product is 182.

Let the first number = x

Therefore, second number = 27 - x

According to Question,

Product = x(27 – x) = 182

27x – x2 = 182

x2 – 27x – 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) – 14(x – 13) = 0

(x – 13)(x – 14) = 0

x - 13 = 0, x – 14 = 027 - x x = 251/10

x = 13, x = 14

Hence, the numbers are 13 and 14.

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

Let the first number = x

Therefore, second number = x + 1

According to Question,

product = x2 + (x + 1)2 = 365

x2 + x2 + 2x + 1 = 365

2x2 + 2x – 364 =0

x2 + x – 182 = 0

x2 – 13x + 14x - 182 = 0

x(x – 13) + 14(x – 13) = 0

x – 13 = 0, x + 14 = 0

x = 13, x = -14 या 14 [positive integer]

Hence, two consecutive positive integers are 13 and 14.

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.

Let base = x cm

Therefore, height = x – 7 cm

Given, hypotenuse = 13 cm

By the Pythagoras theorem,

x2 + (x – 7)2 = 132

x2 + x2 – 14x + 49 = 169

2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0

x2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

x – 12 = 0, x + 5 = 0

x = 0, x = -5

But x ≠ -5 [Since x is a side of the triangle, the side cannot be negative.]

Therefore, x = 12 and other side (height) = x – 7 = 127 = 5

Hence, the other two sides are 12 cm and 5 cm.

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.

Let the number of pottery articles = x

Hence, cost of one article = 2x + 3

According to Question,

total manufacturing cost = x(2x + 3) = 90

2x2 + 3x = 90

2x2 + 3x – 90 = 0

2x2 + 15x – 12x – 90 = 0

x(2x + 15) – 6(2x + 15) = 0

(2x + 15)(x – 6) = 0

2x + 15 = 0, x – 6 =0

x = -15/2, x = 6

But x ≠ -15/2 [since x is the number of articles.]

so x = 6

So, cost of each article = 2x + 3 = 2×6 + 3 = 15

Hence, the number of articles manufactured is 6 and the cost of each article is ₹15.