## NCERT solutions for class 10 Maths

**Chapter 4**

## Quadratic Equations

**Exercise 4.2**

**Ex 4.2 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) √2 x^{2} + 7x + 5√2 = 0
(iv) 2x^{2} – x +1/8 = 0
(v) 100x^{2} – 20x + 1 = 0**

**Solution:**

(i) *x*^{2} – 3*x* – 10 =0

*⇒ x*^{2} – 5*x* + 2*x* – 10

*⇒ x*(*x *– 5) + 2(*x* – 5)

⇒ (*x* – 5)(*x* + 2)

⇒ *x* – 5 = 0 or *x* + 2 = 0

Either *x* = 5 or *x* = -2

(ii) 2*x*^{2} + *x* – 6 = 0

⇒ 2*x*^{2} + 4*x* – 3*x* – 6

⇒ 2*x*(*x* + 2) – 3(*x* + 2)

⇒ (*x* + 2)(2*x* – 3)

⇒ *x* + 2 = 0 or 2*x* – 3 = 0

Either *x* = -2 or *x* = 3/2

(iii) √2 *x*^{2} + 7*x* + 5√2=0

⇒ √2 *x*^{2 }+ 5*x* + 2*x* + 5√2

⇒ *x* (√2*x* + 5) + √2(√2*x* + 5)= (√2*x* + 5)(*x *+ √2)

⇒ √2*x* + 5 = 0 or *x* + √2 = 0

Either *x* = -5/√2 or *x* = -√2

(iv) 2*x*^{2} – *x* +1/8 = 0

⇒ 16*x*^{2 }– 8*x* + 1 = 0

⇒16*x*^{2 }– 4*x* -4*x* + 1 = 0

⇒(4*x* – 1) = 0 or (4*x* – 1) = 0

Either *x* = 1/4 or *x* = 1/4

(v)100x^{2} – 20x + 1=0

⇒ 100x^{2} – 10x – 10x + 1

⇒ 10x(10x – 1) -1(10x – 1)

⇒(10x – 1)^{2}

⇒(10x – 1) = 0 or (10x – 1) = 0

Either x = 1/10 or x = 1/10

**Ex 4.2 Question 2.**

**Solve the problems given in Example 1.**

**Represent the following situations mathematically:**

**(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 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.**

**Solutions:**

(i) Let the number of marbles John had = *x*.

∴The, number of marble Jivanti had = 45 – *x*

After losing 5 marbles each,

Number of marbles John have = *x* – 5

Number of marble Jivanti have = 45 – *x* – 5 = 40 –* x*

Given that the product of their marbles is 124.

∴ (*x *– 5)(40 – *x*) = 124

⇒ *x*^{2} – 45*x* + 324 = 0

⇒ *x*^{2} – 36*x* – 9*x* + 324 = 0

⇒ *x*(*x* – 36) -9(*x* – 36) = 0

⇒ (*x* – 36)(*x* – 9) = 0

*x* – 36 = 0 or *x* – 9 = 0

⇒ *x* = 36 or *x* = 9

Therefore, John and Jivanti have 36 and 9 marbles.

(ii) Let the number of toys produced in a day = *x*.

Then, the cost of production of each toy = ₹(55 – *x*)

∴Total cost of production on that day = x(55 – *x*)

∴ *x*(55 – *x*) = 750

⇒ *x*^{2} – 55*x* + 750 = 0

⇒ *x*^{2} – 25*x* – 30*x* + 750 = 0

⇒* x*(*x* – 25) -30(*x* – 25) = 0

⇒ (*x* – 25)(*x* – 30) = 0

Either *x* -25 = 0 or *x* – 30 = 0

⇒ *x* = 25 or *x* = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

**Ex 4.2 Question 3.**

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

**Solution:**

Let the first number = x and the second number = 27 – x.

∴ The product of two numbers

x(27 – x) = 182

⇒ x^{2} – 27x – 182 = 0

⇒ x^{2} – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Hence, the numbers required are 13 and 14.

**Ex 4.2 Question 4.**

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

**Solution:**

Let the first number = x and the second number = *x* + 1.

*x*^{2} + (*x* + 1)^{2} = 365

⇒ *x*^{2 }+ *x*^{2 }+ 1 + 2*x* = 365

⇒ 2*x*^{2} + 2x – 364 = 0

⇒ *x*^{2 }+ *x *– 182 = 0

⇒ *x*^{2 }+ 14*x* – 13*x* – 182 = 0

⇒ *x*(*x* + 14) -13(*x* + 14) = 0

⇒ (*x* + 14)(*x* – 13) = 0

Either, *x* + 14 = 0 or *x* – 13 = 0,

⇒ *x* = – 14 or *x* = 13 [Integers are positive, so *x* can only be 13]

∴ *x* + 1 = 13 + 1 = 14

Hence, two consecutive positive integers will be 13 and 14.

**Ex 4.2 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 = x cm.

∴ The height = (x – 7) cm

Hypotenuse = 13 cm

Using Pythagoras theorem,

Base^{2} + Altitude^{2} = Hypotenuse^{2}

∴ x^{2 }+ (x – 7)^{2} = 132

⇒ x^{2 }+ x^{2 }+ 49 – 14x = 169

⇒ 2x^{2 }– 14x – 120 = 0

⇒ x^{2 }– 7x – 60 = 0

⇒ x^{2 }– 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

But x ≠ -5 [Since sides are positive, x can only be 12]

Therefore, the base of the triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

**Ex 4.2 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 = *x*.

Therefore, the cost of one article = x(2*x* + 3) = 90

⇒ 2*x*^{2 }+ 3*x* – 90 = 0

⇒ 2*x*^{2 }+ 15*x* -12*x* – 90 = 0

⇒ *x*(2*x* + 15) -6(2*x* + 15) = 0

⇒ (2*x* + 15)(*x* – 6) = 0

Either 2*x* + 15 = 0 or *x* – 6 = 0

⇒ *x* = -15/2 or *x* = 6

But x ≠ -15/2 [As the number of article produce can only be positive integer. ∴ x can only be 6.]

∴ x = 6 and the cost of each article = 2 × 6 + 3 = ₹15.

Hence, the number of article = 6

the cost of each article = ₹15