**Class ten NCERT solutions Quadratic Equations**

**Chapter 4: Quadratic Equations**

**TEXTBOOK’S EXERCISE 4.2 **

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

(i) x^{2} – 3x – 10 = 0 (ii) 2x^{2} + x – 6 = 0

(v) 100x^{2} – 20x +
1 = 0

**Sol.
**(i) x^{2} – 3x – 10 = 0 ⇒ x^{2}
– 5x + 2x – 10 = 0

⇒ x(x – 5) + 2(x – 5) = 0

⇒ (x – 5) (x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

i.e., x = 5 or x = –2

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

(ii) 2x^{2} + x –
6 = 0

⇒ 2x^{2 }+
4x – 3x – 6 = 0

⇒ 2x(x + 2) – 3(x + 2) = 0

⇒ (x + 2) (2x – 3) = 0

⇒ x + 2 = 0 or 2x – 3 = 0,

⇒ x = –2 or x = 3/2

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

are the roots of the given quadratic equation.

Hence, 1/4 and 1/4 are the roots of the given quadratic equation.

(v) 100x^{2} – 20x +
1 = 0

⇒ 100x^{2}
– 10x – 10x + 1 = 0

⇒ 10x(10x – 1) – 1(10x – 1) = 0

⇒ (10x – 1) (10x – 1) = 0

⇒ x = 1/ 10 or x = 1/ 10

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

**2. **Solve the problems:

(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 Rs 750. We would like to find out the number of toys produced on that day.

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

A number of marbles Jivanti had = 45 – x.

The number of marbles left with John, when he lost 5 marbles = x – 5

The number of marbles left with Jivanti, when she lost 5 marbles

= 45 – x – 5 = 40 – x

Then, required product = (x – 5) (40 – x)

= 40x – x^{2} – 200 + 5x

= –x2 + 45x – 200

As per condition: –x^{2} + 45x – 200 = 124

⇒ –x^{2} +
45x – 324 = 0

⇒ x^{2} –
45x + 324 = 0

⇒ x^{2} – 9x – 36x + 324 = 0

⇒ x(x – 9) – 36 (x – 9) = 0

⇒ (x – 36) (x – 9) = 0 ⇒ x = 36 or x = 9

Hence, they had started with 36 and 9 marbles.

**Class ten NCERT solutions Quadratic Equations**

(ii) Let the number of toys produced on a particular day be x.

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

The total cost of production on that day = Rs x(55 – x)

Therefore, x(55 – x) = 750

⇒ 55x – x^{2} = 750

⇒ x^{2} –
55x + 750 = 0

⇒ x^{2} –
25x – 30x + 750 = 0 ⇒ x(x – 25) – 30(x –
25) = 0

⇒ (x – 25)(x – 30) = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced on a particular day can be 25 or 30.

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

**Sol.
**Let first number = x.

And second number = 27 – x

Their product = x(27 – x)

As per condition: x(27 – x) = 182

⇒ 27x – x^{2} = 182

⇒ –x^{2} + 27x –
182 = 0

⇒ x^{2}
– 27x + 182 = 0

⇒ x^{2}
– 13x – 14x + 182 = 0

⇒ x(x – 13) – 14(x – 13) = 0

⇒ (x – 13) (x – 14) = 0

⇒ x = 13 or x = 14

Hence, the required numbers are 13 and 14.

**Class ten NCERT solutions Quadratic Equations**

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

**Sol. **Let first integer = x

Second consecutive integer = x + 1

Sum of their squares = x^{2}
+ (x + 1)^{2}

= x^{2} + x^{2} + 1 + 2x = 2x^{2} + 2x +
1

As per condition: 2x^{2}
+ 2x + 1 = 365

⇒ 2x^{2}
+ 2x = 364 ⇒ x^{2}
+ x = 182

⇒ x^{2}
+ x – 182 = 0

⇒ x^{2}
+ 14x – 13x – 182 = 0

⇒ x(x + 14) – 13 (x + 14) = 0

⇒ (x + 14) (x – 13) = 0

⇒ x = 13 or x = –14

Since, x is a positive integer, therefore x _ −14.

If x = 13 then, 2nd number = 13 + 1 = 14

Hence, the required two positive consecutive 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.

**Sol.
**Let base of the right triangle = x cm

Height of the right triangle = (x – 7) cm

By Pythagoras theorem,

(Hyp.)^{2} =
(Base)^{2} + (Height)^{2}

⇒ (13)^{2} = x^{2}
+ (x – 7)^{2}

⇒ 169 = x^{2}
+ x^{2} + 49 – 14x

⇒ 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

⇒ x = 12 or x = –5

∵ Side cannot be negative, so, x = 12 cm

Now, x – 7 = 12 – 7 = 5 cm.

Hence, the length of the other two sides is 5 cm and 12 cm.

**Class ten NCERT solutions Quadratic Equations**

**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.

**Sol. **Let the number of
articles produced = x

Cost of production of an article = Rs (2x + 3)

Total cost of production = Number of articles produced on that day × Cost of production of each article

As per condition : x(2x + 3) = 90

⇒ 2x^{2} +
3x – 90 = 0

⇒ 2x^{2} +
15x – 12x – 90 = 0

⇒ x (2x + 15) –6(2x + 15) = 0

⇒ (2x + 15) (x – 6) = 0 ⇒ x = 6 or x = − 15/2

As, number of articles cannot be negative. Therefore, x ≠ − 15/ 2

Hence, number of articles produced on that day = 6

So, cost of each article = 2x + 3

= Rs (2 × 6 + 3) = Rs 15