# NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2 – Quadratic Equations

NCERT Solutions for Quadratic Equations Chapter 4 Ex 4.2 Class 10 Maths has all the solutions to the questions provided in the NCERT Book of the latest edition.

Students are advised to practice all the questions to get good marks in the board examination.

Textbook | NCERT |

Class | 10 |

Subject | Mathematics |

Chapter | 4 |

Exercise | 4.2 |

Chapter Name | Quadratic Equations |

## Class 10 Maths Chapter 4 Exercise 4.2 Quadratic Equations NCERT Solution

Find the roots of the following quadratic equations by factorisation:

(i) $$x^{2}– 3x – 10 = 0$$

Answer

$$x^{2}-3x-10=0$$

⇒ $$x^{2}-5x+2x-10=0$$

⇒ $$x(x-5)+2(x-5)=0$$

⇒ $$(x-5)(x+2) = 0$$

$$\therefore x=5,-2$$

(ii) $$2x^{2}+ x – 6 = 0$$

Answer

$$2x^{2}+x-6=0$$

⇒ $$2x^{2}+4x-3x-6=0$$

⇒ $$2x(x+2)-3(x+2)=0$$

⇒ $$(2x-3)(x+2) = 0$$

$$\therefore x=\frac{3}{2},-2$$

(iii) $$\sqrt{2}x^{2}+7x+5\sqrt{2} = 0$$

Answer

$$\sqrt{2}x^{2}+7x+5\sqrt{2} = 0$$

⇒ $$\sqrt{2}x^{2}+2x+5x+5\sqrt{2} = 0$$

⇒ $$\sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2}) = 0$$

⇒ $$(x+\sqrt{2})(\sqrt{2}x+5) = 0$$

$$\therefore x=-\sqrt{2},\frac{-5}{\sqrt{2}}$$

(iv) $$2x^{2} -x+\frac{1}{8} = 0$$

Answer

Multiply by 8, we get

$$16x^{2}-8x+1=0$$

⇒ $$16x^{2}-4x-4x+1=0$$

⇒ $$4x(4x-1)-1(4x-1)=0$$

⇒ $$(4x-1)(4x-1) = 0$$

$$\therefore x=\frac{1}{4},\frac{1}{4}$$

(v) $$100x^{2}-20x+1 = 0$$

Answer

$$100x^{2}-20x+1=0$$

⇒ $$100x^{2}-10x-10x+1=0$$

⇒ $$10x(10x-1)-1(10x-1)=0$$

⇒ $$(10x-1)(10x-1) = 0$$

$$\therefore x=\frac{1}{10},\frac{1}{10}$$

Solve the problems given in Example 1

Answer

Do it yourself

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

Answer

let first no. be x

then second no. = $$27-x$$

A/Q,

$$(27-x)x=182$$

⇒ $$27x-x^{2}=182$$

⇒ $$x^{2}-27x+182=0$$

⇒ $$x^{2}-14x-13x+182=0$$

⇒ $$x(x-14)-13(x-14)=0$$

⇒ $$(x-14)(x-13)=0$$

$$x=14,13$$

$$\therefore$$ The required numbers are 13 & 14

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

Answer

let the smaller positive integer be x

then the larger positive integer = $$x+1$$

A/Q,

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

⇒ $$x^{2}+x^{2}+1+2x=365$$

⇒ $$2x^{2}+2x-364=0$$

⇒ $$x^{2}+x-182=0$$

⇒ $$x^{2}+14x-13x-182=0$$

⇒ $$x(x+14)-13(x+14)=0$$

⇒ $$(x+14)(x-13)=0$$

$$x=-14,13$$

$$\therefore$$ The smallest integer = 13

The larger integer = $$x+1$$ = 14

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer

Let the base of right angled triangle be x cm

then the altitude of right angled triangle = $$(x-7)$$ cm

Hypotenuse = 13 cm

$$\therefore x^{2}+(x-7)^{2}=(13)^{2}$$

⇒ $$2x^{2}+49-14x=169$$

⇒ $$2x^{2}-14x-120=0$$

⇒ $$x^{2}-7x-60=0$$

⇒ $$x^{2}-12x+5x-60=0$$

⇒ $$x(x-12)+5(x-12)=0$$

⇒ $$(x-12)(x+5)=0$$

$$\therefore x = 12, -5$$

$$\because$$ Side of a triangle can be negative no.

$$\therefore$$ Base = x = 12cm

Altitude = x – 7 = 5cm

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.

Answer

Let the no. of articles produced be x

then the cost of production of each article = $$₹(3+2x)$$

$$\therefore x\times(3+2x)=90$$

⇒ $$3x+2x^{2}-90=0$$

⇒ $$2x^{2}+15x-12x-90=0$$

⇒ $$x(2x+15)-6(2x+15)=0$$

⇒ $$(2x+15)(x-6)=0$$

$$\therefore x = \frac{-15}{2}, 6$$

$$\because$$ Number of articles produced can’t be negative

$$\therefore$$ No. of articles = x = 6

Cost of production = $$3+2x$$ = ₹15

Hope NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2, helps you in solving problems. If you have any doubts, drop a comment below and we will get back to you.