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

Q1 (ii) Find the roots of the following quadratic equations by factorisation: $2x^2 + x - 6 = 0$

Answer:

Given the quadratic equation: $2x^2 + x - 6 = 0$

Factorisation gives, $2x^2 +4x-3x - 6 = 0$

$\Rightarrow 2x(x+2) -3(x+2) =0$

$\Rightarrow (x+2)(2x-3) = 0$

$\Rightarrow x= -2\ or\ \frac{3}{2}$

Hence, the roots of the given quadratic equation are

$-2\ and\ \frac{3}{2}$

Q1 (iii) Find the roots of the following quadratic equations by factorisation:$\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Answer:

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

$\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0$

$\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0$

$\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2$

Hence, the roots of the given quadratic equation are

$\frac{-5}{\sqrt 2 }\ and\ -\sqrt 2$

Q1 (iv) Find the roots of the following quadratic equations by factorisation:$2x^2 -x + \frac{1}{8} = 0$

Answer:

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

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

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

$\Rightarrow 4x(4x-1)-1(4x-1) = 0$

$\Rightarrow (4x-1)(4x-1) = 0$

$\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}$

Hence, the roots of the given quadratic equation are

$\frac{1}{4}\ and\ \frac{1}{4}$

Q1 (v) Find the roots of the following quadratic equations by factorisation: $100x^2 -20x +1 = 0$

Answer:

Given the quadratic equation: $100x^2 -20x +1 = 0$

Factorization gives, $100x^2 -10x-10x +1 = 0$

$\Rightarrow 10x(10x-1)-10(10x-1) = 0$

$\Rightarrow (10x-1)(10x-1) = 0$

$\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}$

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

Q2 Solve the problems given in Example 1. (i) $x^2-45x+324 = 0$ (ii) $x^2-55x+750 = 0$

Answer:

From Example 1, we get:

Equations:

(i) $x^2-45x+324 = 0$

Solving by the factorisation method:

Given the quadratic equation: $x^2-45x+324 = 0$

Factorization gives, $x^2-36x-9x+324 = 0$

$\Rightarrow x(x-36) - 9(x-36) = 0$

$\Rightarrow (x-9)(x-36) = 0$

$\Rightarrow x=9\ or\ 36$

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$.

Therefore, John and Jivanti have 36 and 9 marbles, respectively, in the beginning.

(ii) $x^2-55x+750 = 0$

Solving by the factorisation method:

Given the quadratic equation: $x^2-55x+750 = 0$

Factorization gives, $x^2-30x-25x+750 = 0$

$\Rightarrow x(x-30) -25(x-30) = 0$

$\Rightarrow (x-25)(x-30) = 0$

$\Rightarrow x=25\ or\ 30$

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$.

Therefore, the number of toys on that day was $30\ or\ 25.$

Q3 Find two numbers whose sum is 27 and the product is 182.

Answer:

Let two numbers be x and y .

Then, their sum will be equal to 27, and the product equals 182.

$x+y = 27$ ..........(1)

$xy =182$ ...........(2)

From equation (2) we have:

$y = \frac{182}{x}$

Then, putting the value of y in equation (1), we get

$x+\frac{182}{x} = 27$

Solving this equation:

$\Rightarrow x^2-27x+182 = 0$

$\Rightarrow x^2-13x-14x+182 = 0$

$\Rightarrow x(x-13)-14(x-13) = 0$

$\Rightarrow (x-14)(x-13) = 0$

$\Rightarrow x = 13\ or\ 14$

Hence, the two required numbers are $13\ and \ 14$.

Q4 Find two consecutive positive integers, the sum of whose squares is 365.

Answer:

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

. $x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

$\Rightarrow x^2 - 13x+14x+182 = 0$

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$.

Q5 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 length of the base of the triangle be $b\ cm$.

Then, the altitude length will be: $b-7\ cm$.

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem, we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But the length of the base cannot be negative.

Hence, the base length will be $12\ cm$.

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Q6 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.

Answer:

Let the number of articles produced in a day $= x$

The cost of production of each article will be $=2x+3$

Given that the total production on that day was $Rs.90$.

Hence, we have the equation;

$x(2x+3) = 90$

$2x^2+3x-90 = 0$

$\Rightarrow 2x^2+15x-12x-90 = 0$

$\Rightarrow x(2x+15) - 6(2x+15) = 0$

$\Rightarrow (2x+15)(x-6) = 0$

$\Rightarrow x =-\frac{15}{2}\ or\ 6$

But, x cannot be negative as it is the number of articles.

Therefore, $x=6$ and the cost of each article $= 2x+3 = 2(6)+3 = 15$

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