NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2. This exercise teaches students to solve quadratic equations by the factorisation method. Factorisation involves splitting the middle term and expressing the quadratic as a product of two linear factors, then solving each factor independently.
Factorisation Method: To solve $ax^2 + bx + c = 0$, find two numbers whose sum is $b$ and product is $ac$. Split the middle term, group the terms, factorise, and solve each factor = 0.
1. Find the roots of the following quadratic equations by factorisation: #
(i) $x^2 - 3x - 10 = 0$
We need two numbers with sum $-3$ and product $-10$.
The numbers are $-5$ and $2$:
$(-5) + 2 = -3$
$(-5) \times 2 = -10$
Splitting the middle term:
$$x^2 - 3x - 10 = 0$$$$\Rightarrow x^2 - 5x + 2x - 10 = 0$$
$$\Rightarrow x(x - 5) + 2(x - 5) = 0$$
$$\Rightarrow (x - 5)(x + 2) = 0$$
Setting each factor to zero:
$$x - 5 = 0$$$$\Rightarrow x = 5$$
$$x + 2 = 0$$
$$\Rightarrow x = -2$$
∴ The roots are $x = \mathbf{5}$ and $x = \mathbf{-2}$.
(ii) $2x^2 + x - 6 = 0$
Here $a = 2,\ b = 1,\ c = -6$.
We need two numbers with sum $1$ and product $ac = 2 \times (-6) = -12$.
The numbers are $4$ and $-3$:
$4 + (-3) = 1$
$4 \times (-3) = -12$
Splitting the middle term:
$$2x^2 + x - 6 = 0$$$$\Rightarrow 2x^2 + 4x - 3x - 6 = 0$$
$$\Rightarrow 2x(x + 2) - 3(x + 2) = 0$$
$$\Rightarrow (x + 2)(2x - 3) = 0$$
Setting each factor to zero:
$$x + 2 = 0$$$$\Rightarrow x = -2$$
$$2x - 3 = 0$$
$$\Rightarrow 2x = 3$$
$$\Rightarrow x = \dfrac{3}{2}$$
∴ The roots are $x = \mathbf{-2}$ and $x = \dfrac{3}{2}$.
(iii) $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$
We need two numbers with sum $7$ and product $\sqrt{2} \times 5\sqrt{2} = 10$.
The numbers are $5$ and $2$:
$5 + 2 = 7$
$5 \times 2 = 10$
Splitting the middle term:
$$\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$$$$\Rightarrow \sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = 0$$
$$\Rightarrow x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0$$
$$\Rightarrow (\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0$$
Setting each factor to zero:
$$\sqrt{2}\,x + 5 = 0$$$$\Rightarrow \sqrt{2}\,x = -5$$
$$\Rightarrow x = -\dfrac{5}{\sqrt{2}}$$
$$\Rightarrow x = -\dfrac{5\sqrt{2}}{2}$$
$$x + \sqrt{2} = 0$$
$$\Rightarrow x = -\sqrt{2}$$
∴ The roots are $x = -\dfrac{5\sqrt{2}}{2}$ and $x = -\sqrt{2}$.
(iv) $2x^2 - x + \dfrac{1}{8} = 0$
Multiplying both sides by $8$:
$$8\left(2x^2 - x + \dfrac{1}{8}\right) = 8 \times 0$$$$\Rightarrow 16x^2 - 8x + 1 = 0$$
We need two numbers with sum $-8$ and product $16 \times 1 = 16$.
The numbers are $-4$ and $-4$:
$(-4) + (-4) = -8$
$(-4) \times (-4) = 16$
Splitting the middle term:
$$16x^2 - 8x + 1 = 0$$$$\Rightarrow 16x^2 - 4x - 4x + 1 = 0$$
$$\Rightarrow 4x(4x - 1) - 1(4x - 1) = 0$$
$$\Rightarrow (4x - 1)^2 = 0$$
Setting the factor to zero:
$$4x - 1 = 0$$$$\Rightarrow x = \dfrac{1}{4} \quad \text{(repeated root)}$$
∴ The roots are $x = \dfrac{1}{4}$ and $x = \dfrac{1}{4}$.
(v) $100x^2 - 20x + 1 = 0$
We need two numbers with sum $-20$ and product $100 \times 1 = 100$.
The numbers are $-10$ and $-10$:
$(-10) + (-10) = -20$
$(-10) \times (-10) = 100$
Splitting the middle term:
$$100x^2 - 20x + 1 = 0$$$$\Rightarrow 100x^2 - 10x - 10x + 1 = 0$$
$$\Rightarrow 10x(10x - 1) - 1(10x - 1) = 0$$
$$\Rightarrow (10x - 1)^2 = 0$$
Setting the factor to zero:
$$10x - 1 = 0$$$$\Rightarrow x = \dfrac{1}{10} \quad \text{(repeated root)}$$
∴ The roots are $x = \dfrac{1}{10}$ and $x = \dfrac{1}{10}$.
2. Solve the problems given in Example 1. #
(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.
Let John have $x$ marbles.
Together they have 45 marbles, so Jivanti has $(45 - x)$ marbles.
After losing 5 marbles each:
- John has $(x - 5)$ marbles.
- Jivanti has $(40 - x)$ marbles.
The product of their marbles is now 124, so:
$$(x-5)(40-x) = 124$$$$\Rightarrow 40x - x^2 - 200 + 5x = 124$$
$$\Rightarrow -x^2 + 45x - 200 = 124$$
$$\Rightarrow -x^2 + 45x - 324 = 0$$
$$\Rightarrow x^2 - 45x + 324 = 0$$
We need two numbers with sum $45$ and product $324$. The numbers are $36$ and $9$.
Splitting the middle term:
$$x^2 - 45x + 324 = 0$$$$\Rightarrow x^2 - 36x - 9x + 324 = 0$$
$$\Rightarrow x(x-36) - 9(x-36) = 0$$
$$\Rightarrow (x-36)(x-9) = 0$$
Setting each factor to zero:
$$x - 36 = 0$$$$\Rightarrow x = 36$$
$$x - 9 = 0$$
$$\Rightarrow x = 9$$
If $x = 36$, John has 36 marbles and Jivanti has $45 - 36 = 9$ marbles.
If $x = 9$, John has 9 marbles and Jivanti has $45 - 9 = 36$ marbles.
∴ John started with 36 marbles (and Jivanti 9), or John started with 9 marbles (and Jivanti 36).
(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.
Let the number of toys produced be $x$.
The cost of each toy is 55 minus the number of toys, so the cost per toy is $(55 - x)$ rupees.
The total cost is ₹750, so:
$$x(55 - x) = 750$$$$\Rightarrow 55x - x^2 = 750$$
$$\Rightarrow x^2 - 55x + 750 = 0$$
We need two numbers with sum $55$ and product $750$. The numbers are $30$ and $25$.
Splitting the middle term:
$$x^2 - 55x + 750 = 0$$$$\Rightarrow x^2 - 30x - 25x + 750 = 0$$
$$\Rightarrow x(x-30) - 25(x-30) = 0$$
$$\Rightarrow (x-30)(x-25) = 0$$
Setting each factor to zero:
$$x - 30 = 0$$$$\Rightarrow x = 30$$
$$x - 25 = 0$$
$$\Rightarrow x = 25$$
∴ The number of toys produced is 25 or 30.
3. Find two numbers whose sum is 27 and product is 182. #
Answer
Let one number be $x$.
The sum is 27, so the other number is $(27 - x)$.
The product is 182, so:
$$x(27 - x) = 182$$$$\Rightarrow 27x - x^2 = 182$$
$$\Rightarrow x^2 - 27x + 182 = 0$$
We need two numbers with sum $27$ and product $182$. The numbers are $14$ and $13$.
Splitting the middle term:
$$x^2 - 27x + 182 = 0$$$$\Rightarrow x^2 - 14x - 13x + 182 = 0$$
$$\Rightarrow x(x-14) - 13(x-14) = 0$$
$$\Rightarrow (x-14)(x-13) = 0$$
Setting each factor to zero:
$$x - 14 = 0$$$$\Rightarrow x = 14$$
$$x - 13 = 0$$
$$\Rightarrow x = 13$$
∴ The two numbers are 14 and 13.
4. Find two consecutive positive integers, sum of whose squares is 365. #
Answer
Let the two consecutive positive integers be $x$ and $x + 1$.
The sum of their squares is 365, so:
$$x^2 + (x+1)^2 = 365$$$$\Rightarrow x^2 + x^2 + 2x + 1 = 365$$
$$\Rightarrow 2x^2 + 2x - 364 = 0$$
$$\Rightarrow x^2 + x - 182 = 0$$
We need two numbers with sum $1$ and product $-182$. The numbers are $14$ and $-13$.
Splitting the middle term:
$$x^2 + x - 182 = 0$$$$\Rightarrow x^2 + 14x - 13x - 182 = 0$$
$$\Rightarrow x(x+14) - 13(x+14) = 0$$
$$\Rightarrow (x+14)(x-13) = 0$$
Setting each factor to zero:
$$x + 14 = 0$$$$\Rightarrow x = -14$$
$$x - 13 = 0$$
$$\Rightarrow x = 13$$
$x = -14$ is rejected, because the integers must be positive. So $x = 13$ and the next integer is $13 + 1 = 14$.
∴ The 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. #
Answer
Let the base be $x$ cm.
The altitude is 7 cm less than the base, so the altitude is $(x - 7)$ cm.
By Pythagoras’ theorem, base² + altitude² = hypotenuse², so:
$$x^2 + (x-7)^2 = 13^2$$$$\Rightarrow x^2 + x^2 - 14x + 49 = 169$$
$$\Rightarrow 2x^2 - 14x - 120 = 0$$
$$\Rightarrow x^2 - 7x - 60 = 0$$
We need two numbers with sum $-7$ and product $-60$. The numbers are $-12$ and $5$.
Splitting the middle term:
$$x^2 - 7x - 60 = 0$$$$\Rightarrow x^2 - 12x + 5x - 60 = 0$$
$$\Rightarrow x(x-12) + 5(x-12) = 0$$
$$\Rightarrow (x-12)(x+5) = 0$$
Setting each factor to zero:
$$x - 12 = 0$$$$\Rightarrow x = 12$$
$$x + 5 = 0$$
$$\Rightarrow x = -5$$
$x = -5$ is rejected, because a length must be positive. So the base is 12 cm.
Substituting $x = 12$ in the altitude $x - 7$:
$$x - 7 = 12 - 7 = 5$$∴ 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. #
Answer
Let the number of articles produced be $x$.
The cost of each article is 3 more than twice the number of articles, so the cost per article is $(2x + 3)$ rupees.
The total cost is ₹90, so:
$$x(2x+3) = 90$$$$\Rightarrow 2x^2 + 3x = 90$$
$$\Rightarrow 2x^2 + 3x - 90 = 0$$
We need two numbers with sum $3$ and product $2 \times (-90) = -180$. The numbers are $15$ and $-12$.
Splitting the middle term:
$$2x^2 + 3x - 90 = 0$$$$\Rightarrow 2x^2 + 15x - 12x - 90 = 0$$
$$\Rightarrow x(2x+15) - 6(2x+15) = 0$$
$$\Rightarrow (2x+15)(x-6) = 0$$
Setting each factor to zero:
$$2x + 15 = 0$$$$\Rightarrow 2x = -15$$
$$\Rightarrow x = -\dfrac{15}{2}$$
$$x - 6 = 0$$
$$\Rightarrow x = 6$$
$x = -\dfrac{15}{2}$ is rejected, because the number of articles must be positive. So $x = 6$.
Substituting $x = 6$ in the cost per article $2x + 3$:
$$2x + 3 = 2(6) + 3 = 15$$∴ The number of articles is 6 and the cost of each article is ₹15.