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 $2$: $(-5) + 2 = -3$ and $(-5) \times 2 = -10$.
$$x^2 - 5x + 2x - 10 = 0$$$$x(x - 5) + 2(x - 5) = 0$$$$(x - 5)(x + 2) = 0$$$$x - 5 = 0 \implies x = 5$$$$x + 2 = 0 \implies x = -2$$∴ 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 numbers with sum $= 1$ and product $= 2 \times (-6) = -12$.
The numbers are $4$ and $-3$: $4 + (-3) = 1$ and $4 \times (-3) = -12$.
$$2x^2 + 4x - 3x - 6 = 0$$$$2x(x + 2) - 3(x + 2) = 0$$$$(x + 2)(2x - 3) = 0$$$$x = -2 \quad \text{or} \quad x = \frac{3}{2}$$∴ The roots are $x = \mathbf{-2}$ and $x = \dfrac{3}{2}$.
(iii) $\sqrt{2}\,x^2 + 7x + 5\sqrt{2} = 0$
We need numbers with sum $= 7$ and product $= \sqrt{2} \times 5\sqrt{2} = 10$.
The numbers are $5$ and $2$: $5 + 2 = 7$ and $5 \times 2 = 10$.
$$\sqrt{2}\,x^2 + 5x + 2x + 5\sqrt{2} = 0$$$$x(\sqrt{2}\,x + 5) + \sqrt{2}(\sqrt{2}\,x + 5) = 0$$$$(\sqrt{2}\,x + 5)(x + \sqrt{2}) = 0$$$$x = -\frac{5}{\sqrt{2}} = -\frac{5\sqrt{2}}{2} \quad \text{or} \quad x = -\sqrt{2}$$∴ The roots are $x = -\dfrac{5\sqrt{2}}{2}$ and $x = -\sqrt{2}$.
(iv) $2x^2 - x + \dfrac{1}{8} = 0$
Multiply through by $8$:
$$16x^2 - 8x + 1 = 0$$We need numbers with sum $= -8$ and product $= 16 \times 1 = 16$.
The numbers are $-4$ and $-4$.
$$16x^2 - 4x - 4x + 1 = 0$$$$4x(4x - 1) - 1(4x - 1) = 0$$$$(4x - 1)^2 = 0$$$$x = \frac{1}{4} \text{ (repeated root)}$$∴ The roots are $x = \dfrac{1}{4}$ and $x = \dfrac{1}{4}$.
(v) $100x^2 - 20x + 1 = 0$
We need numbers with sum $= -20$ and product $= 100 \times 1 = 100$.
The numbers are $-10$ and $-10$.
$$100x^2 - 10x - 10x + 1 = 0$$$$10x(10x - 1) - 1(10x - 1) = 0$$$$(10x - 1)^2 = 0$$$$x = \frac{1}{10} \text{ (repeated root)}$$∴ 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. Jivanti has $(45 - x)$ marbles.
After losing 5 each:
- John: $(x - 5)$ marbles
- Jivanti: $(40 - x)$ marbles
Factorising: Find numbers with sum $45$ and product $324$: $36$ and $9$.
$$x^2 - 36x - 9x + 324 = 0$$$$x(x-36) - 9(x-36) = 0$$$$(x-36)(x-9) = 0$$$x = 36$ or $x = 9$.
If $x = 36$: John has 36, Jivanti has 9. If $x = 9$: John has 9, Jivanti has 36.
∴ John started with 36 (and Jivanti 9), or John started with 9 (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 number of toys produced = $x$.
Cost per toy = $₹(55 - x)$.
Total cost:
$$x(55 - x) = 750$$$$55x - x^2 = 750$$$$x^2 - 55x + 750 = 0$$Find numbers with sum $55$ and product $750$: $30$ and $25$.
$$x^2 - 30x - 25x + 750 = 0$$$$x(x-30) - 25(x-30) = 0$$$$(x-30)(x-25) = 0$$$x = 30$ or $x = 25$.
∴ Number of toys produced = 25 or 30.
3. Find two numbers whose sum is 27 and product is 182. #
Answer
Let one number = $x$. Other number = $27 - x$.
$$x(27 - x) = 182$$$$27x - x^2 = 182$$$$x^2 - 27x + 182 = 0$$Find numbers with sum $27$ and product $182$: $14$ and $13$.
$$(x-14)(x-13) = 0$$$x = 14$ or $x = 13$.
∴ The two numbers are 14 and 13.
4. Find two consecutive positive integers, sum of whose squares is 365. #
Answer
Let the consecutive positive integers be $x$ and $x+1$.
$$x^2 + (x+1)^2 = 365$$$$x^2 + x^2 + 2x + 1 = 365$$$$2x^2 + 2x - 364 = 0$$$$x^2 + x - 182 = 0$$Find numbers with sum $1$ and product $-182$: $14$ and $-13$.
$$(x+14)(x-13) = 0$$$x = 13$ (taking positive value, since $x = -14$ gives negative integers)
∴ 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 = $x$ cm. Then altitude = $(x - 7)$ cm.
By Pythagoras’ theorem:
$$x^2 + (x-7)^2 = 13^2$$$$x^2 + x^2 - 14x + 49 = 169$$$$2x^2 - 14x - 120 = 0$$$$x^2 - 7x - 60 = 0$$Find numbers with sum $-7$ and product $-60$: $-12$ and $5$.
$$(x-12)(x+5) = 0$$$x = 12$ (since base must be positive, $x = -5$ is rejected)
Base $= 12$ cm and Altitude $= 12 - 7 = 5$ cm.
∴ 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 number of articles produced = $x$.
Cost per article = $₹(2x + 3)$.
$$x(2x+3) = 90$$$$2x^2 + 3x - 90 = 0$$Find numbers with sum $3$ and product $2 \times (-90) = -180$: $15$ and $-12$.
$$2x^2 + 15x - 12x - 90 = 0$$$$x(2x+15) - 6(2x+15) = 0$$$$(2x+15)(x-6) = 0$$$x = 6$ (since number of articles must be positive, $x = -\dfrac{15}{2}$ is rejected)
Cost per article $= 2(6) + 3 = ₹15$.
∴ Number of articles = 6 and Cost per article = ₹15.