NCERT Class 10 Maths – Exercise 4.2 Solutions

NCERT Class 10 Maths

Chapter 4 – Quadratic Equations | Exercise 4.2

(Rationalized Syllabus 2025-26)

Find the roots of the following quadratic equations by factorization:

(i) $x^2 – 3x – 10 = 0$

Split the middle term $-3x$ into two numbers whose sum is $-3$ and product is $-10$. Numbers are $-5$ and $+2$.

$$x^2 – 5x + 2x – 10 = 0$$

$$x(x – 5) + 2(x – 5) = 0$$

$$(x – 5)(x + 2) = 0$$

Either $x – 5 = 0 \Rightarrow x = 5$ or $x + 2 = 0 \Rightarrow x = -2$.

✔️ Roots: 5, -2

(ii) $2x^2 + x – 6 = 0$

Split the middle term $+x$ into two numbers whose sum is $1$ and product is $2 \times (-6) = -12$. Numbers are $+4$ and $-3$.

$$2x^2 + 4x – 3x – 6 = 0$$

$$2x(x + 2) – 3(x + 2) = 0$$

$$(x + 2)(2x – 3) = 0$$

Either $x = -2$ or $x = 3/2$.

✔️ Roots: -2, 3/2

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

Split middle term $7x$. Product $= \sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10$. Sum $= 7$. Numbers are 5 and 2.

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

Note: We can write $2x$ as $(\sqrt{2} \cdot \sqrt{2})x$.

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

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

Either $\sqrt{2}x = -5 \Rightarrow x = -5/\sqrt{2}$ or $x = -\sqrt{2}$.

✔️ Roots: $-\frac{5}{\sqrt{2}}, -\sqrt{2}$

(iv) $2x^2 – x + \frac{1}{8} = 0$

Multiply equation by 8 to remove fraction:

$$16x^2 – 8x + 1 = 0$$

This is a perfect square $(4x – 1)^2 = 0$.

$$(4x – 1)(4x – 1) = 0$$

✔️ Roots: 1/4, 1/4

(v) $100x^2 – 20x + 1 = 0$

Split $-20x$ into $-10x$ and $-10x$.

$$100x^2 – 10x – 10x + 1 = 0$$

$$10x(10x – 1) – 1(10x – 1) = 0$$

$$(10x – 1)(10x – 1) = 0$$

✔️ Roots: 1/10, 1/10

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’s marbles $= x$. Then Jivanti’s marbles $= 45 – x$.

After losing 5 marbles:

John $= x – 5$

Jivanti $= (45 – x) – 5 = 40 – x$

Product is 124:

$$(x – 5)(40 – x) = 124$$

$$40x – x^2 – 200 + 5x = 124$$

$$-x^2 + 45x – 324 = 0 \Rightarrow x^2 – 45x + 324 = 0$$

Split 45 into 36 and 9 (since $36 \times 9 = 324$).

$$x^2 – 36x – 9x + 324 = 0$$

$$(x – 36)(x – 9) = 0$$

✔️ Marbles: 36 and 9

(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. Find the number of toys produced on that day.

Let number of toys $= x$.

Cost of each toy $= 55 – x$.

Total Cost = $x(55 – x) = 750$.

$$55x – x^2 = 750 \Rightarrow x^2 – 55x + 750 = 0$$

Factors of 750 summing to 55 are 30 and 25.

$$x^2 – 30x – 25x + 750 = 0$$

$$(x – 30)(x – 25) = 0$$

✔️ Toys produced: 30 or 25

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

Let first number be $x$. Second number is $27 – x$.

$$x(27 – x) = 182$$

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

We need factors of 182 that sum to 27. Numbers are 13 and 14.

$$x^2 – 13x – 14x + 182 = 0$$

$$(x – 13)(x – 14) = 0$$

✔️ Numbers are 13 and 14

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

Let first integer $= x$, next integer $= x + 1$.

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

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

$$2x^2 + 2x – 364 = 0$$

Divide by 2:

$$x^2 + x – 182 = 0$$

Factors of 182 with difference 1 are 14 and 13.

$$x^2 + 14x – 13x – 182 = 0$$

$$(x + 14)(x – 13) = 0$$

Since integer is positive, $x \neq -14$, so $x = 13$.

✔️ Integers are 13 and 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.

Let base $= x$ cm. Then altitude $= x – 7$ cm.

By Pythagoras theorem:

$$x^2 + (x – 7)^2 = 13^2$$

$$x^2 + x^2 – 14x + 49 = 169$$

$$2x^2 – 14x – 120 = 0$$

Divide by 2:

$$x^2 – 7x – 60 = 0$$

Factors of -60 summing to -7 are -12 and +5.

$$(x – 12)(x + 5) = 0$$

Since length cannot be negative, $x = 12$.

Base = 12 cm, Altitude = $12 – 7 = 5$ cm.

✔️ Base = 12 cm, Altitude = 5 cm

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.

Let number of articles $= x$.

Cost of each article $= 2x + 3$.

Total cost $= x(2x + 3) = 90$.

$$2x^2 + 3x – 90 = 0$$

Find factors of $2 \times -90 = -180$ that sum to 3. Numbers are 15 and -12.

$$2x^2 + 15x – 12x – 90 = 0$$

$$x(2x + 15) – 6(2x + 15) = 0$$

$$(2x + 15)(x – 6) = 0$$

Since quantity cannot be negative, $x = 6$.

Cost per article = $2(6) + 3 = 15$.

✔️ Number of articles = 6, Cost per article = ₹15