**NCERT Solution For Class 10, Maths, Quadratic Equations, Exercise 4.2** has some practical examples of quadratic equations. Chapter 4, Ex. 4.2 mainly has six questions to study.

Table of Contents

Toggle## Class 10, Maths, Chapter 4, Exercise 4.2 Solutions

**Q.1. Find the roots of the following quadratic equations by factorization:**

**(i) x ^{2} – 3x – 10 = 0 **

**(ii) 2x ^{2} + x – 6 = 0**

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

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

**(v) 100 x ^{2} – 20x + 1 = 0**

**Ans:**

**Q.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.**

** **

**(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**

**Ans:**

**(i) Let the number of marbles John had be x.**

Then the number of marbles Jivanti had = 45 –x

When john lost 5 marbles, then number of marble left =x – 5

Therefore, the number of marbles left with Jivanti, when she lost 5 marbles =45 – x – 5 = 40–x

Therefore, their product =(x–5) (40–x)

= 40x–x^{2} – 200 + 5x

= –x^{2} + 45x – 200

Given the product of the number of marbles =124

So, –x^{2} + 45x – 200 = 124

i.e., –x^{2} + 45x – 324 = 0

i.e., x^{2} –45x + 324 = 0

Therefore, the number of marbles John had, satisfies the quadratic equation

⇒ x^{2}–45x + 324 = 0

We have, x^{2} – 45x + 324 = 0

⇒ x^{2}–9x – 36x + 324 = 0

⇒ x(x-9) – 36(x-9) = 0

⇒ (x-9) (x-36) = 0

⇒ x-9 = 0 or x – 36 =0

⇒ x = 9 or x = 36

Thus, x=9 and x=36 are two roots of the equation x^{2}–45x + 324 = 0 or we can say that, John and Jivanti had 9 and 36 or 36 and 9 respectively.

**(ii) Let the number of toys produced on that day be x.**

Therefore, the cost of production (in ₹) of each toy that day = 55 – x

So, the total cost of production (in ₹) that day =x(55–x)

Therefore, x(55–x =750

i.e., 55x–x^{2}=750

i.e., –x^{2} + 55x–750 =0

i.e., x^{2}– 55x + 750 =0

Therefore, the number of toys produced that day satisfies the quadratic equation

⇒ x^{2}–55x + 750 = 0

We have, x^{2}–55x + 750 = 0

⇒ x^{2}–30x – 25x + 750 = 0

⇒ x(x- 30) – 25(x-30)=0

⇒ (x-30) (x-25) = 0

⇒ x – 30 = 0 or x–25 =0

⇒ x = 30 or x =25

Thus, x=30 and x = 25 are two roots of the equation x^{2}–55x + 750 = 0 or

We can say that

If the number of toys produced on that day be 30.

Then, the cost of production (in ₹) of each toy that day would be =55 –x=55-30=25 ₹

Or

If the number of toys produced on that day be 25.

Then, the cost of production (in ₹) of each toy that day would be =55 –x=55-25=35 ₹

**Q.3. Find two numbers whose sum is 27 and product is 182.**

**Ans:**

**Q.4. Find two consecutive positive integers, sum of whose squares is 365.**

**Ans:**

**Q.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.**

**Ans:**

**Q.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. **

**Ans: **Let the number of articles produced in a day = x

Then, cost of production of each article = ₹ (2x + 3) (according to question)

Total cost of production of each article on that day = ₹ 90

Therefore,

i.e., x(2x + 3) = 90

i.e., 2x^{2} + 3x = 90

i.e., 2x^{2} + 3x – 90 = 0

Therefore, the number of pottery articles produced that day satisfies the quadratic equation

⇒ 2x^{2 }+ 3x – 90 = 0

We have, 2x^{2} + 3x – 90 = 0

⇒ 2x^{2} – 12x + 15x – 90 = 0

⇒ 2x(x – 6) + 15(x – 6) = 0

⇒ (x – 6) (2x + 15) = 0

⇒ x – 6 = 0 or 2x + 15 = 0

⇒ x = 6 or x = $-\frac{15}{2}$

Since, the number of articles produced cannot be a negative number. Now, we can say that

If the number of articles produced on that day = x = 6

Then, the cost of production (in ₹) of each article that day would be =(2x + 3) = (2×6 + 3) = ₹ 15