NCERT Class 10 Maths Chapter 4 – Quadratic Equation: Exercise 2

In this exercise, questions related to the first method which is called splitting the middle term. Other than the basic solution of equations in this exercise some word problems based on real-world scenarios are also included, in which we need to formulate the quadratic equation first and then solve it to find the required solution.

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

(i) x²– 3x – 10 = 0 

Solution:

Here, LHS = x²– 3x – 10 

= x² – 5x + 2x – 10

= x(x – 5) + 2(x – 5)

= (x – 5)(x + 2)

The roots of this equation, x² – 3x – 10 = 0 are the values of x for which 

(x – 5)(x + 2) = 0

Hence, x – 5 = 0 or x + 2 = 0

⇒ x = 5 or x = -2

(ii) 2x² + x – 6 = 0

Solution:

Here, LHS = 2x² + x – 6

= 2x² + 4x – 3x – 6

= 2x(x + 2) – 3(x + 2)

= (2x– 3)(x + 2)

The roots of this equation, 2x² + x – 6 = 0 are the values of x for which

(2x– 3)(x + 2) = 0

Hence, 2x– 3 = 0 or x + 2 = 0

⇒ x = 3/2 or x = –2

(iii) √2x² + 7x + 5√2 = 0

Solution:

Here, LHS = √2x² + 7x + 5√2

= √2x² + 5x + 2x + 5√2

= x(√2x + 5) + √2(√2x + 5)

= (√2x + 5) (x +√2)

The roots of this equation, √2x² + 7x + 5√2 = 0 are the values of x for which

(√2x + 5) (x +√2) = 0

Hence, √2x + 5 = 0 or x +√2 = 0

⇒ x = –5/√2 or x = –√2

(iv) 2x² – x + 1/8 = 0

Solution:

Here, LHS = 2x² – x + 1/8

= 1/8(16x² – 8x + 1)

= 1/8(16x² – 4x -4x + 1)

= 1/8(4x(4x-1) -1 (4x-1))

= 1/8 (4x-1) (4x-1) 

The roots of this equation, 2x² – x + 1/8 = 0 are the values of x for which

1/8 (4x-1) (4x-1)  = 0

(4x-1)² = 0

Hence, 4x-1 = 0 or 4x-1 = 0

⇒ x = 1/4 or x = 1/4

(v) 100x² – 20x + 1 = 0

Solution:

Here, LHS = 100x² – 20x + 1

= 100x² – 10x – 10x + 1

= 10x(10x – 1) – 1(10x – 1)

= (10x – 1) (10x – 1)

The roots of this equation, 100x² – 20x + 1 = 0 are the values of x for which

(10x – 1) (10x – 1) = 0

(10x – 1)² = 0

Hence, 10x – 1 = 0 or 10x – 1 = 0

⇒ x = 1/10 or x = 1/10

Question 2. Solve the following situations mathematically:

(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.

Solution:

Let’s say, 

The number of marbles John have = x.

So, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Here, According to the given condition

(x – 5)(40 – x) = 124

x² – 45x + 324 = 0

x² – 36x – 9x + 324 = 0

x(x – 36) -9(x – 36) = 0

(x – 36)(x – 9) = 0

Hence, x – 36 = 0 or x – 9 = 0

x = 36 or x = 9

Therefore,

If, John’s marbles = 36, then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9, then, Jivanti’s marbles = 45 – 9 = 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.

Solution:

 Let us say, 

Number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

So, x(55 – x) = 750

x² – 55x + 750 = 0

x² – 25x – 30x + 750 = 0

x(x – 25) -30(x – 25) = 0

(x – 25)(x – 30) = 0

Hence, x – 25 = 0 or x – 30 = 0

x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

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

Solution:

Let’s say, 

First number be x and the second number is 27 – x.

Therefore, the product of two numbers will be:

x(27 – x) = 182

x² – 27x – 182 = 0

x² – 13x – 14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 13)(x -14) = 0

Hence, x – 13 = 0 or x – 14= 0

x = 13 or x = 14

Hence, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

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

Solution:

Let’s say, 

Two consecutive positive integers be x and x + 1.

Here, According to the given condition,

x² + (x + 1)2 = 365

x² + x² + 1 + 2x = 365

2x² + 2x – 364 = 0

x² + x – 182 = 0

x² + 14x – 13x – 182 = 0

x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Hence, x – 13 = 0 or x + 14= 0

x = 13 or x = – 14

As, here it is said positive integers, so x can be 13, only.

So,

x = 13

and, x + 1 = 13 + 1 = 14

Hence, two consecutive positive integers will be 13 and 14.

Question 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.

Solution:

Let’s say, 

Base of the right triangle be x cm.

So, the altitude of right triangle = (x – 7) cm

Base² + Altitude² = Hypotenuse²                       (Pythagoras theorem)

x² + (x – 7)² = 132

x² + x² + 49 – 14x = 169    (using identity (a-b)² = a² – 2ab + b²)

2x² – 14x – 120 = 0

x² – 7x – 60 = 0               (Dividing by 2)

x² – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Hence, x – 12 = 0 or x + 5= 0

x = 12 or x = – 5

As, here side will be a  positive integers, so x can be 12, only.

Therefore, the base of the given triangle is 12 cm and,

the altitude of this triangle will be (12 – 7) cm = 5 cm.

Question 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.

Solution:

Let’s say,

Number of articles produced be x.

So, cost of production of each article = ₹ (2x + 3)

Here, According to the given condition

x(2x + 3) = 90

2x² + 3x – 90 = 0

2x² + 15x -12x – 90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Hence, 2x +15 = 0 or x – 6= 0

x = –15/2 or x = 6

As the number of articles produced can only be a positive integer, 

So, x = 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = ₹ 15.

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations is a resource created by the team at GFG to help students clear their doubts while solving problems from the NCERT textbook. It helps clear the frustration caused by being stuck on a problem for a while. The NCERT Solutions for Class 10 Maths covers all the questions in the exercise of the NCERT textbook for this chapter.

In this NCERT Class 10 Maths Chapter 4 Solutions, students are introduced to the world of quadratic equations, although quadratic equations are introduced to students in Class 9 under the chapter named “Polynomials” there was no separation of this from polynomials but in reality, there are many applications of quadratic equation, therefore it is studied here in class 10 maths as a stand-alone chapter. In this chapter, students learn about quadratic equations, various methods to solve quadratic equations, the nature of roots based on discriminants for quadratic equations, and many real-life based problems which can be modelled in the form of quadratic equations and then solved to get the required solution.