NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations - Ex 4.2

Refer to page 1 - 7 for Exercise 4.2 in the PDF

1. Find the roots of the following quadratic equations by factorisation:

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans: ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

$\Rightarrow 2\text{x}\left( \text{x+2} \right)-3\left( \text{x+2} \right)$

$\Rightarrow \left( \text{x+2} \right)\left( \text{2x-3} \right)$

Therefore, roots of this equation are –

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$

iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans: $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

Therefore, roots of this equation are –

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow {{\left( \text{4x-1} \right)}^{2}}$

Therefore, roots of this equation are –

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$

v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans: $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

Therefore, roots of this equation are –

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$

2. i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans: Let the number of john’s marbles be $\text{x}$.

Thus, the number of Jivanti’s marbles is $\text{45-x}$.

According to question i.e, 

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And the number of Jivanti’s marble is $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

So now,

Case 1: If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles is $\text{36}$.

Thus, the number of Jivanti’s marbles is $\text{9}$.

Case 2: If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles will be $9$.

Thus, the number of Jivanti’s marbles is $36$.

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 $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy is $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1: If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys will be $25$.

Case 2: If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys will be $30$.

3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$.

Ans: Let the first number be $\text{x}$ ,

Thus, the second number is $\text{27-x}$.

Therefore,

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1: If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number is $\text{14}$.

Case 2: If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number is $\text{14}$.

Thus, the second number is $13$.

4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1: If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because the number is positive.

Case 2: If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number is $\text{13}$.

Thus, the second number is $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.

5. The altitude of a right triangle is $\text{7 cm}$ less than its base. If the hypotenuse is $\text{13 cm}$, find the other two sides.

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude is $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagoras theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1: If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2: If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because the side is always positive.

Hence, the base of the right-angled triangle is $\text{12 cm}$ and its altitude is $\text{5cm}$.

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 $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans:  Let the number of articles produced be $\text{x}$.

Therefore, the cost of production of each article is $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1: If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because the number of articles is always positive.

Case 2: If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced will be $6$.

Therefore, the cost of production of each article is $\text{Rs15}$.

Class 10 Maths Chapter 4 - Exercise 4.2

Chapter 4 Exercise 4.2 Class 10 Mathematics Solutions

This chapter introduces students to how quadratic equations are formed and how to solve them. Students must first learn the basic format of quadratic equations which is ax2+bx+c=0; in this equation, a,b and c are the numbers which stay constant throughout the problem. Here we are trying to figure out the values of x. 

• In this first question of Exercise, 4.2 students must use the factorization method to find the root of each equation. After finding the root of each equation, split the middle term to find the value of x. The value of x is the root of the equation. These are the first basic problems which help students grasp the basic concept of quadratic equations. 

• The second question of this exercise can be a little tricky because students have to form the equations by themselves. A prior understanding of basic algebra is required to form the equation more easily. A hint to solving this problem is using the discriminant ( b2-4ac = 0), this will provide students with a faster way to solve both the equations in the question.

• The third question has provided the sum and its product already, students have to use the basic concepts they learned in quadratic equations to figure the sum out. Here students have to make an assumption on the numbers where one of them can be x and the other "27-x". To better understand the steps involved in solving this problem, you can refer to Chapter 4, Exercise 4.2 Solutions provided by Vedantu. 

• Question 4 and 5 is a culmination of the previous sums, and students must be thorough with knowing how to form a quadratic equation. It is also crucial for students to know the Pythagoras theorem because, in question 4, you have to form an algebraic equation by applying the Pythagoras theorem. Students must note that if answers come in positive and negative, they must consider the positive figure.

• Question 6 is the last question of the exercise and is also the most tricky as it is a situation based problem. Once again, students are required to solve the problem by assuming the values of x. Students have to factorize the equation and split the middle term to find the value. These problems become easier over time with a lot of practice and hard work.
  
  

Benefits of Chapter 4 Exercise 4.2 Class 10 Mathematics Solved Solutions. 

The solved solutions for Chapter 4 in Class 10 can provide students with a number of benefits during last-minute preparations of the subject. The benefits of this are as follows. 

• The solutions are solved in a manner that would be easy for students to interpret. 

• The solutions are solved according to the CBSE guidelines.

• The problems are solved by experts in the field, thereby leaving no space for mistakes.

• The solutions can act as useful revision material for students during last-minute preparations.

NCERT Solutions for Class 10 Maths Chapter 4 Exercises