CBSE Class 10 Maths Chapter 4 Quadratic Equations – Solutions 2025–26

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 $\text{Rs15}$.

Conclusion

Class 10 Maths Ex 4.2 of Chapter 4 - Quadratic Equations, is crucial for a solid foundation in math. Understanding the concept of factoring quadratic expressions is a key takeaway. Vedantu's NCERT solutions can guide you in conquering quadratic equations using the method of factorization. Pay attention to the step-by-step solutions provided, grasp the underlying principles, and ensure clarity on the concepts before moving forward. Consistent practice and understanding will lead to proficiency in Quadratic Equations related problems

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