Class 10 Math Chapter 4 Quadratic Equations Solutions

Here, Class 10 Math Chapter 4 Quadratic Equations Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 4 Quadratic Equations Solutions are very important for board exam. Solving these Quadratic Equations solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Quadratic Equations Solutions

Exercise 4.1

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Exercise 4.2

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Exercise 4.3

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Chapter 3: Pair of Linear Equations in Two Variables Solutions

We have given solutions of all exercise of class 10 math chapter 4. These solutions will be helpful for students understanding core concept of Quadratic Equations.

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation. You can also find other study material resources on our website.

Important Questions for NCERT Chapter 4: Quadratic Equations

Question : Find the roots of the following quadratic equations by factorisation:

100x2−20x+1=0100x^2 – 20x + 1 = 0

Answer :

100x2−20x+1=0100x^2 – 20x + 1 = 0
=100x2−10x−10x+1= 100x^2 – 10x – 10x + 1 \\
=10x(10x−1)−1(10x−1)= 10x(10x – 1) – 1(10x – 1) \\
=(10x−1)2= (10x – 1)^2
Roots of this equation are the values for which (10x−1)2=0\text{Roots of this equation are the values for which } (10x – 1)^2 = 0 \\
Therefore, (10x−1)=0 or (10x−1)=0\text{Therefore, } (10x – 1) = 0 \text{ or } (10x – 1) = 0 \\
x=110 or x=110x = \frac{1}{10} \text{ or } x = \frac{1}{10}

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

Answer :

Let the consecutive positive integers be x and x+1.\text{Let the consecutive positive integers be } x \text{ and } x + 1. \\
Given that x2+(x+1)2=365\text{Given that } x^2 + (x + 1)^2 = 365
⇒x2+x2+1+2x=365\Rightarrow x^2 + x^2 + 1 + 2x = 365 \\
⇒2x2+2x−364=0\Rightarrow 2x^2 + 2x – 364 = 0 \\
⇒x2+x−182=0\Rightarrow x^2 + x – 182 = 0 \\
⇒x2+14x−13x−182=0\Rightarrow x^2 + 14x – 13x – 182 = 0 \\
⇒x(x+14)−13(x+14)=0\Rightarrow x(x + 14) – 13(x + 14) = 0
⇒(x+14)(x−13)=0\Rightarrow (x + 14)(x – 13) = 0
Either x+14=0 or x−13=0, i.e., x=−14 or x=13\text{Either } x + 14 = 0 \text{ or } x – 13 = 0\text{, i.e., } x = -14 \text{ or } x = 13
Since the integers are positive, x can only be 13.\text{Since the integers are positive, } x \text{ can only be } 13. \\
∴x+1=13+1=14 \therefore x + 1 = 13 + 1 = 14

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

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