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

# NCERT Solutions for Exercise 4.2 Class 10 Maths Chapter 4 - Quadratic Equations

Edited By Ramraj Saini | Updated on Nov 16, 2023 11:42 AM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.2

NCERT Solutions for Exercise 4.2 Class 10 Maths Chapter 4 Quadratic Equations are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 4.2 class 10 deals with the concept of solution of the quadratic equation by factorization. The general form of the quadratic equation in the variable x is ax2+ bx + c = 0 where a, b, c are real numbers. Class 10 maths ex 4.2 comprises 6 straightforward questions with sub-questions that are simple to illuminate and additionally explore the concepts and nature of the roots. In exercise 4.2 Class 10 Maths, there are three main methods to solve the quadratic equation i.e. Factorizations, Completing the square and Quadratic Formula.

NCERT solutions for Class 10 Maths exercise 4.2, focused on the concepts of solving quadratic equations and understanding more about the relationship between the roots of the equation and the nature of the equation. These class 10 maths ex 4.2 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Assess NCERT Solutions for Class 10 Maths chapter 4 exercise 4.2

Quadratic Equations class 10 chapter 4 Excercise: 4.2

Given the quadratic equation: $x^2 - 3x - 10 =0$

Factorization gives, $x^2 - 5x+2x - 10 =0$

$\Rightarrow x^2 - 5x+2x - 10 =0$

$\Rightarrow x(x-5) +2(x-5) =0$

$\Rightarrow (x-5)(x+2) =0$

$\Rightarrow x= 5\ or\ -2$

Hence, the roots of the given quadratic equation are $5\ and\ -2$ .

Given the quadratic equation: $2x^2 + x - 6 = 0$

Factorisation gives, $2x^2 +4x-3x - 6 = 0$

$\Rightarrow 2x(x+2) -3(x+2) =0$

$\Rightarrow (x+2)(2x-3) = 0$

$\Rightarrow x= -2\ or\ \frac{3}{2}$

Hence, the roots of the given quadratic equation are

$-2\ and\ \frac{3}{2}$

$\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Given the quadratic equation: $\sqrt2x^2 + 7x + 5\sqrt2 = 0$

Factorization gives, $\sqrt2x^2 + 5x+2x + 5\sqrt2 = 0$

$\Rightarrow x(\sqrt2 x +5) +\sqrt2 (\sqrt 2 x +5)= 0$

$\Rightarrow (\sqrt2 x +5)(x+\sqrt{2}) = 0$

$\Rightarrow x=\frac{-5}{\sqrt 2 }\ or\ -\sqrt 2$

Hence, the roots of the given quadratic equation are

$\frac{-5}{\sqrt 2 }\ and\ -\sqrt 2$

$2x^2 -x + \frac{1}{8} = 0$

Given the quadratic equation: $2x^2 -x + \frac{1}{8} = 0$

Solving the quadratic equations, we get

$16x^2-8x+1 = 0$

Factorization gives, $\Rightarrow 16x^2-4x-4x+1 = 0$

$\Rightarrow 4x(4x-1)-1(4x-1) = 0$

$\Rightarrow (4x-1)(4x-1) = 0$

$\Rightarrow x=\frac{1}{4}\ or\ \frac{1}{4}$

Hence, the roots of the given quadratic equation are

$\frac{1}{4}\ and\ \frac{1}{4}$

$100x^2 -20x +1 = 0$

Given the quadratic equation: $100x^2 -20x +1 = 0$

Factorization gives, $100x^2 -10x-10x +1 = 0$

$\Rightarrow 10x(10x-1)-10(10x-1) = 0$

$\Rightarrow (10x-1)(10x-1) = 0$

$\Rightarrow x=\frac{1}{10}\ or\ \frac{1}{10}$

Hence, the roots of the given quadratic equation are

$\frac{1}{10}\ and\ \frac{1}{10}$ .

(i) $x^2-45x+324 = 0$ (ii) $x^2-55x+750 = 0$

From Example 1 we get:

Equations:

(i) $x^2-45x+324 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-45x+324 = 0$

Factorization gives, $x^2-36x-9x+324 = 0$

$\Rightarrow x(x-36) - 9(x-36) = 0$

$\Rightarrow (x-9)(x-36) = 0$

$\Rightarrow x=9\ or\ 36$

Hence, the roots of the given quadratic equation are $x=9\ and \ 36$ .

Therefore, John and Jivanti have 36 and 9 marbles respectively in the beginning.

(ii) $x^2-55x+750 = 0$

Solving by factorization method:

Given the quadratic equation: $x^2-55x+750 = 0$

Factorization gives, $x^2-30x-25x+750 = 0$

$\Rightarrow x(x-30) -25(x-30) = 0$

$\Rightarrow (x-25)(x-30) = 0$

$\Rightarrow x=25\ or\ 30$

Hence, the roots of the given quadratic equation are $x=25\ and \ 30$ .

Therefore, the number of toys on that day was $30\ or\ 25.$

Let two numbers be x and y .

Then, their sum will be equal to 27 and the product equals 182.

$x+y = 27$ ...............................(1)

$xy =182$ .................................(2)

From equation (2) we have:

$y = \frac{182}{x}$

Then putting the value of y in equation (1), we get

$x+\frac{182}{x} = 27$

Solving this equation:

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

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

$\Rightarrow x(x-13)-14(x-13) = 0$

$\Rightarrow (x-14)(x-13) = 0$

$\Rightarrow x = 13\ or\ 14$

Hence, the two required numbers are $13\ and \ 14$ .

Let the two consecutive integers be $'x'\ and\ 'x+1'.$

Then the sum of the squares is 365.

. $x^2+ (x+1)^2 = 365$

$\Rightarrow x^2+x^2+1+2x = 365$

$\Rightarrow x^2+x-182 = 0$

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

$\Rightarrow x(x-13)+14(x-13) = 0$

$\Rightarrow (x-13)(x-14) = 0$

$\Rightarrow x =13\ or\ 14$

Hence, the two consecutive integers are $13\ and\ 14$ .

Let the length of the base of the triangle be $b\ cm$ .

Then, the altitude length will be: $b-7\ cm$ .

Given if hypotenuse is $13\ cm$ .

Applying the Pythagoras theorem; we get

$Hypotenuse^2 = Perpendicular^2 + Base^2$

So, $(13)^2 = (b-7)^2 +b^2$

$\Rightarrow 169 = 2b^2+49-14b$

$\Rightarrow 2b^2-14b-120 = 0$ Or $b^2-7b-60 = 0$

$\Rightarrow b^2-12b+5b-60 = 0$

$\Rightarrow b(b-12) + 5(b-12) = 0$

$\Rightarrow (b-12)(b+5) = 0$

$\Rightarrow b= 12\ or\ -5$

But, the length of the base cannot be negative.

Hence the base length will be $12\ cm$ .

Therefore, we have

Altitude length $= 12cm -7cm = 5cm$ and Base length $= 12\ cm$

Let the number of articles produced in a day $= x$

The cost of production of each article will be $=2x+3$

Given the total production on that day was $Rs.90$ .

Hence we have the equation;

$x(2x+3) = 90$

$2x^2+3x-90 = 0$

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

$\Rightarrow x(2x+15) - 6(2x+15) = 0$

$\Rightarrow (2x+15)(x-6) = 0$

$\Rightarrow x =-\frac{15}{2}\ or\ 6$

But, x cannot be negative as it is the number of articles.

Therefore, $x=6$ and the cost of each article $= 2x+3 = 2(6)+3 = 15$

Hence, the number of articles is 6 and the cost of each article is Rs.15.

## More About NCERT Solutions for Class 10 Maths Exercise 4.2: Quadratic Equations

Exercise 4.2 Class 10 Maths consists of a question based on the solution of a quadratic equation by factorisation. The most important method in factorization is splitting of the middle term. In order to solve the quadratic equations by Factorisation, we need to arrange the terms in the form of ax2+bx+c=0 . Then we need to find the product of the first and last terms of the given expression. Next, we need to try to split the middle term into two terms but their product should be equal to the product of the first and last terms of the expression. Finally, we need to group the terms to factorize the given expression. The NCERT solutions for Class 10 Maths exercise 4.2 basically centred on the concepts of understanding quadratic conditions. Scientifically, the roots of the condition and the zeroes of the polynomial are the same. There must be at least two roots/zeros for a quadratic condition as given in exercise 4.2 Class 10 Maths. Also students can get access of Quadratic Equations Class 10 Notes to revise all the concepts quickly.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 4.2:

• NCERT solutions for Class 10 Maths exercise 4.2 helps in solving the problems in a step-by-step manner which is essential to avoid confusion for you.
• If you go through the NCERT solution for Class 10th Maths chapter 4 exercise 4.2, we can gain more knowledge about quadratic equations. Since it is an important exercise, you need to practice all the questions at least twice.
• Exercise 4.2 Class 10 Maths, is based on tackling the arrangements of the quadratic equations by factorisation and their roots which are imperative concepts of the chapter.

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## Subject Wise NCERT Exemplar Solutions

### Frequently Asked Questions (FAQs)

1. What is the general form of the quadratic equation?

The common shape of the quadratic condition is ax62+bx+c=0 where a, b, c are real numbers.

2. Whether the roots of the condition and the zeroes of the condition are the same ?

Yes, the roots of the equation and the zeroes of the equation are the same.

3. Factorize the polynomial a^2-10a+24 .

a^2-10a+24=a2-4a-6a+24

=a(a-4)-6(a-4)

=(a-4)(a-6)

4. What is the vital condition for understanding the quadratic conditions by utilizing the strategy of factorization?

In the strategy of factorization, the product of 1st and final terms of a given condition must be broken even with the product of 2nd and 3rd terms of the same given condition.

5. What is the splitting of the middle term?

Splitting of the middle term is nothing but we have to rewrite the middle term of the quadratic expression as the sum or difference of the two terms, that is we have to split the middle term into two parts in terms of sum or difference of the terms.

6. What is the sum product form according to NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 ?

The sum-product form is nothing but in the equation ax^2+bx+c=0 , the product of the middle term after splitting must be equal to a×c and the sum must be equal to b.

7. How numerous questions and what sort of questions are secured inside the NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 ?

NCERT solutions for Class 10 Maths chapter 4 exercise 4.2 comprises of six questions which are based on the factorization strategy.

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### Questions related to CBSE Class 10th

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Hello

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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

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Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

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