NCERT Solutions for Exercise 2.2 Class 10 Maths Chapter 2 - Polynomials

# NCERT Solutions for Exercise 2.2 Class 10 Maths Chapter 2 - Polynomials

Edited By Ramraj Saini | Updated on Nov 02, 2023 03:15 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2 Polynomials

NCERT Solutions for class 10 maths ex 2.2 Polynomials is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This exercise 2.2 Class 10 Maths is an introductory exercise that includes the concept of quadratic polynomials in x with real coefficients. Here we have the quadratic equation in form of ax² + bx + c=0 where a, b, c are real numbers with a cannot equal to zero.

In chapter 2 of Class 10 Mathematics NCERT syllabus the concept of zeroes along their relationship with the coefficients of equations is initialized. The Class 10 Maths chapter 2 exercise 2.2 lists a few practice problems on Polynomials. The Class 10th Maths chapter 2 exercise 2.2 covers the topics like sum and product of zeroes collectively in form of assumed roots/ zeroes. This chapter includes four exercises which are listed below, practice these to get in-depth understanding of concepts.

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## Access Exercise 2.2 Class 10 Maths Answers

Polynomials Class 10 Maths Chapter 2 Excercise: 2.2

x 2 - 2x - 8 = 0

x 2 - 4x + 2x - 8 = 0

x(x-4) + 2(x-4) = 0

(x+2)(x-4) = 0

The zeroes of the given quadratic polynomial are -2 and 4

$\\\alpha =-2\\ \beta =4$

VERIFICATION

Sum of roots:

$\\\alpha+\beta =-2+4=2$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-2}{1}\\ =2\\=\alpha +\beta$

Verified

Product of roots:

$\\\alpha \beta =-2\times 4=-8$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-8}{1}\\ =-8\\=\alpha \beta$

Verified

$\\4s^2 - 4s + 1 = 0 \\4s^2 - 2s - 2s + 1 = 0 \\2s(2s-1) - 1(2s-1) = 0 \\(2s-1)(2s-1) = 0$

The zeroes of the given quadratic polynomial are 1/2 and 1/2

$\\\alpha =\frac{1}{2}\\ \beta =\frac{1}{2}$

VERIFICATION

Sum of roots:

$\\\alpha +\beta =\frac{1}{2}+\frac{1}{2}=1$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-4}{4}\\ =1\\=\alpha +\beta$

Verified

Product of roots:

$\alpha \beta =\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{1}{4}\\ =\alpha \beta$

Verified

6x 2 - 3 - 7x = 0

6x 2 - 7x - 3 = 0

6x 2 - 9x + 2x - 3 = 0

3x(2x - 3) + 1(2x - 3) = 0

(3x + 1)(2x - 3) = 0

The zeroes of the given quadratic polynomial are -1/3 and 3/2

$\\\alpha =-\frac{1}{3}\\ \beta =\frac{3}{2}$

Sum of roots:

$\\\alpha+\beta =-\frac{1}{3}+\frac{3}{2}=\frac{7}{6}$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-7}{6}\\ =\frac{7}{6}\\=\alpha +\beta$

Verified

Product of roots:

$\\\alpha \beta =-\frac{1}{3}\times \frac{3}{2}=-\frac{1}{2}$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-3}{6}\\ =-\frac{1}{2}\\=\alpha \beta$

Verified

Answer: 4u 2 + 8u = 0

4u(u + 2) = 0

The zeroes of the given quadratic polynomial are 0 and -2

$\\\alpha =0\\ \beta =-2$

VERIFICATION

Sum of roots:

$\\\alpha +\beta =0+(-2)=-2$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{8}{4}\\ =-2\\=\alpha +\beta$

Verified

Product of roots:

$\alpha \beta =0\times-2=0$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{0}{4}\\=0\\ =\alpha \beta$

Verified

Answer: t 2 - 15 = 0

$(t-\sqrt{15})(t+\sqrt{15})=0$

The zeroes of the given quadratic polynomial are $-\sqrt{15}$ and $\sqrt{15}$

$\\\alpha =-\sqrt{15}\\ \beta =\sqrt{15}$

VERIFICATION

Sum of roots:

$\\\alpha +\beta =-\sqrt{15}+\sqrt{15}=0$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{0}{1}\\ =0\\=\alpha +\beta$

Verified

Product of roots:

$\alpha \beta =-\sqrt{15}\times \sqrt{15}=-15$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-15}{1}\\=-15\\ =\alpha \beta$

Verified

Answer: 3x 2 - x - 4 = 0

3x 2 + 3x - 4x - 4 = 0

3x(x + 1) - 4(x + 1) = 0

(3x - 4)(x + 1) = 0

The zeroes of the given quadratic polynomial are 4/3 and -1

$\\\alpha =\frac{4}{3}\\ \beta =-1$

VERIFICATION

Sum of roots:

$\\\alpha +\beta =\frac{4}{3}+(-1)=\frac{1}{3}$

$\\-\frac{coefficient\ of\ x}{coefficient\ of\ x^{2}}\\ =-\frac{-1}{3}\\ =\frac{1}{3}\\=\alpha +\beta$

Verified

Product of roots:

$\alpha \beta =\frac{4}{3}\times -1=-\frac{4}{3}$

$\\\frac{constant\ term}{coefficient\ of\ x^{2}}\\ =\frac{-4}{3}\\ =\alpha \beta$

Verified

$\\\alpha +\beta =\frac{1}{4}\\ \alpha \beta =-1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\frac{1}{4}x-1=0\\ 4x^{2}-x-4=0$

$\\\alpha +\beta =\sqrt{2}\\ \alpha \beta =\frac{1}{3}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\sqrt{2}x+\frac{1}{3}=0\\ 3x^{2}-3\sqrt{2}x+1=0$

The required quadratic polynomial is $3x^{2}-3\sqrt{2}x+1$

$\\\alpha +\beta =0\\ \alpha \beta =\sqrt{5}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-0x+\sqrt{5}=0\\ x^{2}+\sqrt{5}=0$

The required quadratic polynomial is x 2 + $\sqrt{5}$ .

$\\\alpha +\beta =1\\ \alpha \beta =1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-1x+1=0\\ x^{2}-x+1=0$

The required quadratic polynomial is x 2 - x + 1

$\\\alpha +\beta =-\frac{1}{4}\\ \alpha \beta =\frac{1}{4}$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-\left (-\frac{1}{4} \right )x+\frac{1}{4}=0\\ 4x^{2}+x+1=0$

The required quadratic polynomial is 4x 2 + x + 1

$\\\alpha +\beta =4\\ \alpha \beta =1$

$\\x^{2}-(\alpha +\beta )+\alpha \beta =0\\ x^{2}-4x+1=0\\$

The required quadratic polynomial is x 2 - 4x + 1

## More About NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

The problems from the concepts of attaining sum and product of roots are covered in ex 2.2 class 10 Maths. The Initial questions of NCERT solutions for Class 10 Maths chapter 2 exercise 2.2 is to represent the problems of finding zeros out of given quadratic equations. The remaining questions in class 10 ex 2.2 is to find the updated equations from given roots along with the special concept of the discriminant. In Class 10th Maths chapter 2 exercise 2.2 questions related to simplification of given expression of polynomial and theory type questions of proofs whether the zeroes belongs to the given quadratic equation or not.

## Key Features of NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

• NCERT 10th class maths exercise 2.2 answers are important as it covers questions from the basic form of equations comprehensively, step by step and according the latest syllabus and pattern of CBSE.
• If students can solve each question of this exercise 2.2 Class 10 Maths they will be able to grasp the hit and trial concept of finding zeros of polynomial and concept of discriminant as given in Class 10th Maths chapter 2 exercise 2.2
• For Class 10 final exams students may get the short answer or long answer questions from the type covered in the Class 10 Maths chapter 2 exercise 2.2. students can also study Real Numbers Class 10 Notes to revise the concepts quickly.

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## Subject Wise NCERT Exemplar Solutions

1. What will be the graph of polynomial F(x) = ax² + bx + c?

The graph of the given polynomial will be Parabola but if it's upward or downward it is decided by the condition given – if a>0 the parabola is upward but if a<0 then the parabola is downward. To understanding how this is happening keep reading 10th class maths exercise 2.2 answers from the pdf provided above in this article.

2. How many exercises are there in Chapter 2?

There are 4 exercises in chapter 2 i.e. Ex 2.1, 2.2, 2.3, 2.4. practice problems from each exercise including ex 2.2 class 10 to command the concepts.

3. What is the formula of the sum of roots of Polynomial ax² + bx + c =0?

The formula for sum of roots = α + β = -(b/a). where b and a are the coefficients of the polynomial. This formula is comprehensively covered in class 10 ex 2.2.

4. What is the expression of finding Equation from sum and product of roots?

Ans: If sum of roots = α + β and Product of roots = α * β then

x² - (α + β)x + α * β = 0

To get deeper understanding keep reading class 10 maths ex 2.2 using the pdf provided above in this article.

5. If 2x^4 + x -3 is divided by g(x) and quotient is x² + 5x + 6, what will be the possible degree of remainder is?

considering the highest degree of f(x) is 4 taking highest degree of g(x) = 2 as mentioned, if you use simple exponent rule then degree of remainder = 4-2 = 2

6. The graph in quadrant of the quadratic polynomial ax² + bx + c?

The graph of the given polynomial will be parabola and its orientation depends on the value of a, if a<0 then it's a downward parabola, and if a>0 then it's an upward parabola.

7. When does quadratic equation become perfect square?

It comes under the ideal case when an Equation becomes an identity, which means When the discriminant becomes zero.

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### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

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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.

hello Zaid,

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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