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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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Download PDF of NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2 Polynomials

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Access Exercise 2.2 Class 10 Maths Answers

Polynomials Class 10 Maths Chapter 2 Excercise: 2.2

Q1 (1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. x ^2 - 2x -8

Answer:

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

Q1 (ii) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4s^2 - 4s + 1

Answer:

\\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

Q1 (3) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 6x ^2 - 3 -7x

Answer:

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

Q1 (4) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4 u^2 + 8u

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

Q1 (5) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. t^2 - 15

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

Q1 (6) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 3 x^2 - x - 4

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

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.

Also see-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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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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.

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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.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} 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

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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