Class 10 Math’s Chapter 2-Polynomials Exercise 2.2
NCERT
Solutions for Class 10 Math’s Chapter 2-Polynomials Exercise 2.2
. Exercise 2.2 of NCERT solutions for Class 10 Maths
Chapter 2 is the second exercise of Polynomials of Class 10 Maths. Polynomials
are introduced in Class 9 and is further discussed in detail in Class 10, by
studying different cases of relationship between Zeroes and Coefficients of a
Polynomial.
- Relationship between Zeroes and
Coefficients of a Polynomial – It includes two questions with six
different cases each.
- These NCERT Solutions helps you
solve and revise all questions of Exercise 2.1.
- Solving these NCERT Solutions
will help you score well in exams.
- These are the best study
resources as they are prepared by Maths subject experts.
- It follows NCERT guidelines,
which help in preparing the students accordingly.
- It contains all the important
questions from the examination point of view.
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Answers to NCERT Class 10 Maths Chapter 2 – Polynomials Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and
verify the relationship between the zeroes and the coefficients
(i) x2–2x –8
(ii) 4s2–4s+1
(iii) 6x2–3–7x
(iv) 4u2+8u
(v) t2–15
(vi)3x2–x–4
Solutions:
(i) x2–2x
–8 = 0
x2– (4-2)x–8
= 0
x2–
4x+2x–8 = 0
x(x–4)+2(x–4) = 0
(x−4)(x+2)
= 0
Here, x - 4 = 0 ; and x+2=0
So, x=4 and x=−2.
Hence, α=4; and β=−2
Verification:
Sum of zeroes (α+β) = 4 + (-2)
=2
= 2/1 =
-(Coefficient of x)/(Coefficient of x2)
And product of zeroes (α×β) =4 × (−2)
=
−8
= -(8)/1 = (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1 = 0
4s2–(2+2)s+1
= 0
4s2–2s–2s+1
= 0
2s (2s–1)–1(2s-1) =0
(2s–1)(2s–1) = 0
Here, (2s−1) = 0 and (2s−1) = 0
So, s = 1/2 and s = 1/2
Verification:
Sum of zeroes (α+β) = ½ + ½
=1
=1/1×4/4 = 1/4 = -(Coefficient of
s)/(Coefficient of s2)
And product of zeroes (α×β) =1/2 ×1/2
=
1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x = 0
⇒ 6x2–7x–3
= 0
6x2 –
(9 − 2)x – 3 = 0
6x2 –
9x + 2x – 3 = 0
3x(2x – 3) +1(2x – 3) = 0
(3x+1)(2x-3)
= 0
Here, (3x+1) = 0 and (2x−3) = 0
So, x = - 1/3 and x = 3/2
Verification:
Sum of zeroes (α+β) = −1/3 +3/2
= 7/6 = − (Coefficient of x)/(Coefficient
of x2)
And product of zeroes (α×β) = −1/3 × 3/2
=
− 3/6 = − (Constant term)/(Coefficient of x2)
(iv) 4u2+8u =
0
⇒ 4u(u+2) = 0
Here, 4u = 0 and u+2 = 0
So, u = 0 ; and u = −2
Verification:
Sum of zeroes (α+β) = 0 + (−2)
= −2
= −2/1× 4/4
= −8/4 = (Coefficient of u)/(Coefficient of u2)
And product of zeroes (α×β) = 0 × (−2)
= 0
= 0/8 = (Constant term)/(Coefficient of u2 )
(v) t2–15 = 0
⇒ t2 = 15
Or t = ±√15
t = −√15 and
+√15
Verification:
Sum of zeroes (α+β) = −√15 + √15
= 0
= 0/1 = − (Coefficient of t) /
(Coefficient of t2)
And product of zeroes (α×β) = −√15 × √15
= −15
=
−15/1 = (Constant term) / (Coefficient of t2)
(vi) 3x2–x–4 = 0
3x2–x(4–3)
= 0
3x2–4x+3x−4
= 0
X(3x−4)+1(3x−4) = 0
(3x−4)(x+1) = 0
Here, (3x−4) = 0 and (x+1) = 0
So, x = 4/3 and x = −1
Verification:
Sum of zeroes (α+β) = 4/3 + (−1)
= 1/3 = − (Coefficient of x) /
(Coefficient of x2)
And product of zeroes (α×β) = 4/3 × (−1)
= − 4/3
= (Constant term) /(Coefficient of x2 )
2. Find a quadratic polynomial each with the
given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1; (ii)√2,1/3 (iii) 0, √5 (iv)1,1 (v)−1/4 , 1/4
Solution:
(i) given,
Sum of zeroes = (α+β) = 1/4
Product of zeroes = (α.β) = −1
So, x2–(α+β)x +αβ = 0
x2–(1/4)x
+(-1) = 0
4x2–x−4
= 0
Thus, 4x2–x–4 is the quadratic
polynomial.
(ii) Given,
Sum of zeroes = (α + β) =√2
Product of zeroes = (α.β) = 1/3
So, x2–(α+β)x +αβ = 0
x2 –(√2)x + (1/3)
= 0
3x2-3√2x+1 =
0
Thus, 3x2-3√2x+1 is
the quadratic polynomial.
(iii) Given,
Sum of zeroes = α+β = 0
Product of zeroes = α β = √5
So, x2 – (α+β)x + αβ = 0
x2–(0)x
+√5= 0
Thus, x2+√5 is
the quadratic polynomial.
(iv) Given,
Sum of zeroes = α+β = 1
Product of zeroes = α β = 1
So, x2–(α+β)x +αβ = 0
x2–x+1
= 0
Thus , x2–x+1is the quadratic
polynomial.
(v) Given,
Sum of zeroes = α+β = − 1/4
Product of zeroes = α β = 1/4
So, x2–(α+β)x +αβ = 0
x2–(-
1/4)x + (1/4) = 0
4x2+x+1
= 0
Thus, 4x2+x+1 is the quadratic
polynomial.
(vi) Given,
Sum of zeroes = α+β =4
Product of zeroes = αβ = 1
So, x2–(α+β)x + αβ = 0
x2–4x+1
= 0
Thus, x2–4x+1 is the quadratic
polynomial.
The NCERT Solutions Maths Class
10 for Chapter 2 Exercise 2.2 Polynomials are
provided here in pdf format. These solutions are created by Maths experts who
have reviewed them from time to time. These NCERT chapter-wise solutions help
the students to study and prepare well for their CBSE first term exams.
The Solutions
of NCERT for Exercise 2.2 Chapter 2- polynomials are given here in
a step-wise and easy to understand method. You will surely be able to solve
these easily, once you go through these NCERT
solutions for Class 10 Maths.
All the important factors like paying attention to NCERT
guidelines while preparing these solutions have been focused upon
here. Exercise 2.2 explains the relationship between Zeroes and
Coefficients of Polynomials
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