NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2 – Polynomials contain 2 questions, for which detailed answers have been provided in this note.
| Category | NCERT Solutions for Class 10 |
| Subject | Maths |
| Chapter | Chapter 2 |
| Exercise | Exercise 2.2 |
| Chapter Name | Polynomials |
Download NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2 – Polynomials
NCERT Solutions For Class 10 Maths Chapter 2 Exercise 2.2 – Polynomials
1. Find the zeros of the quadratic polynomials and verify the relationships between the zeros and the coefficients.
(i) x2-2x-8
Solution:
Let, P(x) = x2-2x-8
Comparing P(x) with a quadratic equation ax2+bx+c
we have b/a = -2 and c/a = -8.
P(x) = x2-2x-8
= x2-4x+2x-8
= x(x-4) + 2(x-4)
= (x-4) (x+2)
Therefore, the value of P(x) will be zero if x-4=0 or x+2=0 i.e., when x=4 or x=-2
Hence the zeros of P(x) are 4, -2
Sum of zeros = 4+(-2) = 2 = -(-2/1) = -(b/a)
Product of the zeros = 4(-2) = -8 = c/a
(ii) 4s2-4s+1
Solution:
Let P(s) = 4s2-4s+1
Comparing P(s) with a quadratic equation as2+bs+c we have b/a = -(4/4)Â = -1 and (c/a) = 1/4
P(s) = 4s2-4s+1
= (2s)2-2. 2s+12
= (2s-1)2
Therefore, the value of P(s) will be zero if (2s-1)2 = 0 i.e., if s = 1/2 ,or 1/2
Hence the zeros of P(s) are 1/2, 1/2
Sum of zeros = 1/2 + 1/2 = (1+1)/2 = 2/2 = 1 = -b/a
Product of the zeros = (1/2).(1/2) = 1/4 = c/a
(iii) 6x2-3-7x
Solution:
Let, P(x) = 6x2-3-7x
Comparing P(x) with a quadratic equation ax2+bx+c we have b/a = -7/6Â and c/a = – 3/6
Now P(x) = 6x2-7x-3
= 6x2-9x+2x-3
=3x(2x-3) + 1(2x-3)
= (2x-3) (3x+1)
Therefore the value of P(x) will be zero if 2x-3=0 or 3x+1=0
Then, 2x = 3
x = 3/2
Or, 3x = -1
x = -(1/3)
Hence the zeros of P(x) are 3/2, -(1/3)
Sum of zeros = 3/2 + (-1/3) = 3/2Â – 1/6 = (9-1)/6 = 7/6Â = -(b/a)
Product of the zeros = (3/2).(-1/3) = -(3/6)Â = c/a
(iv) 4u2+8u
Solution:
Let, P(u) = 4u2+8u
Comparing P(u) with a quadratic equation au2+bu+c we have b/a = 8/4 = 2 and c/a = 0/4 = 0
P(u) = 4u2+8u
= u(4u+8)
Therefore, the value of P(u) will be zero
Then, u = 0
Or, 4u+8 = 0
4u = -8
u = -2
Hence the zeros of P(s) are 0, -2
Now sum of zeros = 0+(-2) =-2 =-(b/a)
And product of the zeros = 0. 2= 0 = c/a
(v) t2-15
Solution:
Let P(t) = t2-15
Comparing P(t) with a quadratic equation at2+bt+c we have b/a = 0 and c/a = -15
Therefore, the value of P(t) will be zero if t2-15 = 0 i.e., t2= 15 so t = ±√15
Hence the zeros of P(s) are √15, – √15
Sum of zeros = √15 – √15 = 0 = -(b/a)
Product of the zeros = √15(-√15) = -15 = c/a
(vi) 3x2– x-4
Solution:
Let P(x) = 3x2-x-4
Comparing P(x) with a quadratic equation ax2+bx+c we have b/a = -(1/3) and (c/a) = -(4/3)
P(x) = 3x2-x-4
= 3x2-4x+3x-4
= x(3x-4)+1(3x-4)
= (3x-4)(x+1)
Therefore the value of P(x) will be zero if 3x-4=0 or x+1=0
Then, 3x = 4
x = 4/3
Or, x = -1
Hence the zeros of P(x) are 4/3, – 1
Sum of zeros = (4/3) – 1= (4-3)/3Â = 1/3 = -(b/a)
Product of the zeros = (4/3).(-1) = -(4/3)Â = c/a
2. Find the quadratic polynomials each with the given numbers as the sum and products of its zeros respectively.
(i) (1/4), -1
Solution: Let the quadratic polynomial be ax2+bx+c and its zeros be α, β.
Then α+β = (1/4) = -(b/a)
And αβ = -1 = (c/a)
Then the quadratic polynomial is given by
x2 –(α+β)x+ αβ = x2– (1/4)x – 1
(ii) √2, 1/3
Solution: Let α, β be the two zeros of the quadratic polynomial.
Then α+β = √2 and αβ = 1/3
Hence the required quadratic polynomial is given by
x2 – (α+β)x + αβ = x2– √2x + (1/3)
(iii) 0, √5
Solution: Let α, β be the two zeros of the quadratic polynomial.
Then α+β = 0 and αβ = √5
Hence the required quadratic polynomial is given by
x2 – ( α+β)x + αβ = x2– 0.x + √5 = x2+ √5
(iv) 1, 1
Solution: Let α, β be the two zeros of the quadratic polynomial.
Then α+β = 1 and αβ = 1
Hence the required quadratic polynomial is given by
x2 – ( α+β)x + αβ = x2– 1.x + 1 = x2– x + 1
(v) -(1/4) , (1/4)
Solution: Let α, β be the two zeros of the quadratic polynomial.
Then α+β = -(1/4) and αβ = 1/4
Hence the required quadratic polynomial is given by
x2 – ( α+β)x + αβ = x2– (-1/4).x + (1/4) = x2+ (1/4)x + 1/4
(vi) 4, 1
Solution: Let α, β be the two zeros of the quadratic polynomial.
Then α+β = 4 and αβ = 1
Hence the required quadratic polynomial is given by
x2 –( α+β)x+ αβ = x2– 4.x + 1 = x2– 4x +1
NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2 – Polynomials, has been designed by the NCERT to test the knowledge of the student on the topic – Relationship between Zeroes and Coefficients of a Polynomial