NCERT Solutions Class 10th 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

Solution: x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)

Therefore, zeroes of polynomial equation x² – 2x – 8 are (4, -2)

Sum of zeroes = 4 – 2 = 2 = – (- 2)/1 = – (Coefficient of x)/(Coefficient of x²)
Product of zeroes = 4 × (-2) = -8 = -(8)/1 = (Constant term)/(Coefficient of x²)

(ii) 4s²– 4s + 1

Solution: 4s² – 4s + 1
= 4s² – 2s – 2s + 1
= 2s(2s – 1) –1(2s – 1)
= (2s – 1)(2s – 1)

Therefore, zeroes of polynomial equation 4s² – 4s + 1 are (1/2, 1/2)

Sum of zeroes = (1/2) + (1/2) = 1 = – (- 4)/4 = – (Coefficient of s)/(Coefficient of s²)
Product of zeros = (1/2) × (1/2) = 1/4 = (Constant term)/(Coefficient of s²)

(iii) 6x²– 3 – 7x

Solution: 6x² – 3 – 7x
= 6x² – 7x –3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1)(2x -3)

Therefore, zeroes of polynomial equation 6x² – 3 – 7x are (-1/3, 3/2)

Sum of zeroes = -(1/3) + (3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x²)
Product of zeroes = -(1/3) × (3/2) = -(3/6) = (Constant term)/(Coefficient of x²)

(iv) 4u²+ 8u

Solution: 4u² + 8u
= 4u(u + 2)

Therefore, zeroes of polynomial equation 4u² + 8u are (0, -2)

Sum of zeroes = 0 + (-2) = -2 = -(8/4) =  – (Coefficient of u)/(Coefficient of u²)
Product of zeroes = 0 × -2 = 0 = 0/4 = (Constant term)/(Coefficient of u²)

(v) t² – 15

Solution: t²– 15
= t² = 15 or t
= ± √15

Therefore, zeroes of polynomial equation t² –15 are (√15, -√15)

Sum of zeroes = √15 + (-√15) = 0 = -(0/1) = – (Coefficient of t)/(Coefficient of t²)
Product of zeroes = √15 × (-√15) = -15 = -15/1 = (Constant term)/(Coefficient of t²)

(vi) 3x² – x – 4

Solution: 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation 3x² – x – 4 are (4/3, -1)

Sum of zeroes = (4/3) + (-1) = (1/3)= -(-1/3) = – (Coefficient of x)/(Coefficient of x²)
Product of zeroes = (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

Solution: From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β
Product of zeroes = αβ
Sum of zeroes = α + β = 1/4Product of zeroes = αβ = -1

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

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) √2, 1/3

Solution: Sum of zeroes = α + β = √2
Product of zeroes = α β = 1/3

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

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) 0, √5

Solution: Given,
Sum of zeroes = α + β = 0
Product of zeroes = αβ = √5

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0
x² – (0)x + √5= 0

Thus, x² + √5 is the quadratic polynomial.

(iv) 1, 1

Solution: Given,
Sum of zeroes = α + β = 1
Product of zeroes = αβ = 1

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0
x² – x + 1 = 0

Thus, x² – x + 1 is the quadratic polynomial.

(v) -1/4, 1/4

Solution: Given,
Sum of zeroes = α + β = -1/4
Product of zeroes = αβ = 1/4

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

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) 4, 1

Solution: Given,
Sum of zeroes = α + β = 4
Product of zeroes = αβ = 1

If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x² – (α + β)x + αβ = 0
x² – 4x + 1 = 0

Thus, x² – 4x + 1 is the quadratic polynomial.