Exercise 2.2 Page: 33

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

Solutions:

(i) 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

⇒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 = (½)+(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

⇒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

⇒ 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

⇒ 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

⇒ 3x²–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)

Therefore, zeroes of polynomial equation3x² – 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 = α β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -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 = α β = √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 = α β = 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+1is the quadratic polynomial.

(v) -1/4, 1/4

Solution:

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 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 =

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.