Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
x2 – 2x – 8 = x2 – 4x + 2x – 8
= x (x – 4) + 2(x – 4)
= (x – 4) (x + 2)
Therefore, zeroes of equation x2 – 2x – 8 are (4, -2)
Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1
i.e. = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1
i.e.= (Constant term) / (Coefficient of x2)
(ii) 4s2 – 4s + 1
4s2 – 4s + 1 = 4s2 – 2s – 2s +1
= 2s(2s – 1) – 1(2s – 1)
= (2s – 1) (2s – 1)
Therefore, zeroes of equation 4s2 – 4s +1 are (1/2, 1/2)
Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4
i.e.= -(Coefficient of s) / (Coefficient of s2)
Product of zeros is equal to [(1/2) × (1/2)] = 1/4
i.e.= (Constant term) / (Coefficient of s2 )
(iii) 6x2 – 3 – 7x
6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1) (2x – 3)
Therefore, zeroes of equation 6x2 – 3 – 7x are (-1/3, 3/2)
Sum of zeroes is equal to -(1/3) + (3/2) = (7/6)
i.e.= -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to -(1/3) × (3/2) = -(3/6)
i.e.= (Constant term) / (Coefficient of x2 )
(iv) 4u2 + 8u
4u2 + 8u = 4u(u + 2)
Therefore, zeroes of equation 4u2 + 8u are (0, -2).
Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)
i.e. = -(Coefficient of u) / (Coefficient of u2)
Product of zeroes is equal to 0 × -2 = 0 = 0/4
i.e. = (Constant term) / (Coefficient of u2 )
(v) t2 – 15
t2 – 15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of equation t2 – 15 are (√15, -√15)
Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1)
i.e.= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes is equal to √15 × (-√15) = -15 = -15/1
i.e. = (Constant term) / (Coefficient of t2 )
(vi) 3x2 – x – 4
3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4) (x + 1)
Therefore, zeroes of equation 3x2 – x – 4 are (4/3, -1)
Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3)
i.e. = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to (4/3) × (-1) = (-4/3)
i.e. = (Constant term) / (Coefficient of x2)
Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given, 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
∴ 4x2 – x – 4 is the quadratic polynomial.
(ii) √2, 1/3
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given Sum of zeroes = α + β =√2
Product of zeroes = αβ = 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
∴ 3x2 – 3√2x + 1 is the quadratic polynomial.
(iii) 0, √5
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ x2 + √5 is the quadratic polynomial.
(iv) 1, 1
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ x2 – x + 1 is the quadratic polynomial.
(v) -1/4, 1/4
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ 4x2 + x + 1 is the quadratic polynomial.
(vi) 4, 1
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ x2 – 4x + 1 is the quadratic polynomial.