Find the zeros of each of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
(i) 
(ii) 
(iii) 
Answers
SOLUTION :
Given : f(x)= x² - 2x - 8
= x² - 4x+ 2x - 8
[By splitting the middle term]
= x(x–4) +2 (x–4)
= (x+2) (x–4)
To find zeroes, put f(x) = 0
(x+2) = 0 or (x–4) = 0
x = - 2 or x = 4
Hence, Zeroes of the polynomials are α = -2 and β = 4.
VERIFICATION :
Sum of the zeroes = −coefficient of x / coefficient of x²
α + β = −coefficient of x / coefficient of x²
-2 + 4 = −(−2)1
2 = 2
Product of the zeroes = constant term/ Coefficient of x²
α β = constant term / Coefficient of x²
-8 = −8/1
-8 = -8
Hence, the relationship between the Zeroes and its coefficients is verified.
(ii) Given : g(s)= 4s² –4s+1
= 4s² –2s–2s + 1
= 2s(2s–1)−1(2s–1)
= (2s–1)(2s–1)
To find zeroes, put g(s) = 0
(2s–1) = 0 or (2s–1) = 0
2s = 1 or 2s = 1
s = ½ or s = 1/2
Hence, Zeroes of the polynomials are α = 1/2 and β = 1/2
Sum of zeroes = −coefficient of s / coefficient of s²
α + β = −coefficient of s / coefficient of s²
½ + ½ = −(−4) / 4
2/2 = 4/4
1 = 1
Product of zeroes = constant term/ Coefficient of s²
α β = constant term / Coefficient of s²
½ × ½ = ¼
¼ = ¼
Hence, the relationship between the Zeroes and its coefficients is verified.
(iii) Given : h(t)= t² –15
=(t)² –(√15)²
= (t + √15)(t - √15)
[a² - b² = (a + b) (a - b)
To find zeroes, put h(t) = 0
(t + √15)(t - √15) = 0
(t + √15) = 0 or (t - √15) = 0
t = - √15 or t = √15
Hence, Zeroes of the polynomials are α = - √15 and β = √15
VERIFICATION :
Sum of the zeroes = −coefficient of t / coefficient of t²
α + β = −coefficient of t / coefficient of t²
(−15+ √15) = 0/1
0 = 0
Product of zeroes = constant term / Coefficient of t²
α β = constant term / Coefficient of t²
−√15 × √15 = −15/1
-15 = -15
Hence, the relationship between the Zeroes and its coefficients is verified.
HOPE THIS ANSWER WILL HELP YOU….
