45-45+ 1 =0 finde the zeros of the following quadratic polynomial and varify the relationship between the zero and coefficient
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Step-by-step explanation:
Answer:
Given quadratic polynomial is
4s2 − 4s + 1
To find the zeros of the quadratic polynomial we consider
4s2 − 4s + 1 = 0
4s2 − 2s – 2s + 1 = 0
2s(2s – 1) -1(2s – 1) = 0
(2s – 1)(2s – 1) = 0
2s – 1 = 0 and 2s – 1 = 0
2s = 1 and 2s = 1
s = 1/2 and s = 1/2
∴ The zeroes of the polynomial = 1/2 and 1/2
Sum of the zeroes = -(coefficient of s)/(coefficient of s2)
Sum of the zeroes = -(-4)/4 = 1
Let’s find the sum of the roots = 1/2 + 1/2 = 1
Product of the zeros = Constant term / Coefficient of s2
Product of the zeros =1 / 4
Let’s find the products of the roots = 1/2 × 1/2 = 1/4
hope it helps
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