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Question
Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:
f(x) = 4x2 + 4x – 3
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Solution
Given: f(x) = 4x2 + 4x – 3
Step-wise calculation:
1. Set f(x) = 0:
4x2 + 4x – 3 = 0
2. Use the quadratic formula
`x = (-b ± sqrt(b^2 - 4ac))/(2a)` with a = 4, b = 4, c = –3:
b2 – 4ac = 42 – 4 × 4 × (–3)
= 16 + 48
= 64
So sqrt = 8.
`x = (-4 ± 8)/8`.
Thus `x_1 = (-4 + 8)/8`
= `4/8`
= `1/2`
And `x_2 = (-4 - 8)/8`
= `(-12)/8`
= `(-3)/2`
3. Sum of zeros: `α + β = 1/2 + (-3/2) = -1`.
Product of zeros: `αβ = (1/2) xx (-3/2) = -3/4`.
4. Compare with coefficients: `-b/a = -4/4 = -1` and `c/a = (-3)/4 = −3/4`.
The general relations sum = `-b/a` and product = `c/a` hold.
The zeros are `x = 1/2` and `x = -3/2`. The sum and product of the zeros equal `-b/a` and `c/a` respectively, so the relationship between zeros and coefficients is verified.
