Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:

f(x) = x² – 2x – 8

Answer 1

Given: f(x) = x² – 2x – 8

Step-wise calculation:

1. Solve f(x) = 0:

x² – 2x – 8 = 0

2. Factor:

x² – 2x – 8 = (x – 4)(x + 2)

3. Set each factor to zero:

x – 4 = 0 or x + 2 = 0 

⇒ x = 4 or x = –2

Verification:

Sum of zeros: α + β = 4 + (–2) = 2. For a quadratic ax² + bx + c, `α + β = −b/a`; here `-b/a = -(-2)/1 = 2`.

Product of zeros: αβ = 4 × (–2) = –8. For a quadratic, `αβ = c/a`; here `c/a = -8/1 = -8`.

The zeros are x = 4 and x = –2 and they satisfy `α + β = -b/a` and `αβ = c/a`.