Question

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

h(t) = t² – 15

Answer 1

Given: h(t) = t² – 15

Step-wise calculation:

1. Set h(t) = 0:

t² – 15 = 0

2. Solve for t:

t² = 15

`t = ±sqrt(15)` 

So the factorization is `h(t) = (t - sqrt(5))(t + sqrt(15))`.

3. Let the zeros be `α = sqrt(15)` and `β = -sqrt(15)`.

Sum of zeros: `α + β = sqrt(15) + (-sqrt(15)) = 0`. 

Product of zeros: `αβ = (sqrt(15))(-sqrt(15)) = -15`.

Verify relationship with coefficients For a quadratic at² + bt + c, sum of zeros = `-b/a` and product = `c/a`. 

Here a = 1, b = 0, c = –15, so `-b/a = -0/1 = 0` (matches α + β), `c/a = -15/1 = -15` (matches αβ).

The zeros of h(t) = t² – 15 are `t = sqrt(15)` and `t = -sqrt(15)` and they satisfy sum = 0 and product = –15, which agree with the coefficients `(-b/a and c/a)`.