Question

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

`p(y) = y^2 + (3sqrt(5))/2 y - 5`

Answer 1

Given: `p(y) = y^2 + (3sqrt(5))/2 y - 5`.

Step-wise calculation:

1. Write coefficients:

`a = 1, b = (3sqrt(5))/2, c = -5`

2. Discriminant: Δ = b² – 4ac

= `((3sqrt(5))/2)^2 - 4(1)(-5)`

= `45/4 + 80/4`

= `125/4`

3. `sqrt(Δ) = (5sqrt(5))/2`.

4. Roots by quadratic formula:

`y = (-b ± sqrt(Δ))/(2a)`

= `(-(3sqrt(5))/2 ± (5sqrt(5))/2)/2` 

= `sqrt(5)(-3 ± 5)/4` 

So `y_1 = sqrt(5)(2)/4 = sqrt(5)/2` and `y_2 = sqrt(5)(-8)/4 = -2sqrt(5)`.

Thus the zeros are `sqrt(5)/2` and `-2sqrt(5)`.

Verify relationship between zeros and coefficients:

Sum of zeros = `(sqrt(5)/2) + (-2sqrt(5))`

= `-(3sqrt(5))/2 = -b/a`, which equals `-(3sqrt(5))/2`

Product of zeros = `(sqrt(5)/2) xx (-2sqrt(5))`

= –5 

= `c/a`

Zeros: `y = sqrt(5)/2` and `y = -2sqrt(5)`. The sum and product match the standard relations sum = `-b/a` and product = `c/a`, so the relationship between zeros and coefficients is verified.