Question

Find a quadratic polynomial whose sum and product respectively of the zeros are given. Also, find the zeros of these polynomials.

`-3/(2sqrt(5)), -1/2`

Answer 1

Given: Sum of zeros `S = -3/(2sqrt(5))`, Product of zeros `P = -1/2`.

Step-wise calculation:

1. For a monic quadratic with zeros having sum S and product P, the polynomial is x² – Sx + P.

Substitute S and P: `x^2 - (-3/(2sqrt(5)))x + (-1/2)`

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

2. Multiply by `2sqrt(5)` to get integer-radical coefficients: 

`2sqrt(5)x^2 + 3x - sqrt(5) = 0`

3. Find the zeros using the quadratic formula on `x^2 + (3/(2sqrt(5)))x - 1/2 = 0`.

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

Discriminant Δ = b² – 4ac 

= `(3/(2sqrt(5)))^2 - 4(1)(-1/2)` 

= `9/20 + 2`

= `49/20`

`sqrt(Δ) = 7/(2sqrt(5))`.

Roots: `x = (-b ± sqrt(Δ))/(2a)`

= `(-3/(2sqrt(5)) ± 7/(2sqrt(5)))/2`

= `(-3 ± 7)/(4sqrt(5))`

4. Evaluate the two cases:

`x_1 = (-3 + 7)/(4sqrt(5))`

= `4/(4sqrt(5))`

= `1/sqrt(5)`

= `sqrt(5)/5`

`x_2 = (-3 - 7)/(4sqrt(5))`

= `(-10)/(4sqrt(5))`

= `(-5)/(2sqrt(5))`

= `-sqrt(5)/2`

A quadratic polynomial (monic) is `x^2 + (3/(2sqrt(5))) x - 1/2` `("equivalently"  2sqrt(5)x^2 + 3x - sqrt(5) = 0)`.

Its zeros are `x = sqrt(5)/5` and `x = -sqrt(5)/2`.