Question

How do you find the roots, real and imaginary, of `y=-x^2 +32x -16` using the quadratic formula?

Answer 1

`x=16+sqrt(240)=31.49` OR `x=16-sqrt(240)=.51`

There are no imaginary roots.

Explanation:

Quadratic formula:

For `y=ax^2+bx+c`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

In your problem, `a=-1, b=32, c=-16`

So, `x=(-32+-sqrt(32^2-4(-1)(-16)))/(2(-1))`

Simplifying:

`x=(-32+-sqrt(1024-64))/(-2)`

`x=(-32+-sqrt(960))/(-2)`

`x=(-32)/-2+-sqrt(4*240)/(-2)`

`x=16+-(2sqrt(240))/(-2)`

So `x=16+sqrt(240)=31.49` OR `x=16-sqrt(240)=.51`

There are no imaginary roots.

Answer 2

Roots are `16-4sqrt15` and `16+4sqrt15`

Explanation:

According to quadratic formula, the roots of `y=ax^2+bx+c` are

`(-b+-sqrt(b^2-4ac))/(2a)`

Hence roots of `y=-x^2+32x-16` are

`(-32+-sqrt(32^2-4xx(-1)(-16)))/(2xx(-1))`

= `(-32+-sqrt(1024-64))/(-2)`

= `16+-sqrt960/2`

= `16+-(8sqrt15)/2`

= `16+-4sqrt15`