Question

How do you find the roots, real and imaginary, of `y=-x^2 +12x -32` using the quadratic formula?

Answer 1

The roots are the points on the curve where `y=0`, so set `-x^2+12x-32=0`.

Solve using the quadratic formula as shown below.

The solutions (roots) are `x=4` and `x=8`.

Explanation:

For a quadratic equation in standard form, `ax^2+bx+c=0`, we know:

`y=(-b+-sqrt(b^2-4ac))/(2a)` (the quadratic formula)

In this instance:

`a=-1`
`b=12`
`c=-32`

So:

`y=(-12+-sqrt(12^2-4xx(-1)xx(-32)))/(2xx(-1))`

`=(-12+-sqrt(144-128))/-2=(-12+-sqrt(16))/-2=(-12+-4)/-2`

`=(-8)/-2=4` or `(-16)/-2=8`

So the roots are 4 and 8.

Answer 2

4 and 8

Explanation:

Use the improved quadratic formula (Socratic Search)
`y = -x^2 + 12x - 32 = 0`
`D = d^2 = b^2 - 4ac = 144 - 128 = 16`--> `d = +- 4`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = -12/-2 +- 4/-2 = 6 +- 2`

x1 = 6 + 2 = 8 and x2 = 6 - 2 = 4