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

May 13, 2016

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

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, $a {x}^{2} + b x + c = 0$, we know:

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ (the quadratic formula)

In this instance:

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

So:

$y = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \times \left(- 1\right) \times \left(- 32\right)}}{2 \times \left(- 1\right)}$

$= \frac{- 12 \pm \sqrt{144 - 128}}{-} 2 = \frac{- 12 \pm \sqrt{16}}{-} 2 = \frac{- 12 \pm 4}{-} 2$

$= \frac{- 8}{-} 2 = 4$ or $\frac{- 16}{-} 2 = 8$

So the roots are 4 and 8.

May 13, 2016

4 and 8

#### Explanation:

Use the improved quadratic formula (Socratic Search)
$y = - {x}^{2} + 12 x - 32 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 144 - 128 = 16$--> $d = \pm 4$
There are 2 real roots:
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = - \frac{12}{-} 2 \pm \frac{4}{-} 2 = 6 \pm 2$

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