# How do you find the roots, real and imaginary, of y=-(x -4 )^2-12x+16 using the quadratic formula?

Dec 21, 2017

The two real roots are:

${x}_{1} \setminus \approx 18.24621 \setminus \quad , \quad {x}_{2} \setminus \approx 1.75379$

#### Explanation:

First let’s simplify the first parenthetical term, $- {\left(x - 4\right)}^{2}$, by distributing the negative sign and FOILing it.

$- {\left(x - 4\right)}^{2}$

$\textcolor{m a \ge n t a}{\setminus \implies} \left(- x + 4\right) \left(- x + 4\right)$

$\textcolor{m a \ge n t a}{\setminus \implies} {x}^{2} - 4 x - 4 x + 16$

$\textcolor{m a \ge n t a}{\setminus \implies} {x}^{2} - 8 x + 16$

Add that to the rest of the original expression:

$\textcolor{red}{{x}^{2}} - \textcolor{b l u e}{8 x} + \textcolor{g r e e n}{16} - \textcolor{b l u e}{12 x} + \textcolor{g r e e n}{16}$

$\textcolor{m a \ge n t a}{\setminus \implies} \textcolor{red}{{x}^{2}} - \textcolor{b l u e}{20 x} + \textcolor{g r e e n}{32}$

This is in standard form, $a {x}^{2} + b x + c$, so we can solve using the quadratic formula. Here,

• $a = 1$

• $b = 20$

• $c = 32$

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{m a \ge n t a}{\setminus \implies} x = \setminus \frac{20 \setminus \pm \setminus \sqrt{{\left(- 20\right)}^{2} - 4 \left(1\right) \left(32\right)}}{2 \left(10\right)}$

$\textcolor{m a \ge n t a}{\setminus \implies} x = \setminus \frac{20 \setminus \pm \setminus \sqrt{400 - 128}}{2}$

$\textcolor{m a \ge n t a}{\setminus \implies} x = \setminus \frac{20 \setminus \pm \setminus \sqrt{272}}{2}$

$\textcolor{m a \ge n t a}{\setminus \implies} x = \setminus \frac{20 \setminus \pm 4 \setminus \sqrt{17}}{2}$

$\textcolor{m a \ge n t a}{\setminus \implies} x = 10 \setminus \pm 2 \setminus \sqrt{17}$

$\textcolor{m a \ge n t a}{\setminus \therefore} {x}_{1} \setminus \approx 18.24621 \setminus \quad , \setminus \quad {x}_{2} \setminus \approx 1.75379$