# How do you find the roots, real and imaginary, of y= -2x^2 - 8x +16 - (x-5)^2  using the quadratic formula?

Nov 9, 2017

See a solution process below;

#### Explanation:

First, we need to expand the term in parenthesis on the right side of the equation using this rule:

${\left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right)}^{2} = \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) = {\textcolor{red}{x}}^{2} - 2 \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{y}}^{2}$

Substituting $x$ for $x$ and $5$ for $y$ gives:

$y = - 2 {x}^{2} - 8 x + 16 - \left({x}^{2} - 10 x + 25\right)$

$y = - 2 {x}^{2} - 8 x + 16 - {x}^{2} + 10 x - 25$

We can next group and combine like terms:

$y = - 2 {x}^{2} - {x}^{2} - 8 x + 10 x + 16 - 25$

$y = - 2 {x}^{2} - 1 {x}^{2} - 8 x + 10 x + 16 - 25$

$y = \left(- 2 - 1\right) {x}^{2} + \left(- 8 + 10\right) x + \left(16 - 25\right)$

$y = - 3 {x}^{2} + 2 x + \left(- 9\right)$

$y = - 3 {x}^{2} + 2 x - 9$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{- 3}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{2}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 9}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{2} \pm \sqrt{{\textcolor{b l u e}{2}}^{2} - \left(4 \cdot \textcolor{red}{- 3} \cdot \textcolor{g r e e n}{- 9}\right)}}{2 \cdot \textcolor{red}{- 3}}$

$x = \frac{- \textcolor{b l u e}{2} \pm \sqrt{4 - 108}}{- 6}$

$x = \frac{- \textcolor{b l u e}{2} \pm \sqrt{- 104}}{- 6}$

$x = \frac{- \textcolor{b l u e}{2} \pm \sqrt{4 \cdot - 26}}{- 6}$

$x = \frac{- \textcolor{b l u e}{2} \pm \sqrt{4} \sqrt{- 26}}{- 6}$

$x = \frac{- \textcolor{b l u e}{2} \pm 2 \sqrt{- 26}}{- 6}$

$x = \frac{- 1 \pm \sqrt{- 26}}{- 3}$

Or

$x = \frac{1 \pm \sqrt{- 26}}{3}$