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

Jul 24, 2017

See a solution process below:

#### Explanation:

First, expand the term squared on the right side of the equation using the rule:

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

Substituting $x$ for $a$ and $2$ for $b$ gives:

$y = - 2 {\left(\textcolor{red}{x} + \textcolor{b l u e}{2}\right)}^{2} + 16 {x}^{2} - x - 5$

$y = - 2 \left({\textcolor{red}{x}}^{2} + \left(2 \cdot \textcolor{red}{x} \cdot \textcolor{b l u e}{2}\right) + {\textcolor{b l u e}{2}}^{2}\right) + 16 {x}^{2} - x - 5$

$y = - 2 \left({x}^{2} + 4 x + 4\right) + 16 {x}^{2} - x - 5$

Now, expand the term in parenthesis, group and combine like items:

$y = \left(- 2 \cdot {x}^{2}\right) + \left(2 \cdot 4 x\right) + \left(2 \cdot 4\right) + 16 {x}^{2} - x - 5$

$y = - 2 {x}^{2} + 8 x + 8 + 16 {x}^{2} - x - 5$

$y = - 2 {x}^{2} + 16 {x}^{2} + 8 x - x + 8 - 5$

$y = \left(- 2 + 16\right) {x}^{2} + \left(8 - 1\right) x + \left(8 - 5\right)$

$y = 14 {x}^{2} + 7 x + 3$

We can now use the quadratic equation to solve this problem. The quadratic formula states:

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

$\textcolor{red}{14}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{7}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{3}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{7} \pm \sqrt{{\textcolor{b l u e}{7}}^{2} - \left(4 \cdot \textcolor{red}{14} \cdot \textcolor{g r e e n}{3}\right)}}{2 \cdot \textcolor{red}{14}}$

$x = \frac{- \textcolor{b l u e}{7} \pm \sqrt{49 - 168}}{28}$

$x = \frac{- \textcolor{b l u e}{7} \pm \sqrt{- 119}}{28}$

Or

$x = - \frac{\textcolor{b l u e}{7}}{28} \pm \frac{\sqrt{- 119}}{28}$

$x = - \frac{1}{4} \pm \frac{\sqrt{- 119}}{28}$