# How do you find the roots, real and imaginary, of y= -2(x-3)^2+(-x-2)^2  using the quadratic formula?

Dec 11, 2017

$x = 4 - \sqrt{2}$, $4 + \sqrt{2}$
Check below for my work!

#### Explanation:

You need to put it this in a form that's usable with the quadratic formula. Generally you want an equation in the form of $a {x}^{2} + b x + c$.

Let's work with the equation a bit. To start, we have two binomials to the power of two. We can expand these with multiplication.

$- 2 \textcolor{b l u e}{{\left(x - 3\right)}^{2}} + \textcolor{red}{{\left(- x - 2\right)}^{2}}$

Binomnial 1
$\textcolor{b l u e}{{\left(x - 3\right)}^{2}}$
$\textcolor{b l u e}{\left(x - 3\right) \left(x - 3\right)}$ Multiply these together using the FOIL method.
$\textcolor{b l u e}{{x}^{2} - 6 x + 9}$

Binomial 2
$\textcolor{red}{{\left(- x - 2\right)}^{2}}$
$\textcolor{red}{\left(- x - 2\right) \left(- x - 2\right)}$ Again, use the FOIL method.
$\textcolor{red}{{x}^{2} - 4 x + 4}$

Great! Now we're left with $- 2 \left({x}^{2} - 6 x + 9\right) + \left({x}^{2} - 4 x + 4\right)$.

Now we can multiply the first polynomial by $- 2$.

$- 2 \left({x}^{2} - 6 x + 9\right)$
$- 2 {x}^{2} + 12 x - 18$

That leaves us with $\left(- 2 {x}^{2} + 12 x - 18\right) + \left({x}^{2} - 4 x + 4\right)$

Now we simply combine like terms.

$\left(- 2 {x}^{2} + {x}^{2}\right) + \left(12 x - 4 x\right) + \left(- 18 + 4\right)$
$- {x}^{2} + 8 x - 14$

Pulling from our simplified polynomial, we can plug these values into the quadratic formula.

$a = - 1$
$b = 8$
$c = - 14$

Finally, we solve!

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

$x = \frac{- \left(8\right) \pm \sqrt{{\left(8\right)}^{2} - 4 \left(- 1\right) \left(- 14\right)}}{2 \left(- 1\right)}$

$x = \frac{- 8 \pm \sqrt{8}}{-} 2$

$x = \frac{- 8 \pm 2 \sqrt{2}}{-} 2$

$x = \frac{- 8 \pm 2 \sqrt{2}}{-} 2$

*I'm splitting the $\pm$ up here. The $+$ equation is in $\textcolor{b l u e}{\text{blue}}$ while the $-$ equation is in $\textcolor{red}{\text{red}}$.

$\textcolor{b l u e}{x = \frac{- 8 + 2 \sqrt{2}}{-} 2}$

$\textcolor{b l u e}{x = 4 - \sqrt{2}}$

$\textcolor{red}{x = \frac{- 8 - 2 \sqrt{2}}{-} 2}$

$\textcolor{red}{x = 4 + \sqrt{2}}$

Since these can't be simplified any further, these are the real roots. There are no imaginary roots.