# How do you find the roots, real and imaginary, of y= (x-3)^2+2x-5  using the quadratic formula?

Mar 27, 2017

In order to use the quadratic equation, we must first get our equation to be in standard form. To do that, we need to simplify everything and then combine like-terms.

$y = {\left(x - 3\right)}^{2} + 2 x - 5$ becomes $y = \left(x - 3\right) \left(x - 3\right) + 2 x - 5$. This becomes ${x}^{2} - 3 x - 3 x + 9 + 2 x - 5$ or ${x}^{2} - 6 x + 9 + 2 x - 5$. That simplifies to ${x}^{2} - 4 x + 4$, which gives us our standard form .

Now we just use the quadratic formula: $\frac{- b}{2 \cdot a} \pm \frac{\sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$

${\textcolor{g r e e n}{x}}^{2} - \textcolor{b l u e}{4} x + \textcolor{red}{4}$
$\textcolor{g r e e n}{1 = a}$
$\textcolor{b l u e}{4 = b}$
$\textcolor{red}{4 = c}$

$x = \frac{- \left(\textcolor{b l u e}{4}\right) \pm \left(\sqrt{{\left(\textcolor{b l u e}{4}\right)}^{2} - 4 \cdot \left(\textcolor{g r e e n}{1}\right) \cdot \left(\textcolor{red}{4}\right)}\right)}{2 \cdot \textcolor{g r e e n}{1}}$

$x = \frac{- 4 \pm \sqrt{16 - 16}}{2}$

$x = \frac{- 4}{2}$

$x = - 2$

This seems to be the only root, but let's double check by graphing
graph{y=(x-3)^2+2x-5}

Yep, our only root is at $x = 2$! Good job!