# How do you find the roots, real and imaginary, of y=3x^2 +4x+(x -4)^2  using the quadratic formula?

Dec 15, 2015

Convert the given expression into standard form and then apply the quadratic formula.
This should give you the (imaginary) roots $\left(\frac{1}{2} + i \frac{\sqrt{15}}{2}\right)$ and $\left(\frac{1}{2} - i \frac{\sqrt{15}}{2}\right)$

#### Explanation:

The quadratic formula says that for a quadratic equation in the form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c}$
the roots are given by
color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4(color(red)(a)color(green)(c))))/(2(color(red)(a))

To apply this to the given equation:
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} + 4 x + {\left(x - 4\right)}^{2}$
we must first convert the equation into standard form by expanding

$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} + 4 x + \left({x}^{2} - 8 x + 16\right)$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{red}{4} {x}^{2} \textcolor{b l u e}{- 4} x + \textcolor{g r e e n}{16}$

We can now apply the quadratic formula:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{4} \pm \sqrt{{\left(\textcolor{b l u e}{- 4}\right)}^{2} - 4 \left(\textcolor{red}{4}\right) \left(\textcolor{g r e e n}{16}\right)}}{2 \left(\textcolor{red}{4}\right)}$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{- \cancel{4} \pm \cancel{4} \sqrt{- 15}}{2 \times \cancel{4}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{1}{2} \pm \frac{i \sqrt{15}}{2}$