# How do you find the roots, real and imaginary, of y= 2 x(x - 4) -(2x-1)^2  using the quadratic formula?

Feb 2, 2016

We have two real roots: $x = - 1 \pm \frac{\sqrt{2}}{2}$.

#### Explanation:

First of all, let's expand and simplify the expression.

$y = 2 x \left(x - 4\right) - {\left(2 x - 1\right)}^{2}$

$\textcolor{w h i t e}{x} = 2 {x}^{2} - 8 x - \left(4 {x}^{2} - 4 x + 1\right)$

... use the formula ${\left(m - n\right)}^{2} = {m}^{2} - 2 m n + {n}^{2}$ ....

$\textcolor{w h i t e}{x} = 2 {x}^{2} - 8 x - 4 {x}^{2} + 4 x - 1$

$\textcolor{w h i t e}{x} = - 2 {x}^{2} - 4 x - 1$

The formula is

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

In our case, $a = - 2$, $b = - 4$ and $c = - 1$. So, we have

$x = \frac{4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \cdot \left(- 2\right) \cdot \left(- 1\right)}}{2 \cdot \left(- 2\right)} = \frac{4 \pm \sqrt{16 - 8}}{- 4} = \frac{4 \pm \sqrt{8}}{- 4} = \frac{4 \pm 2 \sqrt{2}}{- 4} = - 1 \pm \frac{\sqrt{2}}{2}$

So, we have two real solutions:

${x}_{1} = - 1 + \frac{\sqrt{2}}{2} \approx - 0.29$

and

${x}_{2} = - 1 - \frac{\sqrt{2}}{2} \approx - 1.71$