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

Aug 5, 2016

The roots are $x = \frac{1}{2}$ and $x = \frac{1}{3}$

#### Explanation:

We begin with $y = 2 {x}^{2} - x + {\left(2 x - 1\right)}^{2}$. Our first step is to simplify our equation until it cannot be reduced any further. Then we move on to the quadratic formula.

$y = 2 {x}^{2} - x + {\left(2 x - 1\right)}^{2}$ can be changed to $y = 2 {x}^{2} - x + \left(\left(2 x - 1\right) \cdot \left(2 x + 1\right)\right)$. That becomes $y = 2 {x}^{2} - x + 4 {x}^{2} - 2 x - 2 x + 1$ or $y = 2 {x}^{2} - x + 4 {x}^{2} - 4 x + 1$. Now we combine like terms like $2 {x}^{2} + 4 {x}^{2}$ and $- x - 5 x$. That gives us $y = 6 {x}^{2} - 5 x + 1$.

Now we can move on to the quadratic equation, which is $\frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 a}$ where those variables come from $a {x}^{2} + b x + c$. In our case $a = 6$, $b = - 5$, and $c = 1$. Now we just plug in our value to the formulla: $\frac{5 \pm \sqrt{- {5}^{2} - 4 \cdot 6 \cdot 1}}{2 \cdot 6}$. This can be simplified to $\frac{5 \pm \sqrt{25 - 24}}{12}$ or $\frac{5 \pm 1}{12}$. That becomes $\frac{4}{12}$ or $\frac{6}{12}$, which can be sipmlified to $\frac{1}{3}$ and $\frac{1}{2}$, respectivley. Those are our roots, and we can graph the original equation and look at the $x$-intercepts (roots).

graph{y=6x^2-5x+1}

And they are! Nice work!