How do you find the roots, real and imaginary, of y=-5x^2+7x+2(2x-1)^2  using the quadratic formula?

May 15, 2016

The roots are $\frac{1 \pm i \sqrt{23}}{6}$

Explanation:

We begin with $y = - 5 {x}^{2} + 7 x + 2 {\left(2 x - 1\right)}^{2}$. In order to use the quadratic formula, we need to get this equation into standard form, which means that the format needs to look like this: $a {x}^{2} + b x + c$.

So, in order to change the form of our current equation, we first need to expand everything that we can. Let's start with ${\left(2 x - 1\right)}^{2}$. This becomes $\left(2 x - 1\right) \left(2 x - 1\right)$, which is $4 {x}^{2} - 4 x + 1$. Now we have $- 5 {x}^{2} + 7 x + 2 \left(4 {x}^{2} - 4 x + 1\right)$/ By multiplying out the $2$, we arrive at $- 5 {x}^{2} + 7 x + 8 {x}^{2} - 8 x + 2$.

From here, we just need to combine like terms. $- 5 {x}^{2} + 8 {x}^{2}$ gives us $3 {x}^{2}$, while $7 x - 8 x$ equals $- x$. The $2$ is left alone.

So now we have the following: $y = 3 {x}^{2} - x + 2$, which is in standard form, meaning that we are good to go for using the quadratic formula!

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

$\textcolor{b l u e}{a = 3}$
$\textcolor{g r e e n}{b = - 1}$
$\textcolor{red}{c = 2}$

We just need to substitute the $a$s and $b$s and $c$s for their numerical value.

The formula now becomes

$\frac{- \left(\textcolor{g r e e n}{- 1}\right) \pm \sqrt{{\left(\textcolor{g r e e n}{- 1}\right)}^{2} - 4 \cdot \left(\textcolor{b l u e}{3}\right) \cdot \left(\textcolor{red}{2}\right)}}{2 \left(\textcolor{b l u e}{3}\right)}$

Simplifying that we have $\frac{1 \pm \sqrt{1 - 24}}{6}$ or $\frac{1 \pm i \sqrt{23}}{6}$.

That is the imaginary root for $y = - 5 {x}^{2} + 7 x + 2 {\left(2 x - 1\right)}^{2}$. Nice job.