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

Nov 9, 2017

See a solution process below:

#### Explanation:

First, we need to expand the term in parenthesis on the right side of the equation using this rule:

${\left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right)}^{2} = \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right) = {\textcolor{red}{x}}^{2} - 2 \textcolor{red}{x} \textcolor{b l u e}{y} + {\textcolor{b l u e}{y}}^{2}$

Substituting $x$ for $x$ and $1$ for $y$ gives:

$y = 5 {x}^{2} - x - \left({x}^{2} - 2 x + 1\right)$

$y = 5 {x}^{2} - x - {x}^{2} + 2 x - 1$

We can next group and combine like terms:

$y = 5 {x}^{2} - {x}^{2} - x + 2 x - 1$

$y = 5 {x}^{2} - 1 {x}^{2} - 1 x + 2 x - 1$

$y = \left(5 - 1\right) {x}^{2} + \left(- 1 + 2\right) x - 1$

$y = 4 {x}^{2} + 1 x - 1$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

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

Substituting:

$\textcolor{red}{4}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 1}$ for $\textcolor{g r e e n}{c}$ gives:

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

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 - \left(- 16\right)}}{8}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 + 16}}{8}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{17}}{8}$