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

Jul 25, 2017

See a solution process below:

#### Explanation:

We need to convert the equation to standard form by first expanding the terms in parenthesis and then grouping and combining like items. We can use this rule to expand the term in parenthesis:

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

Substituting $x$ for $a$ and $3$ for $b$ gives:

$y = {x}^{2} - 3 x - {\left(\textcolor{red}{x} + \textcolor{b l u e}{3}\right)}^{2}$

$y = {x}^{2} - 3 x - \left({\textcolor{red}{x}}^{2} + \left[2 \cdot \textcolor{red}{x} \cdot \textcolor{b l u e}{3}\right] + {\textcolor{b l u e}{3}}^{2}\right)$

$y = {x}^{2} - 3 x - \left({x}^{2} + 6 x + 9\right)$

$y = {x}^{2} - 3 x - {x}^{2} - 6 x - 9$

$y = {x}^{2} - {x}^{2} - 3 x - 6 x - 9$

$y = 1 {x}^{2} - 1 {x}^{2} - 3 x - 6 x - 9$

$y = \left(1 - 1\right) {x}^{2} + \left(- 3 - 6\right) x - 9$

$y = 0 {x}^{2} + \left(- 9\right) x - 9$

$y = 0 {x}^{2} - 9 x - 9$

We can now use the quadratic formula to solve the problem. The quadratic formula states:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

However, because $2 \cdot \textcolor{red}{a} = 2 \cdot \textcolor{red}{0} = 0$ we cannot use the quadratic because we cannot divide by $0$.