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

May 1, 2017

$x = - 1$

$x = 4$

#### Explanation:

You don't need a quadratic formula for this, since it is in the factored form, you just need to set each expression in the brackets equal to zero, and then find the roots.

$x + 1 = 0$
$\textcolor{red}{x = - 1}$

$x - 4 = 0$
$\textcolor{red}{x = 4}$

May 1, 2017

$x = 4 \mathmr{and} - 1$

#### Explanation:

As we are instructed to use the formula it is perhaps better to change the given equation into the format of $y = a {x}^{2} + b x + c$

Not only that, I suspect that part of the agenda of this question is to see if you can 'handle' expansion of brackets.

Multiplying everything inside the right brackets by everything in the left.

$y = 2 \textcolor{red}{\left(x + 1\right)} \textcolor{g r e e n}{\left(x - 4\right)}$

$y = 2 \textcolor{g r e e n}{\left[\textcolor{w h i t e}{\frac{2}{2}} \textcolor{red}{x} \left(x - 4\right) \textcolor{red}{+ 1} \left(x - 4\right) \text{ }\right]}$

$y = 2 \left({x}^{2} - 4 x + x - 4\right) \text{ "->" } y = 2 {x}^{2} - 6 x - 8$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus we have;$\text{ "a=2"; "b=-6"; } c = - 8$

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

$x = \frac{+ 6 \pm \sqrt{{\left(- 6\right)}^{2} - 4 \left(2\right) \left(- 8\right)}}{2 \left(2\right)}$

$x = \frac{+ 6 \pm \sqrt{36 + 64}}{4}$

As the content of the root (determinate) is not negative there are no imaginary roots. That is, in this case, the graph crosses the axis.

Note: if the determinate is 0 then the x-axis is tangential to the max/min

$x = \frac{3}{2} \pm \frac{5}{2}$

$x = - \frac{2}{2} \to - 1$
$x = \frac{8}{2} \to 4$