# How do you find all zeros of the function x² + 24 = –11x?

Jun 4, 2016

$x = - 3 \textcolor{w h i t e}{\text{XXX")andcolor(white)("XXX}} x = - 8$

#### Explanation:

Re-writing the given equation as
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 11 x + 24 = 0$
and remembering that
$\textcolor{w h i t e}{\text{XXX}} \left(x + a\right) \left(x + b\right) = {x}^{2} + \left(a + b\right) x + a b$

We are looking for two values, $a$ and $b$ such that
$\textcolor{w h i t e}{\text{XXX}} a + b = 11$ and
$\textcolor{w h i t e}{\text{XXX}} a b = 24$

with a bit of thought we come up with the pair $3$ and $8$

So we can factor:
$\textcolor{w h i t e}{\text{XXX}} \left(x + 3\right) \left(x + 8\right) = 0$

which implies either $x = - 3$ or $x = - 8$

Jun 4, 2016

x=-8 or x=-3

#### Explanation:

First you get the equivalent equation
${x}^{2} + 11 x + 24 = 0$
then you solve
$x = - \frac{11}{2} \pm \frac{\sqrt{{11}^{2} - 4 \left(24\right)}}{2}$
$x = - \frac{11}{2} \pm \frac{\sqrt{25}}{2}$
$x = - \frac{11}{2} \pm \frac{5}{2}$
so x=-8 or x=-3