# How do you solve x^2-3x-32>0?

Jan 16, 2017

$x > \frac{3 + \sqrt{137}}{2} \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x < \frac{3 - \sqrt{137}}{2}$

#### Explanation:

If we use the quadratic formula to factor $\textcolor{red}{1} {x}^{2} \textcolor{b l u e}{- 3} x \textcolor{g r e e n}{- 32}$
we find zeros at
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \left(\textcolor{b l u e}{- 3}\right) \pm \sqrt{{\left(\textcolor{b l u e}{- 3}\right)}^{2} - 4 \cdot \textcolor{red}{1} \cdot \left(\textcolor{g r e e n}{- 32}\right)}}{2 \cdot \textcolor{red}{1}}$
$\textcolor{w h i t e}{\text{XXXX}} = \frac{3 \pm \sqrt{137}}{2}$
giving the factors:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 3 x - 32 = \left(x - \frac{3 + \sqrt{137}}{2}\right) \cdot \left(x - \frac{3 - \sqrt{137}}{2}\right)$

${x}^{2} - 3 x - 32$ will be greater than zero if and only if
{: ("both terms are positive","or","both terms are negative"), (x>(3+sqrt(137))/2,,x < (3-sqrt(137))/2) :}

If you like you could use a calculator to evaluate these irrational expressions:
$x > 7.352349955 \textcolor{w h i t e}{\text{XX")orcolor(white)("XX}} x < - 4.35234995$

We could verify that these results are reasonable by examining the graph of ${x}^{2} - 3 x - 32$ and noting the values of $x$ for which the graph ($y$) is above the X-axis: