# How do you solve \frac { 1} { 2} x ^ { 2} - x > 4?

May 16, 2018

x in (–oo, –2)uu(4,oo).
or
SS={x in RR|x<–2 or x>4}.

#### Explanation:

Bring everything to one side.

$\frac{1}{2} {x}^{2} - x - 4 > 0$

Factor that side if possible.
Factor the $\frac{1}{2}$ out of all three terms.

$\frac{1}{2} \left({x}^{2} - 2 x - 8\right) > 0$

Recognize that
–4 xx 2 = 8, and
–4 + 2 = –2.

$\frac{1}{2} \left(x - 4\right) \left(x + 2\right) > 0$

Divide both sides by $\frac{1}{2}$:

$\left(x - 4\right) \left(x + 2\right) > 0$

We now have a product of two factors on the left: $x - 4$ and $x + 2$. We are interested in when this product is positive $\left(> 0\right) .$

Both factors depend on $x$. Therefore, their product will be positive when $x$ is small/large enough to make the factors either both negative or both positive.

The 1st factor, $x - 4$, is negative when $x < 4$.
The 2nd factor, $x + 2$, is negative when x<–2.

So both factors will be negative when x<–2.

Similarly, both factors will be positive when $x > 4$.

Our solution is all $x$ in both these regions: x<–2 uu x>4.