# How do you solve and write the following in interval notation: x - 48/x < -8?

Aug 15, 2017

$x \in \left(- 12 , 4\right)$

#### Explanation:

Put on a common denominator.

${x}^{2} - 48 < - 8 x$

${x}^{2} + 8 x - 48 < 0$

Solve as an equation:

${x}^{2} + 8 x - 48 = 0$

$\left(x + 12\right) \left(x - 4\right) = 0$

$x = - 12 \mathmr{and} 4$

Now select a test point, let it be $x = 1$.

1 - 48/1 <^? -8

$1 - 48 < - 8$

Therefore, the solution is $x \in \left(- 12 , 4\right)$. Note the circular brackets instead of the square brackets. This is because the points $- 12$ and $4$ are not included in the solution;.

Hopefully this helps!

Aug 15, 2017

The solution set is $\left(- \infty , - 12\right) \cup \left(0 , 4\right)$.

#### Explanation:

We'll solve by a sign table (sign chart, sign analysis)

Make one side $0$.

$x - \frac{48}{x} + 8 < 0$

Find the values of $x$ for which the expression is undefined or zero.

$x - \frac{48}{x} + 8 = \frac{{x}^{2} + 8 x - 48}{x} =$

The important point (the partition numbers) are $- 12$, $0$, and $4$.

{: (bb"Interval:",(-oo,-12),(-12,0),(0,4),(4,oo)), (darrbb"Factors"darr,"========","======","=====","======"), (x+12, bb" -",bb" +",bb" +",bb" +"), (x-4,bb" -",bb" -",bb" -",bb" +"), (x,bb" -",bb" -",bb" +",bb" +"), ("==========","========","======","=====","======"), (((x+12)(x-4))/x,bb" -",bb" +",bb" -",bb" +") :}

So the solution set is $\left(- \infty , - 12\right) \cup \left(0 , 4\right)$