Question

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

Answer 1

`x in (-12, 4)`

Explanation:

Put on a common denominator.

`x^2 - 48 < -8x`

`x^2 + 8x - 48 < 0`

Solve as an equation:

`x^2 + 8x - 48 = 0`

`(x + 12)(x - 4) = 0`

`x = -12 or 4`

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

`1 - 48/1 <^? -8`

`1 - 48 < -8`

Therefore, the solution is `x in (-12, 4)`. 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!

Answer 2

The solution set is `(-oo,-12)uu(0,4)`.

Explanation:

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

Make one side `0`.

`x-48/x+8 < 0`

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

`x-48/x+8 = (x^2+8x-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 `(-oo,-12)uu(0,4)`