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

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!

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)#