How do you solve the inequality #abs(x-3)-abs(2x+1)<0# and write your answer in interval notation?

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Answer 1 · 2017-09-13T14:24:14

Assuming x is real, use the alternate form:

#sqrt((x-3)^2)-sqrt((2x+1)^2)< 0#

This implies that:

#(x-3)^2 < (2x+1)^2#

Expand the squares and solve the resulting quadratic inequality.

Starting with the above:

#(x-3)^2 < (2x+1)^2#

Expand the squares

#x^2 - 2x +9 < 4x^2 + 4x + 1#

Combine like terms:

#0 < 3x^2+ 6x - 8#

The quadratic will be greater than 0 to the left and to the right of the roots, therefore, we should find the roots:

#0 = 3x^2+ 6x - 8#

Factor:

#0 = (3x - 2)(x+4)#

#x = 2/3# and #x = -4#

The intervals to the left and to the right of these roots are:

#(-oo, -4)# and #(2/3, oo)#

Here is a graph of #y = |x - 3|- |2x+1|#

www.desmos.com/calculator

Please observe that it drops below the #y = 0# line at #x = -4# and #x = 2/3#