How do you solve the inequality #abs(x-3)-abs(2x+1)

Question

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

Genesis Allen · Accepted Answer

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|#

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

Ethan Adkins · Answer

undefined To solve the inequality |x - 3| - |2x + 1| < 0, first, consider the critical points where the expressions inside the absolute value signs change sign. These critical points are x = -1 and x = 3. Then, test the intervals between these critical points by choosing test points within each interval and checking if the inequality holds true. The solution to the inequality in interval notation is (-1, 3).

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

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