How do you solve # |x – 4| > |3x – 1|#?

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Answer 1 · 2017-11-23T20:20:16

Given:

#|x – 4| > |3x – 1|#

Assuming that #x in RR#, the following is an alternate form:

#sqrt((x – 4)^2) > sqrt((3x – 1)^2)#

Asserting the same inequality on the squares within the radicals:

#(x – 4)^2 > (3x – 1)^2#

Expand the squares:

#x^2 – 8x+16 > 9x^2 – 6x+1#

Combine like terms:

#-8x^2-2x+15>0#

When we multiply both sides by -1, we must change the direction of the inequality:

#8x^2+2x-15<0#

We know that the above quadratic will be less than 0 between the roots, therefore, we shall find the roots:

#x = (-2+-sqrt(2^2-4(8)(-15)))/(2(8))#

#x = (-2+-sqrt(484))/16#

#x = (-2+-22)/16#

#x = -24/16# and #x = 20/16#

#x = -3/2# and #x = 5/4#

The inequality is true between these numbers:

#-3/2 < x < 5/4#