Solve for #x# in #|2x-4| >= -5#? Why doesn't the usual method work in this case?

At first I was blindly solving and got #-0.5<=x<=4.5#. However, the answer is obviously all reals because the absolute value of anything will be positive; therefore, #abs(2x-4)# must be greater than #-5#.

Why doesn't the usual method work in this case?

1 Answer
Nov 18, 2017

#|2x-4| >= -5#

Since all modulus values are bigger or equal to #0#,
#|2x-4| >= 0#

Square both sides which gets rid of the modulus function,
#4x^2-16x+16>=0#
#(x-2)^2>=0#

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#x>=2 or x<=2#

Hence, the solution is all real roots.

All absolute values must be equal or bigger to #0#, and hence, all values of #x# will work.

So, why doesn't the usual method work?
That is because we normally do this:

#|2x-4| >= -5#

Square both sides which gets rid of the modulus function,
#4x^2-16x+16>=25#
#4x^2-16x-9>=0#
#(2x-9)(2x+1)>=0#
#x<=-0.5# or #x>=4.5#

This is because we squared a negative number to make it positive, where in fact, is impossible as all absolute values are positive. Hence, the equation automatically implies that #25# is #5^2# instead of #(-5)^2#, resulting in the solution being (#x<=-0.5# or #x>=4.5#) instead of infinite number of solutions.