#abs(2x-1)=4x +5# has how many numbers in its solution set?

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Answer 1 · 2017-10-04T01:54:01

There is just 1 number in its solution set, #x = -2/3#

Given: #|2x-1|=4x +5#

An alternate form is:

#sqrt((2x-1)^2) = 4x+5#

Remove the radical by square both sides:

#(2x-1)^2 = (4x+5)^2#

Expand both squares:

#4x^2 - 4x + 1 = 16x^2 + 40x + 25#

Combine like terms:

#12x^2 + 44x + 24 = 0#

Divide by 4:

#3x^2 + 11x + 6 = 0#

Factor:

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

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

Check #x = -2/3#:

#|2(-2/3)-1|=4(-2/3) +5#

#|-7/3| = 7/3#

This checks

Check #x = -3#:

#|2(-3)-1|=4(-3) +5#

This cannot possibly check