How do you solve #|x - 5| = | 4x + 1|#?

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Answer 1 · 2017-08-27T14:20:33

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

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

Square the binomials:

#sqrt(x^2-10x + 25)=sqrt(16x^2+8x+1)#

Square both sides:

#x^2-10x + 25=16x^2+8x+1#

Combine like terms:

#0 = 15x^2+18x-24#

Divide both sides by 3:

#0 = 5x^2+6x-8#

Factor:

#(5x-4)(x+2)=0#

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

This method is validated by graphing

#y = |x - 5| and y = | 4x + 1|#

and observing the x values at the intersection points:

graph{(|x - 5|-y)(| 4x + 1|-y)=0 [-9.67, 10.33, -1.16, 8.84]}