How do you solve # |6x + 3| = 6#?

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Answer 1 · 2018-03-12T21:43:05

#x = -3/2,1/2#

One theorem states #absx=sqrt(x^2)#

So

#abs(6x+3)=sqrt((6x+3)^2)=6#

#rArr(6x+3)^2=36#

#rArr 36x^2+36x+9=36#

#rArr 4x^2+4x-3=0#

#rArr (2x-1)(2x+3)=0#

#rArr x=1/2,-3/2#