How do you solve #root3(4x-2) = 2root3(x+6)#?

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Answer 1 · 2016-07-30T23:54:37

#x=-25/2#

First, we need to get rid of the cube roots. We can do this by cubing, or taking each side to the third power.

#root3(4x-2) = 2root3(x+6)#

#(root3(4x-2))^3 = (2root3(x+6))^3#

Now simplify.

#4x-2 = (2^3)((root3(x+6))^3)#

#4x-2 = 8(x+6)#

Distribute the #8# across the parenthesis on the right side of the equal sign.

#4x-2 = 8x+48#

Now isolate the variable by getting all #x# terms to one side. Subtract #8x# from both sides and add #2# to both sides.

#4x-8x cancel(-2+2)=cancel(8x-8x)+48+2#

#4x-8x=48+2#

Combine like terms and finish isolating the variable.

#-4x=50#

#x=- 50/4#

Simplify the fraction.

#x = -25/2#