How do you solve #root3(4x-2) = 2root3(x+6)#?
1 Answer
Jul 30, 2016
Explanation:
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
#4x-2 = 8x+48#
Now isolate the variable by getting all
#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#