How do you solve and check for extraneous solutions in #root3(4x+2) -6 = -10#?

1 Answer
Gió
Aug 1, 2015

I found: #x=-33/2#

Explanation:

I would write it as:
#root3(4x+2)=-10+6#
#root3(4x+2)=-4# apply the power of #3# on both sides:
#4x+2=(-4)^3#
#4x+2=-64#
#4x=-66#
#x=-66/4=-33/2#

Checking the solution we plug it into the original equaton:
#root3(4(-33/2)+2)-6=-10#
#root3(-66+2)=-4#
#root3(-64)=-4# true,
considering that: #-4*-4*-4=-64#

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