Question

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

Answer 1

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`