How do you solve and check for extraneous solutions in root3(4x+2) -6 = -103√4x+2−6=−10?

1 Answer
Gió
Aug 1, 2015

I found: x=-33/2x=−332

Explanation:

I would write it as:
root3(4x+2)=-10+63√4x+2=−10+6
root3(4x+2)=-43√4x+2=−4 apply the power of 33 on both sides:
4x+2=(-4)^34x+2=(−4)3
4x+2=-644x+2=−64
4x=-664x=−66
x=-66/4=-33/2x=−664=−332

Checking the solution we plug it into the original equaton:
root3(4(-33/2)+2)-6=-103√4(−332)+2−6=−10
root3(-66+2)=-43√−66+2=−4
root3(-64)=-43√−64=−4 true,
considering that: -4*-4*-4=-64−4⋅−4⋅−4=−64

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