How do you solve \sqrt { 4x ^ { 2} - 4x - 5} = 2x - 2?

1 Answer
Cri007
Apr 7, 2017

See below.

Explanation:

We start by squaring both sides (because no one likes ugly radicals)

4x^2-4x-5=4x^2-8x+4

Aha! The squared terms cancel!

This simplifies to:

-4x-5=-8x+4

12x=9

x=3/4

We have to be careful now. The right side of this equation cannot be negative, or else the radical does not result in a real solution (extraneous).
We need 2x-2\geq0, or x\geq1, so there are no real solutions to this problem.

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