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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