How do you solve $sqrt(x^2 - 5) = 4$ and find any extraneous solutions?

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Answer 1 · 2016-07-19T16:38:51

$x = +-sqrt(21)$

Square both sides:

$x^2 - 5 = 16$

$x^2 = 21$

$x = +-sqrt(21)$

How do you solve sqrt(x^2 - 5) = 4sqrt(x^2 - 5) = 4 and find any extraneous solutions?