How do you solve and check for extraneous solutions in $sqrt(x^2 - 5) = 4$ ?

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How do you solve and check for extraneous solutions in $sqrt(x^2 - 5) = 4$ ?

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Answer 1 · 2015-08-01T17:30:24

$color(red)(x = sqrt21)$ and $color(red)(x=-sqrt21)$ are solutions.
There are $color(red)("no")$ extraneous solutions.

Explanation:

SOLVE:

$sqrt(x^2-5) = 4$

Square each side.

$x^2-5=16$

Add 5 to each side.

$x^2 = 21$

Take the square root of each side.

$x = ±sqrt21$

$x=-sqrt21$ and $x=sqrt21$

CHECK FOR EXTRANEOUS SOLUTIONS

$sqrt(x^2-5) = 4$

If $x=sqrt21$ ,

$sqrt((sqrt21)^2-5) = 4$

$sqrt(21-5) = 4$

$sqrt16 = 4$

$4=4$

$x=sqrt21$ is a solution.

If $x=-sqrt21$ ,

$sqrt((-sqrt21)^2-5) = 4$

$sqrt(21-5)=4$

$sqrt16=4$

$4 = 4$

$x=-sqrt21$ is a solution.

There are no extraneous solutions.

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How do you solve and check for extraneous solutions in $sqrt(x^2 - 5) = 4$ ?

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How do you solve and check for extraneous solutions in $sqrt(x^2 - 5) = 4$ ?