How do you solve and check for extraneous solutions in $sqrt(4n) =2$ ?

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Answer 1 · 2015-08-01T17:38:29

$color(red)(n=1)$ is a solution.
There are $color(red)("no")$ extraneous solutions.

SOLVE:

$sqrt(4n) = 2$

Square each side.

$4n=4$

Divide each side by 4.

$n=1$

CHECK FOR EXTRANEOUS SOLUTIONS

$sqrt(4n) = 2$

If $n=1$ ,

$sqrt(4(1)) = 2$

$sqrt(4) = 2$

$2=2$

$n=1$ is a solution.

There are no extraneous solutions.

How do you solve and check for extraneous solutions in sqrt(4n) =2sqrt(4n) =2?