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.