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.

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.