How do you solve #5(2x-5)^2=55# and find any extraneous solutions?

1 Answer
May 30, 2017

Answer:

#x = (5+-sqrt11)/2#, no extraneous solutions

Explanation:

First, solve this algebraically.

Divide by 5

#(2x-5)^2 = 11#

Now take the square root. Don't forget the #+-# sign.

#2x-5 = +-sqrt11#

Add 5

#2x = 5+-sqrt11#

And divide by 2

#x = (5+-sqrt11)/2#

So we have 2 solutions:

#x = (5+sqrt11)/2 and x=(5-sqrt11)/2#

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Now to test for extraneous solutions, plug both solutions back into the original problem and see if they work.

#5(2((5+sqrt11)/2)-5)^2=^?55#

#5((5+sqrt11)-5)^2=^?55#

#5sqrt11^2=^?55#

#55=^sqrt() 55#


#5(2((5-sqrt11)/2)-5)^2=^?55#

#5((5-sqrt11)-5)^2=^?55#

#5(-sqrt11)^2=^?55#

#55=^sqrt() 55#

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So both solutions work when plugged back in. There are no extraneous solutions.