Question

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

Answer 1

`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`

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`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.