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

May 30, 2017

$x = \frac{5 \pm \sqrt{11}}{2}$, no extraneous solutions

#### Explanation:

First, solve this algebraically.

Divide by 5

${\left(2 x - 5\right)}^{2} = 11$

Now take the square root. Don't forget the $\pm$ sign.

$2 x - 5 = \pm \sqrt{11}$

$2 x = 5 \pm \sqrt{11}$

And divide by 2

$x = \frac{5 \pm \sqrt{11}}{2}$

So we have 2 solutions:

$x = \frac{5 + \sqrt{11}}{2} \mathmr{and} x = \frac{5 - \sqrt{11}}{2}$

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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.