Question 600693
1. SQR(x+20)=x
2.SQR(10x)-(2)SQR(5x-25)=0


Solve each of the following. Check for extraneous solutions.

What I know is;  1. x+20=x^2
That's as far as I know. I don't know how to check for extraneous solutions.
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x+20=x^2
{{{x^2 - x - 20 = 0}}}
(x-5)*(x+4) = 0
x = -4
x = 5
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Check each answer to see if it works.
They both work --> no extraneous solutions.
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2.SQR(10x)-(2)SQR(5x-25)=0
SQR(10x) = (2)SQR(5x-25)
Square both sides
10x = 4*(5x - 25)
etc