SOLUTION: What is the extraneous solution of sqrt(x-3)= (x-5) also sqrt of(x+4)=sqrt(3x)

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First we find solutions. Then we determine which, if any, are extraneous.

Square both sides:

Squaring the left side is as easy as it looks. Squaring an expression like the right side is where a lot of mistakes are made. It is not x%5E2-5%5E2

!! To square a two-term expression like (x-5) correctly we must use FOIL on (x-5)(x-5) or use the %28a-b%29%5E2+=+a%5E2-2ab%2Bb%5E2

pattern. I prefer using the pattern:
x-3=+%28x%29%5E2-2%28x%29%285%29%2B5%5E2

which simplifies to:
x-3=+x%5E2-10x%2B25

Now that x is out of the square root we can solve for it. This is a quadratic equation so we want one side to be zero. Subtracting x and adding 3 we get:
0+=+x%5E2-11x%2B28

Next we factor (or use the Quadratic Formula). This factors fairly easily:
0+=+%28x-7%29%28x-4%29

From the Zero Product Property:
x-7 = 0 or x-4 = 0
Solving these we get:
x = 7 or x = 4

Now we check. Use the original equation:
sqrt%28x-3%29=+%28x-5%29

Checking x = 7:
sqrt%287-3%29=+%287-5%29

Check!
Checking x = 4:
sqrt%284-3%29=+%284-5%29

Check fails!
So x = 7 is actually a solution and x = 4 is extraneous.

sqrt%28x%2B4%29=sqrt%283x%29

Square both sides:
%28sqrt%28x%2B4%29%29%5E2=%28sqrt%283x%29%29%5E2

x+4 = 3x
Subtract x:
4 = 2x
Divide by 2:
2 = x
Check.
sqrt%28x%2B4%29=sqrt%283x%29

Checking x = 2:
sqrt%282%2B4%29=sqrt%283%282%29%29

Check!
x = 2 is the solution to this equation. There are no extraneous solutions.