SOLUTION: Solve 2-x=square root of x+4. I squared both sides to give me: (2-x)^2=x=4 distribute and solve to get 0 on right side x^2-5x=0 hence no solution. Wondering if I am r

Algebra ->  Radicals -> SOLUTION: Solve 2-x=square root of x+4. I squared both sides to give me: (2-x)^2=x=4 distribute and solve to get 0 on right side x^2-5x=0 hence no solution. Wondering if I am r      Log On


   

Question 222305: Solve 2-x=square root of x+4.
I squared both sides to give me:
(2-x)^2=x=4
distribute and solve to get 0 on right side
x^2-5x=0 hence no solution.
Wondering if I am right.
Thank you
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Solve 2-x+=+sqrt%28x%2B4%29
I squared both sides to give me:
%282-x%29%5E2+=+x%2B4
distribute
>Not the right word to use. Exponents do not distribute. But since your following equation is correct I have to assume that you did the right thing even though you describe it incorrectly.
and solve to get 0 on right side
x%5E2-5x=0 hence no solution.
> right equation, wrong conclusion. Why would you think this means no solution?

Now that we have a simplified quadratic equation with one side zero, we solve it by factoring or by using the quadratic formula. This one factors very easily since it has Greatest Common Factor (GCF) of x:
x%28x-5%29+=+0
Now we use the Zero Product Property (a product can be zero only if one of the factors is zero). So
x+=+0 or x-5+=+0
Add both 5 to both sides of the second equation:
x+=+0 or x+=+5
Usually checking our answer(s) is an optional, good idea. But when you square both sides of an equation like we did at the beginning, solutions called "extraneous" solutions may appear. So we must check answers, using the original equation, after we square both sides of an equation.

Checking 0:
2-%280%29+=+sqrt%28%280%29%2B4%29
2+=+sqrt%284%29
2+=+2 Check!

Checking 5:
2-%285%29+=+sqrt%28%285%29%2B4%29
-3+=+sqrt%289%29
-3+=+3 NO!

We have to reject x=5 as an "extraneous" solution. So the only true solution is x=0
Thank you