SOLUTION: x+7 with the root sign =x-5

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Question 241849: x+7 with the root sign =x-5
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%2B7%29+=+x-5
To solve problems like this we need to isolate the square root and square both sides. You square root is already all by itself on the left side so we can go straight to squaring both sides:
%28sqrt%28x%2B7%29%29%5E2+=+%28x-5%29%5E2
x+%2B+7+=+x%5E2+-10x+%2B+25
Now we have no square roots. We have a quadratic equation to solve. So we get one side equal to zero (by subtracting x and 7 from each side:
0+=+x%5E2+-11x+%2B+18
At this point we can either factor or use the Quadratic Formula. This factors pretty easily:
0+=+%28x-9%29%28x-2%29
According to the Zero Product Property, this product can be zero only if one (or more) of the factors is zero. So:
x-9+=+0 or x+-+2+=+0
Solving these we get:
x+=+9 or x+=+2

Whenever you square both sides of an equation like we did here, it is more than just a good idea check your answers. It is important. This is so because when you square both sides of an equation you might introduce what are called "extraneous solutions". Extraneous solutions are "solutions" don't actually work.

Always use the original equation when checking:
sqrt%28x%2B7%29+=+x-5
Checking x = 9:
sqrt%28%289%29%2B7%29+=+%289%29-5
sqrt%2816%29+=+4
4+=+4 Check!

Checking x = 2:
sqrt%28%282%29%2B7%29+=+%282%29-5
sqrt%289%29+=+-3
3+=+-3 This is not true!! So x=2 is an extraneous solution and must be rejected.

So there is just one solution: x = 9.