SOLUTION: square root of x+40-(minus)sqrare root of x =4 square root of y+10-(minus) square root of y-2 =2 thanks!

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Question 78786: square root of x+40-(minus)sqrare root of x =4
square root of y+10-(minus) square root of y-2 =2
thanks!

Answer by mathdoc314(58) About Me  (Show Source):
You can put this solution on YOUR website!
This is two problems. Please ask the second problem separately or work it yourself after you see the technique.
+sqrt%28x%2B40%29+-+sqrt%28x%29+=+4+
To solve algebraically, it's necessary to get rid of the square roots in the expression. I tried to do this with tricks like: rewrite it with
+sqrt%28+%28x%2B20%29+%2B+20+%29+-+sqrt%28+%28x%2B20%29+-+20+%29+=+4+
but this did not seem to simplify very well. So I tried this instead and it worked.
Let a = +sqrt%28x%29+ and b = +sqrt%28x%2B40%29+
(I did this so a will be smaller than b)
Then solve for a and b like this
a^2 = x
b^2 = x + 40
therefore b^2 = a^2 + 40.
The problem stated +sqrt%28+%28x%2B20%29+%2B+20+%29+-+sqrt%28+%28x%2B20%29+-+20+%29+=+4+
which now is: b - a = 4.
b^2 = a^2 + 40
b - a = 4
Substitute in for a:
a = b - 4
b^2 = (b-4)^2 + 40
b^2 = (b^2 - 8 b + 16) + 40
The b^2 cancels out on both sides
0 = -8 b + 56
8b = 56
b = 7
a = b-4 = 7-4 = 3
Don't forget to find x
Evaluate it twice to show it's right:
+sqrt%28x%29+=+3+ --> +x+=+9+
+sqrt%28x-40%29+=+7 --> +x-40+=+49 --> +x=9++