SOLUTION: √(x)+√(x-7)=7 I solved this equation in two different ways and I got two different answers but they both work when I check them? (X=28 and x=16) I did: √(x)+&#8

Algebra ->  Radicals -> SOLUTION: √(x)+√(x-7)=7 I solved this equation in two different ways and I got two different answers but they both work when I check them? (X=28 and x=16) I did: √(x)+&#8      Log On


   

Question 675861: √(x)+√(x-7)=7 I solved this equation in two different ways and I got two different answers but they both work when I check them? (X=28 and x=16)
I did:
√(x)+√(x-7)=7 (then I squared everything)
x+x-7=49
2x=56 x=28
Then I did:
√(x)+√(x-7)=7
√(x-7)=7-√(x) (moved the √(x) to the other side)
x-7=49-14√(x)+x (squared both sides)
-56=-14√(x)
56/14=√(x)
√(x)=4
x=16

Answer by ReadingBoosters(3246) About Me  (Show Source):
You can put this solution on YOUR website!
28 doesn't work
sqrt%2828%29+%2B+sqrt%2828-7%29+=+5.29+%2B+4.58+=+9.87%3C%3E7
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16 does, however, work
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sqrt%2816%29+%2B+sqrt%2816-7%29+=+4+%2B+3+=+7
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Here are the looong, somewhat convoluted steps, which should have been taken when you arrived at 28.
Obviously, your second solution was much easier to achive than this...
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sqrt%28x%29+%2B+sqrt%28x-7%29+=+7
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%28sqrt%28x%29%2Bsqrt%28x-7%29%29%5E2+=+7%5E2
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%28sqrt%28x%29%2Bsqrt%28x-7%29%29%28sqrt%28x%29%2Bsqrt%28x-7%29%29+=+49
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x+%2B+2sqrt%28x%28x-7%29%29+%2B+x-7+=+49
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x+%2B+2sqrt%28x%5E2-7x%29+%2B+x-7+=+49
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2x+%2B+2sqrt%28x%5E2-7x%29+=+56
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2sqrt%28x%5E2-7x%29+=+56+-+2x
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sqrt%28x%5E2-7x%29+=+28+-+x
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sqrt%28x%5E2-7x%29%5E2+=+%2828-x%29%5E2
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x%5E2-7x+=+784+-+56x+%2B+x%5E2
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-7x + 56x = 784
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49x = 784
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x = 16