<PRE><B>Question 609979</B>
solve the root of x minus the root x minus 5=1
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sqrt(x) - sqrt(x-5) = 1
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Square both sides to get:
x - 2sqrt(x(x-5))+ (x-5) = 1
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2x - 6 = 2sqrt(x^2-5x)
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Divide thru by 2:
x - 3 = sqrt(x^2-5x)
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Square both sides to get:
x^2-5x = x^2-6x+9
x = 9
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Check for extraneous root:
solve the root of x minus the root x minus 5=1
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sqrt(9) - sqrt(9-5) = 1
3 - 2 = 1
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Solution: x = 9
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Cheers,
Stan H.</PRE>