Noshad wrote:
For how many real values of x is p \(\sqrt{120 − √ x}\) an integer?
(A) 3
(B) 6
(C) 9
(D) 10
(E) 11
Bunuelis the question p=\(\sqrt{120 − √ x}\). Or it is correct as it is.I feel there is missing "=" sign.
My approachFor \(\sqrt{120 − √ x}\) to be an integer. The 120-√x must be a perfect square. Less than 120
So the value of the expression can be 100,81,64,49,36,25,16,9,4,1,0 total 11 values. For each of which there will be a corresponding value of x
x will have values √400,\(√39^2\),\(√56^2\) ..... \(√120^2\) . 11 values .Hence ans (E)
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39% (01:51) correct 
