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Noshad
For how many real values of x is \(\sqrt{120 − √ x}\) an integer?

(A) 3

(B) 6

(C) 9

(D) 10

(E) 11


total square values upto 120 ; 11 viz. 0,1,4,9,16,25,36,49,64,81,100
IMO E ; 11
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but at x = 0 , this is not an integer


Am I missing something here ?

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IMO D
At x=0, sqrt(120) is not an integer, so we have only 10 real values of X for P to be an integer

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shuvodip04
Noshad
For how many real values of x is p \(\sqrt{120 − √ x}\) an integer?

(A) 3

(B) 6

(C) 9

(D) 10

(E) 11

Bunuel

is the question p=\(\sqrt{120 − √ x}\). Or it is correct as it is.I feel there is missing "=" sign.

My approach

For \(\sqrt{120 − √ x}\) 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)

I also selected E. However, the question only says that x has to be real, not positive. Hence, x can also be (-39)^2, (-56)^2 ... Hence, we will have 22 possible values. Isn't it?
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Hi Guys,

 
At first, I also thought that there are only 10 values of X,
However, If we examine the the equation closely, there can be more possibilities. For eg, when X=-1

Which will then be, √120 -(√-1)= √120+1=√121= 11, an integer.

There are many such numbers that will satisfy the equation for example -2401, which √-2401= -49 which would result in √120+49=√169 = 13.

So all the possibilities are, -1, 1, 2, 3, 4, 5,6 ,7,8 9, 10, -2401 and there could be more.
But since in options 11 is the highest, So we will have to go with that.

Answer choice: E­
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