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For how many real values of x is sqrt(120 - sqrt(x^2)) an integer?

 Jul 8, 2022
 #1
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I bet many people are falling for this:

sqrt(x^2) does not necessarily equal x.

 

I repeat, sqrt(x^2) does NOT Necessasrliy equal x. 

What if x is negative? :) 

 

We'll have to take 2 cases:
1. x >= 0; 

This gives us sqrt(120 - x) is an integer. 

 

This means that 120-x is a perfect square. Since x is nonnegative, the amount of values of x is the amount of perfect squares less than 120.
This ranges from 1^2.... to 10^2.. Therefore there are 10 values where x is nonnegative. 

 

2. x < 0 

This gives us that sqrt(120 + x)  is an integer. However, note that x is negative and there will be the same values of x before because x in our first case is like -x in our case here. Hence, there are 10 values where x is negative.

 

Adding up these two cases, gives us 20 values. 

 Jul 9, 2022

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