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Let
f(x) = \sqrt{x - \sqrt{x}}.
Find the largest three-digit value of $x$ such that $f(x)$ is an integer.

 Jun 27, 2024

Best Answer 

 #1
avatar+1230 
+1

Let's set the total value to be r. We have that

\(r = \sqrt{x - \sqrt{x}}\)

 

Squaring both sides, we get 

\({r}^{2} = x - \sqrt{x}\)

 

Now, let's note something important. Since both must be integers, we find that x must be a perfect square. 

Let's set \(x={a}^{2}, a\ge0\)

 

Subsituting this in, we find that

\({r}^{2}={a}^{2}-a\)

 

Let's notice that this isn't possible. There isn't a three digit value of x that satisfies this equation. 

Thus, our answer is none. 

 

I can elaborate if needed. 

 

Thanks! :)

 Jun 27, 2024
edited by NotThatSmart  Jun 27, 2024
 #1
avatar+1230 
+1
Best Answer

Let's set the total value to be r. We have that

\(r = \sqrt{x - \sqrt{x}}\)

 

Squaring both sides, we get 

\({r}^{2} = x - \sqrt{x}\)

 

Now, let's note something important. Since both must be integers, we find that x must be a perfect square. 

Let's set \(x={a}^{2}, a\ge0\)

 

Subsituting this in, we find that

\({r}^{2}={a}^{2}-a\)

 

Let's notice that this isn't possible. There isn't a three digit value of x that satisfies this equation. 

Thus, our answer is none. 

 

I can elaborate if needed. 

 

Thanks! :)

NotThatSmart Jun 27, 2024
edited by NotThatSmart  Jun 27, 2024

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