Let

f(x) = \sqrt{x - \sqrt{x}}.

Find the largest three-digit value of $x$ such that $f(x)$ is an integer.

gnistory Jun 27, 2024

#1**+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! :)

NotThatSmart Jun 27, 2024

#1**+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