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# Algebra

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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

#1
+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
+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! :)

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