+354 Let
f(x) = \sqrt{x - \sqrt{x}}.
Find the largest three-digit value of $x$ such that $f(x)$ is an integer.
+953 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! :)
+953 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! :)
