2008 Mock ARML 1 Problems/Problem 1

Problem

Compute all real values of $x$ such that $\sqrt {\sqrt {x + 4} + 4} = x$ .

Solution

Let $f(x) = \sqrt{x+4}$ ; then $f(f(x)) = x$ . Suppose that $f(x) = x \Longleftrightarrow x^2 - x - 4 = 0 \Longrightarrow x = \frac{1 \pm \sqrt{17}}{2}$ . However, since $f(x) > 0$ , it follows that the negative root is extraneous, and thus we have $x = \boxed{\frac{1+\sqrt{17}}{2}}$ . The other roots we can verify are not real. Template:Incomplete

2008 Mock ARML 1 (Problems, Source)
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2008 Mock ARML 1 Problems/Problem 1

Problem

Solution

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