2008 Mock ARML 1 Problems/Problem 1

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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

See also

2008 Mock ARML 1 (Problems, Source)
Preceded by
First question
Followed by
Problem 2
1 2 3 4 5 6 7 8

square both sides twice leaving:

{x+4}=(x-4)^2

then subtract x-4 to set to 0 (from x^2-8x^2+16)

using the rational roots theorem, we get the quadratics:

(x^2-x-4)(x^2+x-3)

Solve: -1+/-sqrt{13}/2 1+/-sqrt{17}/2

Seeing that negative roots are extraneous we have:

1+sqrt{17}/2 and -1+sqrt{13}/2 as the answers.