2008 Mock ARML 1 Problems/Problem 1

Revision as of 14:42, 28 November 2010 by Monkeythyme (talk | contribs) (Solution)

Problem

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

Solution

Let $f(x) = \sqrt{x+4}$; then $f(f(x)) = x$. Because $f(x)$ is increasing on $-4<x<\infty$, $f(f(x))=x\Longrightarrowf(x)=x$ (Error compiling LaTeX. Unknown error_msg). Using this we can show $x^2 - x - 4 = 0$. Using your favorite method, solve for $x = \frac{1 \pm \sqrt{17}}{2}$. However, since $f(x) =x$, and because the Square Root function's range does not include negative numbers, 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.

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.