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 $f(f(x)) = x$ . Because $f(x)$ is increasing on $-4<x<\infty$ , $f(f(x))=f(x)=x$ . 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 $x = \boxed{\frac{1+\sqrt{17}}{2}}$ .

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

Page

Toolbox

Search

2008 Mock ARML 1 Problems/Problem 1

Problem

Solution

See also