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Difference between revisions of "2008 Mock ARML 1 Problems/Problem 1"

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== Solution ==
 
== Solution ==
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Let <math>f(x) = \sqrt{x+4}</math>; then <math>f(f(x)) = x</math>. Because <math>f(x)</math> is increasing on <math>-4<x<\infty</math>, <math>f(f(x))=x\Longrightarrowf(x)=x</math>. Using this we can show <math>x^2 - x - 4 = 0</math>. Using your favorite method, solve for <math>x = \frac{1 \pm \sqrt{17}}{2}</math>. However, since <math>f(x) =x</math>, 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 <math>x = \boxed{\frac{1+\sqrt{17}}{2}}</math>. The other roots we can verify are not real.
+
Let <math>f(x) = \sqrt{x+4}</math>; then <math>f(f(x)) = x</math>. Because <math>f(x)</math> is increasing on <math>-4<x<\infty</math>, <math>f(f(x))=f(x)=x</math>. Using this we can show <math>x^2 - x - 4 = 0</math>. Using your favorite method, solve for <math>x = \frac{1 \pm \sqrt{17}}{2}</math>. However, since <math>f(x) =x</math>, 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 <math>x = \boxed{\frac{1+\sqrt{17}}{2}}</math>. The other roots we can verify are not real.
  
 
== See also ==
 
== See also ==

Revision as of 14:43, 28 November 2010

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 $x = \boxed{\frac{1+\sqrt{17}}{2}}$. 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.

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