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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>. Suppose that <math>f(x) = x \Longleftrightarrow x^2 - x - 4 = 0 \Longrightarrow x = \frac{1 \pm \sqrt{17}}{2}</math>. However, since <math>f(x) > 0</math>, 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. {{incomplete|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))=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>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
  
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square both sides twice leaving:
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Square both sides twice leaving:
  
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{x+4}=(x-4)^2
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<math>{x+4}=(x-4)^2</math>
  
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then subtract x-4 to set to 0 (from x^2-8x^2+16)
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Then, subtract <math>x-4</math> to set to <math>0</math> (from <math>x^2-8x^2+16</math>)
  
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using the rational roots theorem, we get the quadratics:
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Using the rational roots theorem, we get the quadratics:
  
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(x^2-x-4)(x^2+x-3)
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<math>(x^2-x-4)(x^2+x-3)</math>
  
 
Solve:
 
Solve:
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-1+/-sqrt{13}/2 1+/-sqrt{17}/2
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<math>-1+-\sqrt{13}/2 1+-\sqrt{17}/2</math>
  
 
Seeing that negative roots are extraneous we have:
 
Seeing that negative roots are extraneous we have:
  
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1+sqrt{17}/2 and -1+sqrt{13}/2 as the answers.
+
<math>1+\sqrt{17}/2</math> and <math>-1+\sqrt{13}/2</math> as the answers.

Latest revision as of 21:49, 4 December 2016

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

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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