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Difference between revisions of "Euc20197/Sub-Problem 2"

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Consider the function <math>f(x) = x^2 - 2x</math>. Determine all real numbers <math>x</math> so that <math>x</math> satisfy <math>f(f(f(x))) = 3</math>
 
Consider the function <math>f(x) = x^2 - 2x</math>. Determine all real numbers <math>x</math> so that <math>x</math> satisfy <math>f(f(f(x))) = 3</math>
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==Solution 1==
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Let's start with the outermost <math>f(x)</math>. If <math>f(x)=3</math>, then <math>x^2-2x-3=0</math>, so <math>x=-1</math> or <math>3</math>. Now, let's do the middle <math>f(x)</math>. Here, <math>f(x)=-1</math> or <math>3</math>. If <math>f(x)=3</math>, then <math>x=3</math> or <math>-1</math>. If <math>f(x)=-1</math>, then <math>x^2-2x=-1</math>, so <math>x^2-2x+1=0</math>. Here, <math>x=1</math> is the only solution. Now, let's do the innermost <math>f(x)</math>. Here, because from the middle <math>f(x)</math> we have the possibilities of <math>x=1, -1,</math> or <math>3</math>, so we have <math>f(x)=1, -1,</math> or <math>3</math>. If <math>f(x)=3</math>, then <math>x=-1</math> or <math>3</math>. If <math>f(x)=-1</math>, then <math>x=1</math>. If <math>f(x)=1</math>, then we have <math>x^2-2x=1</math>, so <math>x^2-2x-1=0</math>. Here, after applying the quadratic formula, will give us <math>x=1-\sqrt2</math> or <math>1+\sqrt2</math>, so our possibilities are <math>3, -1, 1, 1+\sqrt2, and 1-\sqrt2</math>.
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~Yuhao2012
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 23:52, 9 October 2025

Problem

Consider the function $f(x) = x^2 - 2x$. Determine all real numbers $x$ so that $x$ satisfy $f(f(f(x))) = 3$

Solution 1

Let's start with the outermost $f(x)$. If $f(x)=3$, then $x^2-2x-3=0$, so $x=-1$ or $3$. Now, let's do the middle $f(x)$. Here, $f(x)=-1$ or $3$. If $f(x)=3$, then $x=3$ or $-1$. If $f(x)=-1$, then $x^2-2x=-1$, so $x^2-2x+1=0$. Here, $x=1$ is the only solution. Now, let's do the innermost $f(x)$. Here, because from the middle $f(x)$ we have the possibilities of $x=1, -1,$ or $3$, so we have $f(x)=1, -1,$ or $3$. If $f(x)=3$, then $x=-1$ or $3$. If $f(x)=-1$, then $x=1$. If $f(x)=1$, then we have $x^2-2x=1$, so $x^2-2x-1=0$. Here, after applying the quadratic formula, will give us $x=1-\sqrt2$ or $1+\sqrt2$, so our possibilities are $3, -1, 1, 1+\sqrt2, and 1-\sqrt2$.

~Yuhao2012

Video Solution

https://www.youtube.com/watch?v=M4gzTG8HnQ4

~North America Math Contest Go Go Go

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