Difference between revisions of "2008 Mock ARML 1 Problems/Problem 1"
Monkeythyme (talk | contribs) (→Solution) |
Monkeythyme (talk | contribs) m (→Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | 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))= | + | 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 
such that 
.
Solution
Let 
; then 
. Because 
is increasing on 
, 
. Using this we can show 
. Using your favorite method, solve for 
. However, since 
, 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 
. 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.
