Difference between revisions of "2008 Mock ARML 1 Problems/Problem 1"
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== 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))=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> | + | 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 == | ||
| Line 10: | Line 10: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| − | + | Square both sides twice leaving: | |
| − | {x+4}=(x-4)^2 | + | <math>{x+4}=(x-4)^2</math> |
| − | + | Then, subtract <math>x-4</math> to set to <math>0</math> (from <math>x^2-8x^2+16</math>) | |
| − | + | Using the rational roots theorem, we get the quadratics: | |
| − | (x^2-x-4)(x^2+x-3) | + | <math>(x^2-x-4)(x^2+x-3)</math> |
Solve: | Solve: | ||
| − | -1+ | + | <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: | ||
| − | 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 20:49, 4 December 2016
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 
.
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:
Then, subtract 
to set to 
(from 
)
Using the rational roots theorem, we get the quadratics:
Solve:
Seeing that negative roots are extraneous we have:
and 
as the answers.
