Difference between revisions of "2008 Mock ARML 1 Problems/Problem 1"
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then subtract x-4 to set to 0 (from x^2-8x^2+16) | then subtract x-4 to set to 0 (from x^2-8x^2+16) | ||
| − | using the | + | using the rational roots theorem, we get the quadratics: |
(x^2-x-4)(x^2+x-3) | (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: | Seeing that negative roots are extraneous we have: | ||
1+sqrt{17}/2 and -1+sqrt{13}/2 as the answers. | 1+sqrt{17}/2 and -1+sqrt{13}/2 as the answers. | ||
Revision as of 13:55, 27 November 2009
Problem
Compute all real values of 
such that 
.
Solution
Let 
; then 
. Suppose that 
. However, since 
, it follows that the negative root is extraneous, and thus we have 
. The other roots we can verify are not real. Template:Incomplete
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.
