Difference between revisions of "1995 AIME Problems/Problem 5"
m (→Solution 2) |
(→Solution 2) |
||
| Line 28: | Line 28: | ||
- AlexLikeMath | - AlexLikeMath | ||
| + | |||
- Corrections by VSPuzzler | - Corrections by VSPuzzler | ||
Revision as of 17:41, 2 August 2021
Contents
Problem
For certain real values of 
and 
the equation 
has four non-real roots. The product of two of these roots is 
and the sum of the other two roots is 
where 
Find 
Solution 1
Since the coefficients of the polynomial are real, it follows that the non-real roots must come in complex conjugate pairs. Let the first two roots be 
. Since 
is not real, are not conjugates, so the other pair of roots must be the conjugates of . Let 
be the conjugate of 
, and 
be the conjugate of 
. Then,
![\[m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.\]](http://latex.artofproblemsolving.com/d/a/b/dab8020a8532d068d90e04c999b3b3959a13e140.png)
By Vieta's formulas, we have that 
.
Solution 2
Let's assume that the 2 roots multiplied together are p+qi, and r+si, and the two roots added together are the conjugates of the previous roots. Using Vieta, we get
![\[b = (p+qi)(r+si) + (p+qi)(r-si) + (p-qi)(r+si) + (p-qi)(r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]](http://latex.artofproblemsolving.com/5/b/1/5b101e969952811662ecf13fb54801bfc03e6559.png)
![\[(p+qi+p-qi)(r+si+r-si) + (p+qi)(p-qi) + (r+si)(r-si) =\]](http://latex.artofproblemsolving.com/3/0/4/30412234e98160106154f666ec62cd0a55cc1549.png)
![\[(2p)(2r) + p^2 + q^2 + r^2 + s^2 = 4pr + p^2 + q^2 + r^2 + s^2 = (p+r)^2 + 2pr + q^2 + s^2\]](http://latex.artofproblemsolving.com/f/4/0/f408a38674c98d3a0b1bf79377a9a75058029c20.png)
We are now stuck. We can't simplify further. But, we look back to the problem and see that the two roots that are multiplied together give a product of , and the two roots that are added give 
. This gets three equations necessary for solving the problem. ![\[p+r = 3\]](http://latex.artofproblemsolving.com/6/f/c/6fcd6733a7d967c2b01e748667a4533f2a39790f.png)
![\[pr-qs = 13\]](http://latex.artofproblemsolving.com/1/c/2/1c217d9450cf65c5cf88b10edf3d3cdfe69e82dc.png)
![\[-q-s = 4\]](http://latex.artofproblemsolving.com/f/a/9/fa95b5e24532453b38032f308ab7f27fa613e2f5.png)
So, alright. Let's use the first equation to get that 
, and substitute that in. Now, the equation becomes:
![\[b = 9 + 2pr + q^2 + s^2\]](http://latex.artofproblemsolving.com/0/b/9/0b9274750320212298ecfcb643aea0d36ae651d3.png)
We wish that we can turn the 2pr into 2qs. Fortunately, we can do that. By using the second equation, we can manipulate it to be 
, and substitute that in.
![\[b = 9 + 2(13+qs) + q^2 + s^2 = 9 + 26 + 2qs + q^2 + s^2 = 35 + (q+s)^2\]](http://latex.artofproblemsolving.com/1/f/4/1f4f4d69d2d25f00d8b5a52dc50f897d09482050.png)
We can square both sides of the third equation, and get 
We substitute that in and we get
![\[b = 35+16 = \boxed{051}\]](http://latex.artofproblemsolving.com/3/e/f/3efe5b4d3f60c80b85a8951a60fbcebc895ec337.png)
- AlexLikeMath
- Corrections by VSPuzzler
See also
| 1995 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
