Difference between revisions of "2002 Indonesia MO Problems/Problem 3"
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<cmath>(a-2)^3 = 0</cmath> | <cmath>(a-2)^3 = 0</cmath> | ||
Since <math>a = 2</math> is a triple root to the polynomial, the only solution to the system of equations is <math>\boxed{(2,2,2)}</math>, and plugging the values back in satisfies the system. | Since <math>a = 2</math> is a triple root to the polynomial, the only solution to the system of equations is <math>\boxed{(2,2,2)}</math>, and plugging the values back in satisfies the system. | ||
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+ | ==Solution 2== | ||
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+ | We can use Newton's Sums (https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums) to solve this problem -- we can say the three variables are roots to a cubic monic polynomial (so <math>a_n = a_3 = 1</math>). From the problem we have <math>S_1 = 6, S_2 = 12, S_3 = 24</math> and using Newton's Sums we have <cmath>6 + a_2 = 0\ 12 + 6a_2 + 2a_1 = 0\ 24 + 12a_2 + 6a_1 + 3a_0 = 0</cmath>We can find <math>a_2</math>, then <math>a_1, a_0</math> respectively to get the polynomial <cmath>x^3 - 6x^2 + 12x - 8 = 0</cmath> Using the Rational Root Theorem (or trial and error) we can easily find one of the roots -- <math>2</math>, and see that the other two roots are <math>2</math> as well (eg by factoring out <math>x-2</math>) yielding the only solution <math>\boxed{(2, 2, 2)}</math>. | ||
==See Also== | ==See Also== |
Latest revision as of 22:38, 13 July 2022
Contents
[hide]Problem
Find all real solutions from the following system of equations:
Solution
Square the first equation to get Subtract the second equation from the result to get Multiply the second equation by the first equation to get Subtract the third equation to get Cube the first equation to get If , , and , the solution triplet is the roots of the polynomial Factor the polynomial to get Since is a triple root to the polynomial, the only solution to the system of equations is , and plugging the values back in satisfies the system.
Solution 2
We can use Newton's Sums (https://artofproblemsolving.com/wiki/index.php/Newton%27s_Sums) to solve this problem -- we can say the three variables are roots to a cubic monic polynomial (so ). From the problem we have and using Newton's Sums we have We can find , then respectively to get the polynomial Using the Rational Root Theorem (or trial and error) we can easily find one of the roots -- , and see that the other two roots are as well (eg by factoring out ) yielding the only solution .
See Also
2002 Indonesia MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 • 7 | Followed by Problem 4 |
All Indonesia MO Problems and Solutions |