Difference between revisions of "2023 WSMO Team Round Problems/Problem 6"
(Created page with "==Problem== A quartic real polynomial <math>f(x)</math> satisfying <math>f(3+2i) = 0</math> has 3 distinct roots. If the sum of the three roots is <math>12,</math> find their...") |
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| + | From the conjugate root theorem, since <math>3+2i</math> is a root, <math>3-2i</math> must also be a root. Since there are only three distinct roots, let <math>x</math> be the value of the remaining two roots. We have <cmath>x+(3+2i)+(3-2i) = 12\implies x+6 = 12\implies x = 6.</cmath> So, the product of the four roots is <cmath>(3+2i)(3-2i)(6)(6) = (3^2+2^2)(6)(6) = \boxed{468}.</cmath> | ||
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| + | ~pinkpig | ||
Latest revision as of 14:02, 13 September 2025
Problem
A quartic real polynomial 
satisfying 
has 3 distinct roots. If the sum of the three roots is 
find their product.
Solution
From the conjugate root theorem, since 
is a root, 
must also be a root. Since there are only three distinct roots, let 
be the value of the remaining two roots. We have ![\[x+(3+2i)+(3-2i) = 12\implies x+6 = 12\implies x = 6.\]](http://latex.artofproblemsolving.com/b/d/0/bd0f97ff68deeef21cbacb3663b954234e6e68e4.png)
So, the product of the four roots is ![\[(3+2i)(3-2i)(6)(6) = (3^2+2^2)(6)(6) = \boxed{468}.\]](http://latex.artofproblemsolving.com/f/9/4/f948b5a4721f12af47401195bdd97532b51babd6.png)
~pinkpig
