1983 AIME Problems/Problem 5

Problem

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value of $x + y$ can have?

Solution

The best way to solve this problem seems to be by brute force.

$w^2+z^2=(w+z)^2-2wz=7$ and $w^3+z^3=(w+z)(w^2+z^2)-wz(w+z)=(w+z)(7)-wz(w+z)=(7-wz)(w+z)=10$

Because we are only left with $w+z$ and $wz$ , substitution won't be too bad. Let $x=w+z$ and $y=wz$ .

We get $x^2-2y=7$ and $x(7-y)=10$

Because we want the largest possible $x$ , let's find an expression for $y$ in terms of $x$ . $x^2-7=2y \implies y=\frac{x^2-7}{2}$ .

Substituting, $x^3-21x+20=0$ . Factored, $(x-1)(x+5)(x-4)=0$

The largest possible solution is therefore $4$ .

1983 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
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1983 AIME Problems/Problem 5

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