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.

$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=(x+y)(7)-xy(x+y)=(7-xy)(x+y)=10$

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

We get $w^2-2z=7$ and $w(7-z)=10$

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

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

The largest possible solution is therefore $x+y=w=\box{4}$ (Error compiling LaTeX. Unknown error_msg).

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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