Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.2 Junior Find the answer

Given $x^2+y^2+4x-6y+13=0$ , find the value of $x^y$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Transform the given equation to get: $(x+2)^2+(y-3)^2=0$ ,
thus $x+2=0$ , $y-3=0$ , which means $x=-2$ , $y=3$ ,
therefore, $x^y=(-2)^3=-8$ .
So, the value of $x^y$ is $\boxed{-8}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.