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What is the minimum value of the expression $x^2+y^2-6x+4y+18$ for real $x$ and $y$?

User StoneBird
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2 Answers

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Final answer:

The minimum value of the expression x^2 + y^2 - 6x + 4y + 18 is 5.

Step-by-step explanation:

The given expression is a quadratic equation in terms of x and y.

To find the minimum value of the expression, we can complete the square for both x and y. Let's start by completing the square for x:

x^2 - 6x = (x^2 - 6x + 9) - 9 = (x - 3)^2 - 9

Now, let's complete the square for y:

y^2 + 4y = (y^2 +4y + 4) - 4 = (y + 2)^2 - 4

Substituting back into the original expression, we get:

(x - 3)^2 + (y + 2)^2 - 9 - 4 + 18 = (x - 3)^2 + (y + 2)^2 + 5

The expression (x - 3)^2 + (y + 2)^2 + 5 represents the equation of a circle with center (3, -2) and radius sqrt(5).

Since the minimum value of a sum of squares is 0, the minimum value of the expression is 5.

User Jacob
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x^2+y^2-6x+4y+18=0
x^2-6x+y^2+4y+18=0
x^2-6x+9+y^2+4y+4 =-18+9+4
(x-3)^2+(y+2)^2=-5
-5 is the minimum value of the given circle
User Malaguna
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