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# Help!

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What is the minimum value of the expression $$x^2+y^2+2x-4y+8$$ for real x and y?

Apr 28, 2019

#1
+6045
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just complete the square in x and y

$$x^2 + y^2 + 2x - 4y + 8\\ (x^2+2x +1-1) + (y^2 -4y +4 -4) + 8\\ (x+1)^2 - 1+(y-2)^2 - 4 + 8 \\ (x+1)^2 +(y-2)^2 +3\\ \text{The minimum value is clearly seen to be }3$$

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Apr 28, 2019

#1
+6045
+1
$$x^2 + y^2 + 2x - 4y + 8\\ (x^2+2x +1-1) + (y^2 -4y +4 -4) + 8\\ (x+1)^2 - 1+(y-2)^2 - 4 + 8 \\ (x+1)^2 +(y-2)^2 +3\\ \text{The minimum value is clearly seen to be }3$$