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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

Best Answer 

 #1
avatar+5226 
+1

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
avatar+5226 
+1
Best Answer

just complete the square in x and y

 

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

Rom Apr 28, 2019

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