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

 

I can't figure it out help PLEASE.

Thanks in advance.

 Aug 12, 2019
edited by Guest  Aug 12, 2019
edited by Guest  Aug 12, 2019
 #1
avatar+105195 
+2

x^2 + y^2 + 2x - 4y - 8      complete the square on x and y

 

x^2 +2x + 1  + y^2 - 4y + 4   - 1   - 4      factor

 

(x + 1)^2   + (y - 2)^2   -   5

 

The minimum value of  -5    will be  achieved when x = -1  and y  = 2

 

cool cool cool

 Aug 12, 2019
 #2
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THX FOR the help

laugh

 Aug 12, 2019
 #3
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nvm, It does not work. it says it is wrong. SRRY

Guest Aug 12, 2019
 #4
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it is correct. I read your answer wrong. 

Guest Aug 13, 2019

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