# Completing the square for $2x - x^2$

How do you complete the square for $2x-x^2$?

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$2x - x^2 = -(x^2 - 2x + 1) + 1$... – The Chaz 2.0 Feb 22 '12 at 4:56
It may help the PO to understand the objective first. Completing the square is about writing the expression in the form: $a(x+d)^2 + e$. The trick is to find the value for a, d and e. This page may help you further: mathsisfun.com/algebra/completing-square.html – NoChance Feb 22 '12 at 7:12
@Emmad Nice link; I've never seen that diagram before and now know how to answer the question "why is it called completing the square"? – David Mitra Feb 22 '12 at 12:50
@DavidMitra, I hope you mean this in a good way! – NoChance Feb 22 '12 at 14:58
I vote against closing this question. If anyone would like to vote to close this question, instead cancel your vote against mine. Saying this question is too localized makes no sense. Completing the square is an important tool. The OP doesn't know how to do it and is asking for help. That's a perfectly legitimate question. Saying it is too localized is completely wrong, as other people also may not know how to complete the square and could learn from the answers below. – Graphth Oct 1 '12 at 22:03

## 2 Answers

In case you want the general method,

Start with $$ax^2+bx+c,\qquad a\ne0$$

Factor out the $a$ (which may be negative, as in $2x-x^2$) from the first two terms: $$a(x^2+(b/a)x)+c$$

Add and subtract the square of half the coefficient of $x$: $$a(x^2+(b/a)x+(b/(2a))^2-(b/(2a))^2)+c=a(x^2+(b/a)x+(b/(2a))^2)+c-(b^2/(4a))$$

Factor: $$a(x+(b/(2a)))^2+c-(b^2/(4a))$$ Voila! The square, she is completed. When you've had as much experience as Ross and The Chaz, you'll be able to just look at a quadratic and write down the completion, but, until then, there's the utter and complete procedure.

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The Chaz has it, but so we have an answer, $2x−x^2=−(x^2−2x+1)+1$

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