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Can someone please explain how to complete the square? Is there a specific formula you have to put the given equation in, or something? Every time I search how to complete the square each equation seems to employ a different method and it just confuses me. I'm not even understanding the concept at all. Can someone please help me? It feels like one of those things which I'll never be able to understand.

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  • $\begingroup$ ELI5? What does that mean? $\endgroup$ – Umberto Jan 17 '14 at 13:23
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Suppose you have a quadratic expression $ax^2+bx+c$ with $a\ne 0$.

$ax^2+bx+c$

$=a(x^2+\frac{b}{a}x)+c$

$=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c$

$=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$

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  • $\begingroup$ Wait so we just add (b/2)^2 to both sides? $\endgroup$ – Michalo Jan 17 '14 at 11:06
  • $\begingroup$ Here's how you can re-discover (in case you forget) that your add $\left(\frac{b}{2}\right)^2$ to both sides: You want the result to have the form $(x+R)^2$ for some number $R.$ Now look at the expanded form of $(x+R)^2$ (square first term, multiply the terms and double the result, square second term), which is $x^2 + 2Rx + R^2.$ Notice that the constant term, which is $R^2,$ is the square of half the coefficient of $x.$ That is, you can get $R^2$ from $2R$ (the coefficient of $x$) by taking half of $2R$ and then squaring the result. $\endgroup$ – Dave L. Renfro Jan 17 '14 at 14:55
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A slightly different view, without dividing by $a$, but multiplying by $4a$, so the degree two term is easily a square and we also get the degree one term in good form to be twice a product: \begin{gather} ax^2+bx+c=0\\[2ex] 4a^2x^2+4abx+4ac=0\\[2ex] 4a^2x^2+4abx+b^2-b^2+4ac=0\\[2ex] (2ax+b)^2=b^2-4ac \end{gather}

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  • $\begingroup$ An $x^2$ missing in the second and third lines. $\endgroup$ – bubba Jan 17 '14 at 12:28
  • $\begingroup$ @bubba Thanks, fixed. $\endgroup$ – egreg Jan 17 '14 at 13:20

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