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From the last term in completing the square from the quadratic $ax^2+bx+c$, I was just wondering how $$-\left(\frac{b}{2a}\right)^2+c = \left(c-\frac{b^2}{4a}\right)$$ I would have gotten $$-\left(\frac{b}{2a}\right)^2+c=\left(c-\frac{b^2}{(2a)^2}\right)=\left(c-\frac{b^2}{4a^2}\right)$$ Books final answer was the following when completing the square $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$

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  • $\begingroup$ Where is the problem? $\endgroup$ – George Law Jan 6 '17 at 5:28
  • $\begingroup$ It's the last term in completing the square from the quadratic $ax^2+bx+c$ $\endgroup$ – Alexander John Jan 6 '17 at 5:29
  • $\begingroup$ Um... I believe you're right. The denominator is supposed to be $4a^2$ $\endgroup$ – Frank Jan 6 '17 at 5:30
  • $\begingroup$ The books final answer was $$a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$ @Frank $\endgroup$ – Alexander John Jan 6 '17 at 5:34
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You forgot to multiply the $\left(\frac{b}{2a}\right)^2$ by $a$.

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  • $\begingroup$ OOOOH! Yessss! Thanks! @Robert $\endgroup$ – Alexander John Jan 6 '17 at 5:39

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