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 $$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$
Question on completing the square obtaining the form $a(x+p)^2+q$
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$\begingroup$
algebra-precalculus
quadratics
completing-the-square
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0Where is the problem? – 2017-01-06
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0It's the last term in completing the square from the quadratic $ax^2+bx+c$ – 2017-01-06
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0Um... I believe you're right. The denominator is supposed to be $4a^2$ – 2017-01-06
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0The books final answer was $$a\left(x+\frac{b}{2a}\right)^2+\left(c-\frac{b^2}{4a}\right)$$ @Frank – 2017-01-06
1 Answers
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You forgot to multiply the $\left(\frac{b}{2a}\right)^2$ by $a$.
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0OOOOH! Yessss! Thanks! @Robert – 2017-01-06