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We have a quadratic inequality

$$Ax^2+Bx+C>0$$

After solving it for cases where $B^2-4AC > 0$, my textbook turns to cases where $B^2-4AC < 0$:

Using the perfect square method, let's write down this inequality as
$$A\left[\left(X+\frac{B}{2A}\right)^2-\frac{B^2-4AC}{4A^2}\right]>0$$

But how did they get to that result? When I use the formula for completing the square (wikipedia), I get

$$A\left(X+\frac{B}{2A}\right)^2-\frac{B^2}{4A}+C>0;$$ $$A\left(X+\frac{B}{2A}\right)^2-\frac{B^2+4AC}{4A}>0$$

Clearly if I factor out $A$ I would not get those $A$'s the textbook has in the second term within the square brackets (and one even an $A$ squared!). Could the texbook be wrong?

P.S. It's just occurred to me - could I just multiply the $\frac{B^2+4AC}{4A}$ by another $A$ to get to that result?


Here's the excerpt from the texbook:

enter image description here

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    $\begingroup$ sorry, what exactly are you asking? how you go from $Ax^2 + Bx + C$ to $A\left[\left(X+\frac{B}{2A}\right)^2-\frac{B^2-4AC}{4A^2}\right]$? $\endgroup$ – Chester Aug 9 '15 at 20:01
  • $\begingroup$ @Chester - yes, it's not very clear to me. $\endgroup$ – CopperKettle Aug 9 '15 at 20:01
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    $\begingroup$ That's the way one proves the resolution formulae in high school! $\endgroup$ – Bernard Aug 9 '15 at 20:09
  • $\begingroup$ @Bernard - resolution of quadratic equations? $\endgroup$ – CopperKettle Aug 9 '15 at 20:14
  • $\begingroup$ @CopperKetle: Exactly. $\endgroup$ – Bernard Aug 9 '15 at 20:16
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\begin{align*} &\phantom{{}={}}Ax^2 + Bx + C\\ &=A\left(x^2+ \frac{B}{A} x + \frac{C}{A}\right)\\ &= A\left(x^2+ \frac{B}{A} x + \frac{B^2}{4A^2} - \frac{B^2}{4A^2} +\frac{C}{A}\right)\\ &= A\left(\left(x+ \frac{B}{2A}\right)^2 - \frac{B^2}{4A^2} +\frac{C}{A}\right)\\ &= A\left(\left(x+ \frac{B}{2A}\right)^2 + \frac{4AC-B^2}{4A^2}\right)\\ \end{align*}

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\begin{align*} AX^2 + BX + C&=\frac{1}{4A}\left(4A^2X^2+ 4ABX + 4AC\right)\\ &=\frac{1}{4A}\left((2AX)^2+ 2B(2AX)+ 4AC\right)\\ &=\frac{1}{4A}\left((2AX)^2+ 2B(2AX)+B^2-B^2+4AC\right)\\ &=\frac{1}{4A}\left((2AX+B)^2-B^2+4AC\right)\\ &=\frac{1}{A}\left((AX+\frac{B}{2})^2-\frac{B^2-4AC}{4})\right)\\ &= A\left(\left(x+ \frac{B}{2A}\right)^2 + \frac{4AC-B^2}{4A^2}\right)\\ \end{align*}

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