# an unclear step in a textbook solution of 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:

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