Proving an alternate quadratic formula

Question

It is well known that the quadratic formula for $ax^2+bx+c=0$ is given by$$x=\dfrac {-b\pm\sqrt{b^2-4ac}}{2a}\tag1$$ Where $a\ne0$. However, I read somewhere else that given $ax^2+bx+c=0$, we have another solution for $x$ as$$x=\dfrac {-2c}{b\pm\sqrt{b^2-4ac}}\tag2$$ Where $c\ne0$. In fact, $(2)$ gives solutions for $0x^2+bx+c=0$!

Question:

1. How would you prove $(2)$?
2. Why is $(2)$ somewhat similar to $\dfrac 1{(1)}$ but with $2a$ replaced with $-2c$?

Answer 1

You can write $ax^2+bx+c=0$ as $a+b(\frac 1x)+c(\frac 1x)^2=0,$ solve for $\frac 1x$, then invert it. You can have the minus sign top or bottom as you like by multiplying top and bottom by $-1$

Answer 2

Take (2), and rationalize the denominator:

$$\frac{-2c}{b \pm \sqrt{b^2-4ac}} = \frac{-2c}{b \pm \sqrt{b^2-4ac}}\frac{b \mp \sqrt{b^2-4ac}}{b \mp \sqrt{b^2-4ac}} = $$ $$\frac{-2c(b\mp\sqrt{b^2-4ac})}{b^2-(b^2-4ac)} = \frac{-b \mp \sqrt{b^2-4ac}}{2a}$$

Answer 3

There are a few errors in the question as posed.

First, the quadratic formula is $$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a}$$ Note the factor of $a$ in the denominator, missing from the OP. [EDIT -- I see this has been fixed now.]

Second, you write:

In fact, (2) gives solutions for $0x^2+bx+c=0$!

This is not true, and does not even make sense. The equation $0x^2+bx+c=0$ is equivalent to $bx = -c$, whose only solution is $x=-c/b$ (assuming $b\ne 0$).

In any case, to turn (1) into (2), just multiply by $-b \mp \sqrt{b^2-4ac}$ in both the numerator and the denominator:

$$x = \frac{-b \pm \sqrt {b^2-4ac} }{2a} \cdot \frac{-b \mp \sqrt{b^2-4ac}}{-b \mp \sqrt{b^2-4ac}}$$ The numerator now has the form $(X \pm Y)(Y \pm X)$, which simplifies to just $X^2-Y^2$. So we have $$x = \frac{1}{2a} \frac{ \left( -b \right)^2 - \left(b^2-4ac\right)}{-b \mp \sqrt{b^2-4ac}}$$ Cleaning up, the $(-b)^2-b^2$ in the numerator cancels out; the $2a$ in the denominator reduces against the $4ac$ in the numerator, leaving just $2c$; and you can move a negative sign out of the denominator and into the numerator, giving the alternative form of the quadratic formula you wanted.

Finally, you ask

Why is (2) somewhat similar to $\frac{1}{(1)}$ but with $2a$ replaced with $−2c$?

As Ross Millikan says in his answer, if in the original equation $ax^2 + bx + c =0$ we assume that $x=0$ is not a solution (which is equivalent to assuming that $c\ne 0$), then we can divide the entire equation through by $x^2$, obtaining $$c \left(\frac{1}{x}\right)^2 + b\left(\frac1x\right) + a = 0$$ If we set $u=\frac1x$, then this is $cu^2 + bu + a = 0$, and the quadratic equation tells us $$u = \frac{ - b \pm \sqrt{b^2 - 4ac} }{2c}$$ Finally we get $$x = \frac{2c}{-b \pm \sqrt{b^2-4ac}}$$ and you can pull the negative sign out of the denominator into the numerator. So the reason the equations are so similar is because of a duality in the equation: interchanging $a$ with $c$ and simultaneously replacing $x$ with $1/x$ changes one quadratic equation into an equivalent one.

Answer 4

Just for the fun trivia, here are some other "quadratic formulas" $(a>0):$

for $|\frac{b^2}{2a}-2c-1|\le1:$

$$x=-\frac b{2a}\pm\frac1{\sqrt a}\cos\left(\frac12\arccos\left(\frac{b^2}{2a}-2c-1\right)\right)$$

for $\frac{b^2}{2a}-2c-1>1:$

$$x=-\frac b{2a}\pm\frac1{\sqrt a}\cosh\left(\frac12\operatorname{arccosh}\left(\frac{b^2}{2a}-2c-1\right)\right)$$

You can play around with some graphs here if you want.