Well the formula for solving a Quadratic equation is :

$$\text{If }\space ax^2+bx+c=0$$

then

$$x=\dfrac{-b \pm \sqrt{b^2 -4ac} }{2a}$$

But looking at this : [Wolfram Mathworld] (And also in other places)

They give An Alternate Formula:

$$x=\dfrac{2c}{-b \pm \sqrt{b^2 -4ac}}$$

How does one get this?

Also in the first formula (the one we know) , $$a \neq 0$$ ... but here is it still the case?

• consider multiplying the numerator and denominator by the conjugate of the numerator (that is -b /mp sqrt(b^2-4ac)) – Winston Jun 26 '15 at 15:49
• @Winston Edit your $\LaTeX$ and ... Thanks! – NeilRoy Jun 26 '15 at 15:51
• Something seems wrong with the new formula. If $c=0$, we get one valid root, and another with denominator $0$.?? – ghosts_in_the_code Jun 26 '15 at 15:55
• In the new formula, you have the two restrictions $a \neq 0$, $c \neq 0$, because otherwise the denominator is $0$. – Théophile Jun 26 '15 at 15:57
• @ghosts_in_the_code Haha I know...that's why i am asking the validity of this formula! – NeilRoy Jun 26 '15 at 15:57

Consider the quadratic $ax^2 + bx + c$. If $a = 0$, then finding the root is easy and boring, so we only look at the case when $a \neq 0$.

One way to get at the original quadratic formula is to complete the square. That is, note that \begin{align} 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{4ac - b^2}{4a^2} \right). \end{align} So when looking for solutions to $ax^2 + bx + c = 0$, we are looking at solutions to $$0 = a \left( \left( x + \frac{b}{2a} \right)^2 + \frac{4ac - b^2}{4a^2} \right) = a\left( x + \frac{b}{2a}\right)^2 + a\frac{4ac - b^2}{4a^2},$$ or equivalently $$\left(x + \frac{b}{2a} \right)^2 = \frac{b^2 - 4ac}{4a^2}.$$ Taking square roots and solving for $x$ gives the original quadratic formula.

Now let us derive the alternate form. Notice that if $c = 0$ then $x = 0$ is a root and $x = -b/a$ is the other root. So let us now consider $c \neq 0$. We now know that $x = 0$ is not a root. This means that if $r$ is a root of $ax^2 + bx + c = 0$, then $\frac{1}{r}$ will be a root of $a \frac{1}{x^2} + b \frac{1}{x} + c = 0$. Multiplying through by $x^2$ (which makes sense because we are not interested in $x = 0$), we are interested in reciprocals of the roots of $$a + bx + cx^2 = 0.$$ The original quadratic formula applies, and we know the roots are $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2c}.$$ Since we are interested in the reciprocals, we find the roots to be $$x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}.$$

This isn't necessarily the best derivation, but I like it. $\diamondsuit$

• Thanks man! : But three more proofs [goo.gl/F36dd8 , goo.gl/r8kGVQ , goo.gl/4C3k6Q] ... !!! – NeilRoy Jun 26 '15 at 16:39
• But what about my $2^{\text{nd}}$ question? That is what if $a=0$ and $c=0$ ? – NeilRoy Jun 26 '15 at 16:41
• Then you're looking at $bx = 0$, which is a linear polynomial with solution $x = 0$. – davidlowryduda Jun 26 '15 at 16:42
• No what i meant was if $c=0$ then by the new formula $x=0$ which is not the case with the familiar formula. Also in the familiar formula if $a=0$ then $x$ is undefined but that does not seem to be the case in the new one...So..??? – NeilRoy Jun 26 '15 at 16:44
• I stated explicitly what assumptions were necessary where. You do not use any quadratic formula if $a = 0$ because you do not have a quadratic. If $c = 0$, then read the paragraph including Notice that if $c = 0$... – davidlowryduda Jun 26 '15 at 16:46

I would like to add some point to answer posted user @mixedmath.

For equation $$ax^2 + bx + c = 0$$

roots are

$$x = \frac {-b\ \pm \ \sqrt {b^2 - 4ac} }{2a}$$

Lets put $$y = \frac {1}{x}$$

Then we get converted equation as, $$cy^2 + by + a = 0$$

Equivalent roots are, $$y = \frac {-b\ \mp \ \sqrt {b^2 - 4ac} }{2c}$$

And thus if you invert to get values of $x$ then those are,

$$x = \frac{1}{y} = \frac{2c}{-b \mp \sqrt{b^2 -4ac}}$$

Do note the difference between the signs $\pm$ and $\mp$.

You can verify that by trying to equate both sides as follows:

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

Lets take one sign at a time. Lets choose $+$ from left side, this $-$ from right. It becomes,

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

Thus things are in correct place.

Lets verify with one quick example.

Consider a quadratic equation as follows,

$$f(x) = (x-3)(2x-5) = 2x^2 - 11 x + 15 = 0$$

According to first formula we get values as follows, $$x = \frac{11 \pm 1}{4}$$

$$\{x_+, x_-\} = \{3, \frac{5}{2}\}$$

And with second formula we get, $$x = \frac{30}{11 \mp 1}$$ $$\{x_+, x_-\} = \{\frac{5}{2}, 3\}$$