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## The Alternate Form of the Quadratic Formula

It’s almost sacred in some classrooms, given $ax^2+bx+c=0$, then:

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

It’s proven generally by completing the square; even Wolfram Alpha can do it.

But did you know there was an alternate form?

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

CHALLENGE: Prove it.

### 17 Responses

1. R the N. An often overlooked skill!

Nice one.

2. The plus-or-minus threw me for a loop at first (so I digressed with other methods), but of course you can just rationalize for each root individually and they switch which one is plus and which one is minus.

3. This isn’t nearly as useful as the more commonly-known formula. For the commonly-known one, the only requirement is that a≠0, which isn’t a big deal because if a=0 then the equation isn’t meaningfully a quadratic. But for this given one, if c=0 it breaks down. Consider x²–x=0; this clearly can be factored as x(x-1)=0 and so has solutions x=0 and x=1. Plugging in a=1, b=-1, c=0 to the given formula results in x=0/(1±1); i.e., x=0/2=0 and x=0/0. That’s no good.

• It does get used for some numerical root evaluation when precision issues come up. If c = 0 is a problem the calculation can be split into two parts.

4. You can also derive it from scratch by completing the square.

Divide through by x squared at the beginning [c/xx + b/x + a = 0]. Then solve as normal and you get the 1/x = the normal formula with a and c switched, i.e.

1/x = (-b +- sqrt(bb-4ca))/2c.

Then flip it over and you’re done. I like that because it plays on the fact that the quadratic equation is essentially symmetrical, and anything you can do one way around will also work the other.

5. Jason,
I first found this in an old 1895 book by Simon Newcomb. I have a copy of the page, as well as one other unusual variation of the more common form at my web page at http://www.pballew.net/arithm20.html#quadform

Not sure how far back it goes… but this is the earliest I have found.

• I like how the $\pm$ symbol gets flipped in the alternate form.

• I see the flipped symbol most often (and use it) for factoring $(x\pm y)^3$

Jonathan

6. = $(x \pm y)(x^2 \mp xy + y^2)$

(the latex is \ mp instead of \ pm, if I got it right)

• I confess I’ve never seen that before… very interesting!

• You can work out how to use it usefully in the sine of a sum/difference as well. I learned it long, long ago (1980?) from my 11th grade Analysis teacher (that’s what we called precalc w/trig course back then).

Jonathan

7. Very interesting. Never thought quadratics has alternative formula for the roots.

8. […] I recently wrote about the alternate form of the quadratic formula I found Wolfram Alpha was able to derive the standard formula from scratch. I was then curious if […]

9. Have you ever seen the quadratic formula proved in the following way???

After moving c to the other side, multiply both sides by 4a. Then complete the square, but not in the usual way. Intuitively add b^2 to both sides. The rest is easy. It avoids fractions in the derivation.

10. did you know that you can use

[b+sqrt(b^2+4ac)]/(2a)

notice positive b and positive 4ac this is NOT a typographical error

• Actually both formulas work even when the divisor is zero.
Example ax^2+bx+c= 0 when a = 0
you take derivative with respect to a for case of 0/0 of both parts
that is when (b – sqrt(b^2-4ac))/2a when a =0 rewriting as
it becomes (b – (b^2-4ac)^(.5))/(2a) then taking derivatives with respect to a you get
(-.5(b^2 -4ac)^(-,5)(-4c))/2 at a = 0 you get -c/b you actually also
get c/b since sqrt has two solutions so when you plug back in
only one works. The other case when you take + sqrt at start
give a nonzero number divided by zero which is a singular and
in algebra is not considered a number. So again with a = 0 the standard formula gives the correct answer even if a = 0

In the alternate formula for solutions when a = 0 there is again the
same problem when a=0 but the aquation is not in indeterminate form
so no derivatives needed in its evaluation. One leads directly to
-c/b the other to a the singularity which is not counted as a solution.

So at least for a = 0 or near zero the second solution is usually the better one to implement. However the thing looks nasty at c= 0 and
have for the divisor b-sqrt(b^2+4ac) at the plus sqrt you see x = 0
as one of the solutions. for the current case do as in first example
take derivatives set c=0 and you can get the other root no problem.

11. The alternate form is better written as:
-2c/(b +-sqr(b^2 – 4ac))
Where (-b +-sqr(b^2 – 4ac))/(2a) = -2c/(b +-sqr(b^2 – 4ac))

So when the b value is +
x1 = -2c/(b + sqr(b^2 – 4ac)) this form is more accurate here
x2 = (-b – sqr(b^2 – 4ac))/(2a)
And when the b value is –
x1 = (-b + sqr(b^2 – 4ac))/(2a)
x2 = -2c/(b – sqr(b^2 – 4ac)) this form is more accurate here