It’s almost sacred in some classrooms, given , then:

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

But did you know there was an alternate form?

**CHALLENGE:** Prove it.

Filed under: Education, Mathematics

Posted on May 24, 2009 by Jason Dyer

It’s almost sacred in some classrooms, given , then:

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

But did you know there was an alternate form?

**CHALLENGE:** Prove it.

Filed under: Education, Mathematics

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jonathan, on May 24, 2009 at 7:07 pm said:R the N. An often overlooked skill!

Nice one.

Jason Dyer, on May 24, 2009 at 7:42 pm said: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.

jonah, on May 24, 2009 at 7:42 pm said: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.

Jason Dyer, on May 24, 2009 at 7:51 pm said: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.

Alex, on May 25, 2009 at 3:08 am said: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.

pat ballew, on May 25, 2009 at 10:47 am said: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.

Jason Dyer, on May 25, 2009 at 3:33 pm said:I like how the symbol gets flipped in the alternate form.

jd2718, on May 25, 2009 at 5:28 pm said:I see the flipped symbol most often (and use it) for factoring

Jonathan

jd2718, on May 25, 2009 at 5:30 pm said:=

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

Jason Dyer, on May 25, 2009 at 6:00 pm said:I confess I’ve never seen that before… very interesting!

jd2718, on May 28, 2009 at 4:59 am said: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

Pavel Holoborodko, on June 2, 2009 at 11:32 pm said:Very interesting. Never thought quadratics has alternative formula for the roots.

Teaching to the Limits of Wolfram Alpha « The Number Warrior, on June 13, 2009 at 9:17 am said:[…] 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 […]

phil hartlaub, on November 1, 2009 at 9:48 pm said: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.

jimmy zotos, on April 17, 2011 at 10:11 pm said: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

david, on September 25, 2012 at 7:22 am said: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.

Paul S, on January 13, 2014 at 7:20 pm said: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