December 6th, 2007, 10:04 PM  #1 
Newbie Joined: Dec 2007 Posts: 1 Thanks: 0  alternate quadratic formula
Hi, I have a question that's been bugging me for quite a while... Is there an alternate quadratic formula? I know the standard one... However, I know there has to be another one! I used to be actively involved in Kumon, a rigorous Japanese Mathematics program. While in Kumon, I remember learning this alternate quadratic formula. I am no longer in the Kumon program, so I don't remember this alternate quadratic formula. What I do remember is that it was shorter and simpler. Does anybody know this alternate quadratic formula??? If anyone could please tell me that would be great... it would make my math homework A LOT easier. Thanks!!!! 
December 7th, 2007, 06:43 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Any such formula would have to give the same answers, so you can manipulate the formula into whatever kind of formula you like. But in the end all formulas will be of the form A +/ B where A = 1/2 b/a and B = sqrt(b^2/4  ac)/a though they may look different. 
December 9th, 2007, 02:45 PM  #3 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
If a formula is a solution to an equation and is able to give ALL the solutions to the equation, then it is THE ONLY such formula. It would be contradictory otherwise.

December 18th, 2007, 08:04 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
x = 2c/(b ± √(b²  4ac)) (provided a and c are nonzero)
Last edited by skipjack; May 23rd, 2018 at 06:08 AM. 
January 15th, 2012, 10:23 AM  #5 
Newbie Joined: Jan 2012 Posts: 1 Thanks: 0  Re: alternate quadratic formula
I go to Kumon. I think the answer you're looking for is the one that you can do provided that b in ax2 + bx + c is replaced by an even number (2b). It becomes (b±√(b²ac))/a.
Last edited by skipjack; May 23rd, 2018 at 06:06 AM. 
February 6th, 2012, 07:51 PM  #7 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: alternate quadratic formula
If $\displaystyle ax^2 = bx + c$ then $\displaystyle x = \frac{b \pm\sqrt {b^2 + 4ac}}{2a}$ Last edited by skipjack; May 23rd, 2018 at 05:46 AM. 
June 23rd, 2016, 02:41 AM  #8 
Newbie Joined: Jun 2016 From: Kenya Posts: 1 Thanks: 0  I used to visit kumon. I had the same question as I forgot it too.
HERE IS THE ANSWER. check the attachment Last edited by skipjack; May 23rd, 2018 at 05:49 AM. 
May 22nd, 2018, 08:53 PM  #9 
Newbie Joined: May 2018 From: New york Posts: 1 Thanks: 0 
Haha! I had just learned about the other forms of the quadratic formula in kumon and was looking to see if I could do some research on it!

May 23rd, 2018, 06:40 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,104 Thanks: 1907 
When calculating the roots using the standard formula, first choose a sign to replace "±" so that both terms in the numerator have the same sign. If the numerator then has value f, the roots are f/(2a) and 2c/f. Doing it this way avoids excessive rounding error in certain cases.


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