My Math Forum I discovered a new formula to solve 2nd degree equations

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 November 18th, 2010, 02:00 PM #1 Newbie   Joined: Nov 2010 Posts: 3 Thanks: 0 I discovered a new formula to solve 2nd degree equations Hi! I've been trying to develop a formula to solve a 3 degree equation (a easier and more practical formula). While doing this, I started to study better second degree equation, and I discover 2 new ways to do it. One of them I believe it is already known, but the other one, I think no one has discovered it before. What should I do now? I believe that I can't publish it now, because someone is going to steal my ideia.
 November 18th, 2010, 03:03 PM #2 Global Moderator   Joined: May 2007 Posts: 6,236 Thanks: 498 Re: I discovered a new formula to solve 2º degree equations You could post it here. I suspect you may be naive in thinking that you have discovered a new way that no one has thought of in all these years.
 November 18th, 2010, 04:20 PM #3 Newbie   Joined: Nov 2010 Posts: 3 Thanks: 0 Re: I discovered a new formula to solve 2º degree equations Well, my alternate way to solve a second degree equation: 1st - Derivate the expression 2nd - Find the root of the derivated expression (much easier) 3rd - Substitute that value on the expression 4ºth - Make the higher term equal to the simetric result of the third step 5ºth - The root is equal to the sum of the fourth step and the 2nd step. EXAMPLE: 3x^2 + 4x - 3 = F(x) 1st: F'(x) = 3.2x + 4.1 = 6x + 4 2nd: 6x + 4 = 0 <=> x = -4/6 <=> x = -2/3 3rd: F(-2/3) = 3(-2/3)^2 + 4(-2/3) - 3 = 3*4/9 - 8/3 - 3 = 4/3 - 8/3 - 9/3 = -13/3 4th: 3x^2 = 13/3 <=> x^2 = 13/9 <=> x = (13/9)^(1/2) 5th 3x^2 + 4x - 3 = 0 <=> x = (13/9)^(1/2) - 2/3 Finished. I discovered it while I was trying to find a way to solve a third degree equation.
 November 18th, 2010, 04:57 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: I discovered a new formula to solve 2ï¿½ degree equations Let's try your method on the general quadratic, and see what results. $f(x)=ax^2+bx+c=0$ 1.) Differentiate with respect to x. $f'(x)=2ax+b$ 2.) Equate f'(x) to zero, solve for x. $2ax+b=0$ $x=-\frac{b}{2a}$ 3.) Find $f\left(-\frac{b}{2a}\right)$ $a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=\frac{b^2}{4a}-\frac{b^2}{2a}+c=\frac{4ac-b^2}{4a}$ 4.) Equate the squared term from f to the negative of the result from the 3rd step and solve for x. $ax^2=\frac{b^2-4ac}{4a}$ $x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm\sqrt{b^2-4ac}}{2a}$ 5.) Add the results from steps 2 and 4 to get the roots. $x=-\frac{b}{2a}+\frac{\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Thus, your method is valid, but I can't see anyone who knows differential calculus going through all those steps, when the quadratic formula is most likely permanently emblazoned on their mind. In my opinion, completing the square is an easier way to derive the quadratic formula. But, when working on specific quadratics, perhaps your method is faster for you. And there's certainly nothing wrong with investigation. Last edited by skipjack; December 29th, 2016 at 04:36 PM.
 March 6th, 2011, 10:39 AM #5 Senior Member   Joined: Jul 2010 Posts: 103 Thanks: 0 Re: I discovered a new formula to solve 2º degree equations Hi TGPT I Rick.I checked your method works well.I appreciate you... it's not the first time that new methods have been found for the same question.you may work in a higher degree equation "I am sure you will discover more as you try more"
 March 11th, 2011, 07:55 PM #6 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 I think this is a little too advanced for 1st graders.
March 12th, 2011, 03:56 PM   #7
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Re:

Quote:
 Originally Posted by johnny I think this is a little too advanced for 1st graders.
You are perfectly right! I believe quadratic equations don't come up until around 7th or 8th grade.

 March 12th, 2011, 06:08 PM #8 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 True, but we at least borderline mathematical geniuses are expected to learn quadratic equations around 1st grade.
 May 28th, 2011, 04:04 AM #9 Senior Member   Joined: Feb 2010 Posts: 324 Thanks: 0 Re: I discovered a new formula to solve 2º degree equations They had me hummin' the quadratic formula to the tune of "pop goes the weasel" in pre-k. I don't know where yall went to school.... :P
 July 17th, 2011, 10:17 AM #10 Newbie   Joined: Nov 2010 From: Richmond BC Canada Posts: 16 Thanks: 0 Re: I discovered a new formula to solve 2º degree equations Too cumbersome, as compared to factor method or quadratic formula method. It is a good effort to find some thing new. But, sorry to say, I don't think as a math teacher, I can ever use this method to teach quadratic equations to kids in grade 10. Because, I have to teach calculus to grade nine or ten kids before using this method.

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