My Math Forum Alternative Quadratic Formula

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August 18th, 2017, 11:20 AM   #31
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Quote:
 Originally Posted by cjem Okay, sure. My point was that it's necessary to account for the minus sign somehow (be that in the formula itself, like is standard, or in the computation of $b$, as in your version). But I do agree that accounting for it in the formula itself is unnecessarily slow - it involves more computation on average, as you've explained.
I think my argument is convincing.

Oh yeah , don't forget about the derivation using the Tschirnhausen substitution I presented also.

 August 18th, 2017, 11:22 AM #32 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,444 Thanks: 946 thread closed!!
 August 18th, 2017, 06:12 PM #33 Global Moderator   Joined: Dec 2006 Posts: 19,865 Thanks: 1833 I see there's no response to my initial question.
August 19th, 2017, 05:33 AM   #34
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Quote:
 Originally Posted by skipjack I see there's no response to my initial question.
As it was pretty much the only interesting thing in the thread, it was bound to get ignored!

Just to nitpick, surely you want the $\pm$ sign to be the opposite in each particular case? By which I mean

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

and

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

August 19th, 2017, 07:43 AM   #35
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Quote:
 Originally Posted by skipjack I see there's no response to my initial question.
I thought your initial question was directed at the OP

Quote:
 Originally Posted by skipjack It can be useful to use both $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ and $\displaystyle \frac{2c}{-b\pm\sqrt{b^2-4ac}}$, choosing the Â± sign in the same particular way in each case. Can you see why?
I agree that the alternative formula with the radical in the denominator is useful. You just have to make sure that zero is not a root because it will produce a $\ \ \frac{0}{0}$

I would also like to note that if one uses

$ax^2 = bx + c$

Then the above alternative formula becomes

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

And this may also be useful

August 22nd, 2017, 03:13 PM   #36
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Quote:
 Originally Posted by agentredlum cjem , 'equivalent' and 'exactly' do not mean the same thing. Just because you see the destination and you have an idea about how to get there does not prove I took the route you think I did. You are still trying to defend your original incorrect assumptions about what I did. Anyway , here's a derivation using a Tschirnhausen substitution. For all $\ \ \ a \ne 0$ If $\ \ \ ax^2 = bx + c$ Then $\ \ \ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a}$ Proof by using a Tschirnhausen Transformation $ax^2 = bx + c \ \ \ \$ , Let $\ \ \ \ x = y + \frac{b}{a}$ $a(y + \frac{b }{a})^2 = b( y + \frac{b }{a}) + c$ $(y + \frac{b}{a})^2 = \frac{by}{a} + \frac{b^2}{a^2} + \frac{c}{a}$ $y^2 + \frac{2by}{a} + \frac{b^2}{a^2} = \frac{by}{a} + \frac{b^2}{a^2} + \frac{c}{a}$ $y^2 + \frac{by}{a} = \frac{c}{a}$ $y^2 + \frac{by}{a} + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} + \frac{c}{a}$ $(y + \frac{b}{2a} )^2 = \frac{b^2 + 4ac}{4a^2}$ $y + \frac{b}{2a} = \pm \frac{ \sqrt{b^2 + 4ac}}{2a}$ But $\ \ \ x = y + \frac{b}{a} \ \ \$ , $\ \ \ y = x - \frac{b}{a}$ So ... $x - \frac{b}{a} + \frac{b}{2a} = \pm \frac{\sqrt{b^2 + 4ac}}{2a}$ $x - \frac{b}{2a} = \pm \frac{\sqrt{b^2 + 4ac}}{2a}$ $x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a}$ Q.E.D.
You did something wrong. If you test your formula on x^2 - 3x + 2 = 0, it does not produce the correct roots of 1 and 2.

August 22nd, 2017, 04:34 PM   #37
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Quote:
 Originally Posted by EvanJ You did something wrong. If you test your formula on x^2 - 3x + 2 = 0, it does not produce the correct roots of 1 and 2.
It works. Following agent's setup, we have $\displaystyle a = 1$, $\displaystyle b = 3$ and $\displaystyle c = -2$. Plugging these values into the formula, we find the roots as

$\displaystyle \frac{(3)\pm\sqrt{(3)^2+4 \times (1) \times (-2)}}{2 \times (1)}= 1,2$

as expected.

 August 22nd, 2017, 04:47 PM #38 Global Moderator   Joined: Dec 2006 Posts: 19,865 Thanks: 1833 What makes you think that, EvanJ? I think, agentredlum, that you didn't understand what can make the formulas I suggested particularly useful. Give it some deeper thought. Thanks from agentredlum
August 24th, 2017, 12:01 AM   #39
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Quote:
 Originally Posted by skipjack What makes you think that, EvanJ? I think, agentredlum, that you didn't understand what can make the formulas I suggested particularly useful. Give it some deeper thought.
I think any alternative version may be useful given particular circumstances. Wolfram says the version with the radical in the denominator is helpful if $\ \ b^2 >> 4ac \ \$ ( $\ \ b^2 \ \$ is much greater than 4ac )

But I'm on the fence about that. Maybe you can convince me?

 August 24th, 2017, 12:38 AM #40 Global Moderator   Joined: Dec 2006 Posts: 19,865 Thanks: 1833 Try using an extreme example to test that.

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