My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree11Thanks
Reply
 
LinkBack Thread Tools Display Modes
August 18th, 2017, 10:20 AM   #31
Math Team
 
agentredlum's Avatar
 
Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 233

Quote:
Originally Posted by cjem View Post
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.

agentredlum is offline  
 
August 18th, 2017, 10:22 AM   #32
Math Team
 
Joined: Oct 2011
From: Ottawa Ontario, Canada

Posts: 13,112
Thanks: 911

thread closed!!
Denis is offline  
August 18th, 2017, 05:12 PM   #33
Global Moderator
 
Joined: Dec 2006

Posts: 19,511
Thanks: 1743

I see there's no response to my initial question.
skipjack is offline  
August 19th, 2017, 04:33 AM   #34
Senior Member
 
Joined: Aug 2017
From: United Kingdom

Posts: 219
Thanks: 70

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
Originally Posted by skipjack View Post
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

$\frac{-b - \sqrt{b^2-4ac}}{2a} = \frac{2c}{-b + \sqrt{b^2-4ac}}$
cjem is offline  
August 19th, 2017, 06:43 AM   #35
Math Team
 
agentredlum's Avatar
 
Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 233

Quote:
Originally Posted by skipjack View Post
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 View Post
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 ± 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

agentredlum is offline  
August 22nd, 2017, 02:13 PM   #36
Senior Member
 
Joined: Oct 2013
From: New York, USA

Posts: 608
Thanks: 82

Quote:
Originally Posted by agentredlum View Post
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.
EvanJ is offline  
August 22nd, 2017, 03:34 PM   #37
Senior Member
 
Joined: Aug 2017
From: United Kingdom

Posts: 219
Thanks: 70

Math Focus: Algebraic Number Theory, Arithmetic Geometry
Quote:
Originally Posted by EvanJ View Post
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.
Thanks from agentredlum
cjem is offline  
August 22nd, 2017, 03:47 PM   #38
Global Moderator
 
Joined: Dec 2006

Posts: 19,511
Thanks: 1743

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
skipjack is offline  
August 23rd, 2017, 11:01 PM   #39
Math Team
 
agentredlum's Avatar
 
Joined: Jul 2011
From: North America, 42nd parallel

Posts: 3,372
Thanks: 233

Quote:
Originally Posted by skipjack View Post
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?

agentredlum is offline  
August 23rd, 2017, 11:38 PM   #40
Global Moderator
 
Joined: Dec 2006

Posts: 19,511
Thanks: 1743

Try using an extreme example to test that.
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
alternative, formula, quadratic



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
I need alternative formula life24 Elementary Math 2 November 15th, 2016 04:29 AM
Quadratic formula Opposite Algebra 7 November 22nd, 2014 03:17 PM
Derivation of Alternative Vite Formula (digits of pi) STxAxTIC Algebra 0 December 14th, 2013 06:29 PM
Quadratic Formula yoman360 Algebra 2 May 12th, 2009 07:16 AM
Alternative quadratic formula? Jacob15728 Calculus 1 January 24th, 2009 09:13 AM





Copyright © 2018 My Math Forum. All rights reserved.