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August 18th, 2017, 07:41 AM   #21
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The standard formula has an unnecessary minus sign in front of b. To me that is interesting.

The standard derivation sets the equation equal to zero unnecessarily. To me that is interesting.
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Last edited by agentredlum; August 18th, 2017 at 07:50 AM.
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August 18th, 2017, 07:55 AM   #22
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Originally Posted by agentredlum View Post
The standard formula has an unnecessary minus sign in front of b. To me that is interesting.
That's because the usual formula tells you the solutions to $ax^2 + bx + c = 0$ while yours tells you the solutions to $ax^2 = bx + c$.

So, for example, say you wanted to find the solutions of the equation $6x^2 + 7x + 4 = 0$. With the normal formula, we'd plug in $a = 6, b = 7, c = 4$. To use your formula, we'd rearrange the equation to get it into the relevant form: $6x^2 = -7x - 4$. Then with your system, we'd have $a = 6, b = -7, c = -4$. Notice that the minus sign you've got rid of in the formula has just been reintroduced here.

Furthermore, when using either formula version, you'd still be writing be writing the minus:

$\dfrac{- 7 \pm blah}{2 \times 6}$

Last edited by cjem; August 18th, 2017 at 08:43 AM.
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August 18th, 2017, 08:06 AM   #23
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Quote:
Originally Posted by cjem View Post
That's because the usual formula tells you the solutions to $ax^2 + bx + c = 0$ while yours tells you the solutions to $ax^2 = bx + c$.

So, for example, say you wanted to find the roots of the polynomial $6x^2 + 7x + 4$. With the normal formula, we'd have $a = 6, b = 7, c = 4$ but with yours we'd have $a = 6, b = -7, c = -4$; you've just moved the minus sign from the formula itself to your labeling of the coefficients.
The minus sign in the standard formula is unnecessary. next time you do physics would you use

$-F = -ma $


as your standard formula?


get it?
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Last edited by agentredlum; August 18th, 2017 at 08:10 AM.
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August 18th, 2017, 08:18 AM   #24
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Quote:
Originally Posted by agentredlum View Post
The minus sign in the standard formula is unnecessary. next time you do physics would you use

$-F = -ma $


as your standard formula?


get it?
Unlike in $-F = -ma$, the minus sign here is necessary. All you've done is moved where it comes into the picture - you've accounted for it in your labeling of the coefficients of the polynomial rather than in the formula itself, as my example above shows.

Last edited by cjem; August 18th, 2017 at 08:24 AM.
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August 18th, 2017, 08:25 AM   #25
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The minus sign is necessary. All you've done is moved where it comes into the picture - you've accounted for it in your labeling of the coefficients of the polynomial rather than in the formula itself, as my example above shows.
That's incorrect cjem. Both formulas give the same correct answer. Therefore the minus sign in front of b is unnecessary.

You're back to talking about 'labeling' , I thought we already covered that and you agreed it was a poor choice of wording on your part.
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August 18th, 2017, 08:29 AM   #26
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Originally Posted by agentredlum View Post
That's incorrect cjem. Both formulas give the same correct answer. Therefore the minus sign in front of b is unnecessary.
Yes, they both give the same correct answer because you're still accounting for the minus sign. It is necessary to have the minus sign either in the formula itself, or in your labelling of the coefficients.

Quote:
Originally Posted by agentredlum View Post
You're back to talking about 'labeling' , I thought we already covered that and you agreed it was a poor choice of wording on your part.
I said "relabeling" was a poor choice of words; "labeling" is fine. I don't see what's so hard to understand: normally we label the quadratic, linear and constant coefficients of a quadratic polynomial by $a$, $b$ and $c$, respectively. You've instead labelled them as $a$, $-b$ and $-c$, respectively. If you don't see what I mean by this, I refer you back to my explicit example above - I've added some extra details.

Last edited by cjem; August 18th, 2017 at 08:50 AM.
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August 18th, 2017, 08:38 AM   #27
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Quote:
Originally Posted by cjem View Post
Yes, they both give the same correct answer because you're still accounting for the minus sign. It is necessary to have the minus sign either in the formula itself, or in your labelling of the coefficients.



I said "relabelling" was a poor choice of words; "labeling" is fine. I don't see what's so hard to understand: normally we label the quadratic, linear and constant coefficients of a quadratic polynomial by $a$, $b$ and $c$, respectively. You've instead labelled them as $a$, $-b$ and $-c$, respectively. If you don't see what I mean by this, I refer you back to my explicit example above - I've added some extra details.
I did not do that

$ ax^2 = bx + c $

No labeling or relableng here , just proceed to complete tha square

Get it?
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August 18th, 2017, 08:45 AM   #28
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Originally Posted by agentredlum View Post
I did not do that

$ ax^2 = bx + c $

No labeling or relableng here , just proceed to complete tha square

Get it?
By "labeling" I mean the symbols you are choosing to denote parts of the equation by. For example, you've labeled the coefficient of $x^2$ by the symbol $a$.

Anyway, I've edited post #22 one more time to make my point as clear as I can.

Last edited by cjem; August 18th, 2017 at 08:50 AM.
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August 18th, 2017, 09:31 AM   #29
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Your example is very particular and therefore not relevant because the linear term can be given on either side of the equal sign.

You cannot escape the fact that half the time you must compute $ \ \ -(-b_{1}) $ while I only must compute $ \ \ (-b_{1}) $

The minus sign in the standard formula is unnecessary and it is there because in the standard derivation the equation has been set equal to zero.

Here is your example , $6x^2 + 7x + 4 = 0 $

$ b = 7$

you must compute

$-(7) $

I must compute

$ (-7) $

Now consider

$ b = -7$

$6x^2 + 4 = 7x $

You must compute

$ -(-7) $

I must compute $ (7) $

You cannot escape the fact that the standard formula has an unnecessary minus sign cjem.

There is no guarantee that the equation will be given to you with a positive linear term AND already set equal to zero. All possible arrangements of the linear term appearing on either side of the equals sign are equally likely.

Last edited by skipjack; August 18th, 2017 at 05:04 PM.
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August 18th, 2017, 10:02 AM   #30
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Quote:
Originally Posted by agentredlum 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.
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