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August 15th, 2017, 07:42 AM   #1
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Alternative Quadratic Formula

I stumbled upon something interesting. I don't think it's anything new, but I've never been shown this.

When deriving the quadratic formula there is a part where it is necessary to take the square root of $\displaystyle 4a^2$. Previously I've always automatically written $\displaystyle 2a$. However, I now realize that it could just as well have been $\displaystyle -2a$. When we do write $\displaystyle -2a $, it changes our final formula to $\displaystyle \frac{b\pm\sqrt{b^2-4ac}}{-2a}$. Is this an alternative quadratic formula?

Last edited by skipjack; August 15th, 2017 at 08:39 AM.
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August 15th, 2017, 08:15 AM   #2
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That's exactly the same as the one you're used to.
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August 15th, 2017, 08:53 AM   #3
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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?
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August 15th, 2017, 07:22 PM   #4
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Quote:
Originally Posted by Antoniomathgini View Post
I stumbled upon something interesting. I don't think it's anything new, but I've never been shown this.

When deriving the quadratic formula there is a part where it is necessary to take the square root of $\displaystyle 4a^2$. Previously I've always automatically written $\displaystyle 2a$. However, I now realize that it could just as well have been $\displaystyle -2a$. When we do write $\displaystyle -2a $, it changes our final formula to $\displaystyle \frac{b\pm\sqrt{b^2-4ac}}{-2a}$. Is this an alternative quadratic formula?
$a \ne 0 \text { and } ax^2 + bx + c = 0 \implies x^2 + 2 * \dfrac{b}{2a} * x = -\ \dfrac{c}{a} = -\dfrac{4ac}{4a^2} \implies$

$x^2 + 2 * \dfrac{b}{2a} + \left ( \dfrac{b}{2a} \right )^2 = \left ( \dfrac{b}{2a} \right )^2 - \dfrac{4ac}{4a^2} \implies$

$\left ( x + \dfrac{b}{2a} \right )^2 = \dfrac {b^2 - 4ac}{4a^2} \implies x + \dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}} = \dfrac{\pm \sqrt{b^2 - 4ac}}{2a} \implies$

$x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} \ne \dfrac{b \pm\sqrt{b^2 - 4ac}}{-\ 2a}$

$\text {UNLESS } b = 0 = c.$

What you mean is that

$x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} * \dfrac{-\ 1}{-\ 1} = \dfrac{b \mp \sqrt{b^2 - 4ac}}{-\ 2a}.$

There are not two different formulas. Rather there are two different ways of expressing the same formula. I am being persnickety, but math's strength comes from being precise.
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August 16th, 2017, 02:00 AM   #5
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Hi , your formula works and does not look the same as the standard formula so yes it is an alternative version.

Now , regarding 'things' you've never been shown , perhaps you may find the following interesting.


For all $ \ \ \ a \ne 0$


If $ \ \ \ \ \ ax^2 = bx + c$


Then $ \ \ \ \ \ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $


Proof :


$ax^2 = bx + c$


$ax^2 - bx = c$


$4a^2x^2 - 4abx = 4ac$


$4a^2x^2- 4abx + b^2 = b^2 + 4ac$


$ (2ax - b)^2 = b^2 + 4ac$


$ 2ax - b = \pm \sqrt{b^2 + 4ac} $


$ 2ax = b \pm \sqrt{b^2 + 4ac} $


$ x = \frac{b \pm \sqrt{b^2 + 4ac} }{2a} $


Q.E.D.


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Last edited by agentredlum; August 16th, 2017 at 02:07 AM.
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August 16th, 2017, 05:46 AM   #6
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Quote:
Originally Posted by agentredlum View Post
Now , regarding 'things' you've never been shown , perhaps you may find the following interesting.


For all $ \ \ \ a \ne 0$


If $ \ \ \ \ \ ax^2 = bx + c$


Then $ \ \ \ \ \ x = \frac{b \pm \sqrt{b^2 + 4ac}}{2a} $


Proof :


...
The "normal" formula gives the solutions to the equation $ax^2 + bx + c = 0$. All you've done is relabeled two of the coefficients (you've called the linear coefficient $-b$ rather than $b$ and the constant one $-c$ rather than $c$) and done the standard proof. That does not make it different in any significant way.
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August 16th, 2017, 06:22 AM   #7
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Quote:
Originally Posted by JeffM1 View Post
$x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} \ne \dfrac{b \pm\sqrt{b^2 - 4ac}}{-\ 2a}$

What you mean is that

$x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} * \dfrac{-\ 1}{-\ 1} = \dfrac{b \mp \sqrt{b^2 - 4ac}}{-\ 2a}.$

There are not two different formulas. Rather there are two different ways of expressing the same formula. I am being persnickety, but math's strength comes from being precise.
The issue is that the $\pm$ and $\mp$ symbols are ambiguous - their use depends heavily on context - so you must be careful when trying to be precise. In the context of the quadratic formula, it seems inappropriate to distinguish between the statements "$x = A \pm B$" and "$x = A \mp B$"; we want them to both mean "$x = A + B$ or $x = A - B$" (which is equivalent to "$x = A - B$ or $x = A + B$"). To treat these as different statements makes it seem like the order we express the roots is somehow significant.
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August 16th, 2017, 07:00 AM   #8
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The proof stands alone.

"Different in any significant way"

Is just your opinion
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August 16th, 2017, 07:20 AM   #9
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Quote:
Originally Posted by cjem View Post
The issue is that the $\pm$ and $\mp$ symbols are ambiguous - their use depends heavily on context - so you must be careful when trying to be precise. In the context of the quadratic formula, it seems inappropriate to distinguish between the statements "$x = A \pm B$" and "$x = A \mp B$"; we want them to both mean "$x = A + B$ or $x = A - B$" (which is equivalent to "$x = A - B$ or $x = A + B$"). To treat these as different statements makes it seem like the order we express the roots is somehow significant.
I do not disagree with that. What you are saying is that the two statements are the same in meaning because plus/minus does not work like a normal operation sign.

Fine. Consider the case where the discriminant is zero. Using the normal formula we get

$x = \dfrac{-\ b}{2a}.$

That has the exact same meaning as

$x = \dfrac{b}{-\ 2a}.$

To view them as having some sort of mysterious difference is not recognizing the ambiguity in plus/minus but being engulfed in it. The distinction is obviously trivial.
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August 16th, 2017, 07:55 AM   #10
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Quote:
Originally Posted by agentredlum View Post
The proof stands alone.

"Different in any significant way"

Is just your opinion
You have copied the usual proof verbatim - you've made exactly the same logical steps in exactly the same order. The only difference is that you've decided to change the symbols for two of the coefficients. To pass this off as something different or interesting is to miss the point - in mathematics, it doesn't matter what symbols we use for things...

Last edited by cjem; August 16th, 2017 at 08:03 AM.
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