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 August 15th, 2017, 06:42 AM #1 Member   Joined: Apr 2017 From: Canada Posts: 32 Thanks: 2 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 07:39 AM. August 15th, 2017, 07:15 AM #2 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 312 Thanks: 111 Math Focus: Number Theory, Algebraic Geometry That's exactly the same as the one you're used to. August 15th, 2017, 07:53 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,635 Thanks: 2080 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? Thanks from Antoniomathgini August 15th, 2017, 06:22 PM   #4
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 Originally Posted by Antoniomathgini 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. August 16th, 2017, 01:00 AM #5 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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. Thanks from Antoniomathgini Last edited by agentredlum; August 16th, 2017 at 01:07 AM. August 16th, 2017, 04:46 AM   #6
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 Originally Posted by agentredlum 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. August 16th, 2017, 05:22 AM   #7
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 Originally Posted by JeffM1 $x = \dfrac{-\ b \pm \sqrt{b^2 - 4ac}}{2a} \ne \dfrac{b \pm\sqrt{b^2 - 4ac}}{-\ 2a}$ 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. August 16th, 2017, 06:00 AM #8 Math Team   Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 The proof stands alone. "Different in any significant way" Is just your opinion August 16th, 2017, 06:20 AM   #9
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 Originally Posted by cjem 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. August 16th, 2017, 06:55 AM   #10
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 Originally Posted by agentredlum 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 07:03 AM. Tags alternative, formula, quadratic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post life24 Elementary Math 2 November 15th, 2016 04:29 AM Opposite Algebra 7 November 22nd, 2014 03:17 PM STxAxTIC Algebra 0 December 14th, 2013 06:29 PM yoman360 Algebra 2 May 12th, 2009 07:16 AM Jacob15728 Calculus 1 January 24th, 2009 09:13 AM

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