How can you simplify the quadratic formula using completing the square?

[agentredlum]

All right, here's the derivation using a Tschirnhaus transformation, a clever algebraic substitution invented by this man. I bring to the table the idea of invariance and 'tweak' his clever algebraic substitution just a little bit.
Remember...'a' not equal to 0

Consider ax^2 = bx + c

Let x = b/(2a) + v

a and b are EXACTLY as represented in ax^2 = bx + c

a(b/(2a) + v)^2 = b(b/(2a) + v) + c

a(b^2/(4a^2) + bv/(2a) + bv/(2a) + v^2) = b^2/(2a) + bv + c

combine the middle term inside parenthesis on the left hand side of equation.

a(b^2/(4a^2) + bv/a + v^2) = b^2/(2a) + bv + c

Distribute a on the left hand side of equation

b^2 /(4a) + bv + av^2 = b^2/(2a) + bv + c

Cancel bv from both sides and isolate av^2

av^2 = b^2/(2a) - b^2/(4a) + c

get common denominator

av^2 = (2b^2 - b^2 + 4ac)/(4a)

av^2 = (b^2 + 4ac)/(4a)

Divide by a

v^2 = (b^2 + 4ac)/(4a^2)

Extract the root

v = (+-sqrt(b^2 + 4ac))/(2a)

Now remember that x = b/(2a) + v

x = b/(2a) + (+-sqrt(b^2 + 4ac))/(2a)

x = (b +-sqrt(b^2 + 4ac))/(2a)

Which is the same result as post #1

 

[agentredlum]

Tschirnhaus would start with ax^2 + bx + c = 0 and use x = v - b/(2a) to eliminate the bx term. Then he would multiply it all out, simplify, isolate the av^2 term, get common denominator, divide by 'a', extract the root and substitute back. If he did that, he would derive the textbook definition of the quadratic formula. Try it! It is a nice little excercise in algebra!

I 'tweaked' his substitution by using x = b/(2a) + v and using the idea of 'invariance'.

Of course, I wasn't there so I don't know what Tschirnhaus actually did but i have read accounts by others of his methods.

I'm not sure if Tschirnhaus did this for a quadratic because his major interest in this area was in finding formulas for higher degree, cubic, quartic, quintic, etc.

He believed his techniques should work for any degree but others proved him wrong. Abel, Ruffini, Galios?

His technique works for all 2nd, 3rd, and 4th for sure but as soon as you hit 5th degree exceptions can be found where his technique fails, but it does not fail for ALL 5th degree.

Also, the substitution changes for higher degrees and the higher the degree the more difficult the algebra. My proof for degree 2 is pretty complicated in algebra and degree 3 is more complicated.

Look it up but i think the substitution is x = v - b/(na) where n is the degree, a is the coefficient of x^n and b is the coefficient of x^(n-1).

This will have the effect of eliminating the bx^(n-1) term but of course more work must be done for degree higher than 2

For degree 2 it eliminates the bx term and algebraic rearangement allows one to derive quadratic formula quite easily, not so easy for cubic, and much harder for degree 4

 

[agentredlum]

Here, check out WA

Standard result using Tschirnhaus substitution

http://www.wolframalpha.com/input/?...+=+v+-+b/(2a)&asynchronous=false&equal=Submit

My result using 'tweaked' Tschirnhaus substitution and idea of 'invariance'

http://www.wolframalpha.com/input/?i=ax^2+=+bx+++c,+x+=+v+++b/(2a)&asynchronous=false&equal=Submit

Here's how to do the cubic, for some strange reason the article doesn't mention Tschirnhaus although his technique is used by the poster to eliminate the squared term. As you can see, it is an algebraic nightmare. He also insists that the coefficient of x^3 be 1. That is not necessary and introduces fractions at the VERY FIRST STEP of a long and arduous derivation.

http://mathworld.wolfram.com/CubicFormula.html

 

[Robert1986]

Can someone please explain to me why agentredlum's "method" is anything other that what we have been doing all our lives? Let me see if I get this he says to take the quadratic equation:

ax^2 -bx -c = 0

then the solutions are given by:

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

Now, where is the "short cut" in this? This is the SAME FLIPPING THING that we do now. There is absolutely NO DIFFERENCE other than a trivial arithmetic thing.

Take x^2 +3x +2. The solutions are given using the "regular" quadratic formula (or factoring, but let's stay with the quadratic formula we all know) x=-1,-2. Now, the "trick" agentredlum proposes, is to solve the quadratic: x^2 -3x -2 = 0 with a new quadratic formula. But, plugging everything into the quadratic formula we get: x = 3 +- sqrt(b^2 - 4(1)(c))/2a.

Am I completely missing something, here? Cause to me, this "trick" is no trick at all; at the very least it is a pointless algebraic manipulation.

 

[micromass]

Am I completely missing something, here?

No, you're not. The formula's are exactly the same

 

[gb7nash]

If it's possible, you want to transform a given problem to a simpler problem, solve the simpler problem, then transform the answer to get the correct answer. Rarely do I see a method that makes the original problem more complicated to solve. It's cool I guess, but I don't see any practicality here.

 

[Mark44]

Robert1986, micromass, and gb7nash, those were my thoughts, too.

If you start with your quadratic as ax² = bx + c
then b and c will have the opposite signs as in the usual form of the Quadratic Formula, so the revised form becomes
\frac{b \pm \sqrt{b^2 + 4ac}}{2a}

The -b of the original becomes b, and the -4ac of the original becomes + 4ac. IMO, not that big of an improvement.

 

[ArcanaNoir]

It would be nicer if it meant you could work with positive numbers after putting actual values for a, b, and c more often, but if b and c are positive in ordinary set up then they become negative in the regular way and in agentredlum's formula. if they're negative, both formulae make for positive calculations. So the only value is in ease of memorizing the formula. And maybe it's value is lost when you consider the usefulness of setting equations equal to zero and the methods of factoring, roots, critical points, and inflection points, etc. Conceptually everything is tied to zero.

I don't see why so many professors had trouble accepting this modification, it only took me one real example to see the nature of it. I think real examples are sometimes neglected since mathematicians' purpose seems mostly to prove things. Sometimes a proof obscures something that's obvious.

It's cool though. Super cool.

 

[agentredlum]

Take x^2 +3x +2. The solutions are given using the "regular" quadratic formula (or factoring, but let's stay with the quadratic formula we all know) x=-1,-2. Now, the "trick" agentredlum proposes, is to solve the quadratic: x^2 -3x -2 = 0 with a new quadratic formula. But, plugging everything into the quadratic formula we get: x = 3 +- sqrt(b^2 - 4(1)(c))/2a.

Am I completely missing something, here?

Thank you for the response, please keep the criticism comming.

Here is a difference you missed.

Look at the example x^2 + 3x + 2 you changed this to x^2 - 3x - 2 = 0 and used the OLD formula that has un-necessary minus signs. You also made a mistake somewhere cause your solutions don't work.

x = 3 +- sqrt(b^2 - 4(1)(c))/2a. Does Not give -1 and -2 for many reasons, many mistakes.

But your biggest mistake is the following subtle point...

The original was not set equal to anything but you transformed it into something else you thought i would do and set it equal to zero, which i would NEVER do.

It is not easy to attempt a different way of thinking when everything you have learned screams against the attempt.

I don't know how many of you can really follow a derivation cause if you think it is not correct then you should point out the flaw in the derivation instead of mixing up stuff and coming up with your own versions that do not follow from my derivation.

For the example x^2 + 3x + 2 = 0

I propose to solve x^2 = -3x - 2 by isolating ax^2 term and using my method NOT x^2 - 3x - 2 = 0 as you claim i would.

Now, if you read the posts carefully you would see it cause i worked it out in post #195

 

[micromass]

It is not easy to attempt a different way of thinking when everything you have learned screams against the attempt.

That's the point. It's not a different way of thinking. It's the exact same formula and the old way of thinking. There's nothing novel going on here.

 

[agentredlum]

It would be nicer if it meant you could work with positive numbers after putting actual values for a, b, and c more often, but if b and c are positive in ordinary set up then they become negative in the regular way and in agentredlum's formula. if they're negative, both formulae make for positive calculations. So the only value is in ease of memorizing the formula. And maybe it's value is lost when you consider the usefulness of setting equations equal to zero and the methods of factoring, roots, critical points, and inflection points, etc. Conceptually everything is tied to zero.

I don't see why so many professors had trouble accepting this modification, it only took me one real example to see the nature of it. I think real examples are sometimes neglected since mathematicians' purpose seems mostly to prove things. Sometimes a proof obscures something that's obvious.

It's cool though. Super cool.

Thank you for the response, thank you for the support.

I think you got it! Coefficients can come with different signs so one formula is not superior to the other in that sense. Thank you for pointing out this subtle point.

Let me add that FORMS of the equation can come in 4 ways, here's what i mean...

Form1: ax^2 + bx + c = 0

Form2: ax^2 + bx = c

Form3: ax^2 + c = bx

Form4: ax^2 = bx + c

Every single one of these forms, by itself, describes the same infinite set of 2nd degree equations applicable to the quadratic formula so no form is superior to another in terms of APPLICABILITY.

However, form1 is INFERIOR to the other 3 cause it uses more, un-necessary, symbols. Zero is an extra symbol and there is an extra addition.

So yes my version using form4 is easier to memorize as you point out but also easier to write down cause you use less ink, less chalk, less keystrokes, etc.

Now, if you use form 1 with more symbols in the form and more symbols in the quadratic formula then you are performing more operations than if you use form4 which does NOT require you to set = 0 and has a simpler formula.

To put it another way, you will run out of ink faster than me, or run out of memory faster than me. So perhaps it might be worthwhile to use my derivation?

 

[agentredlum]

That's the point. It's not a different way of thinking. It's the exact same formula and the old way of thinking. There's nothing novel going on here.

So you dismiss it?

The old way of thinking is 'you have to set it = 0 to find the roots'

The new way of thinking is 'the old way of thinking is a myth'

This point of mine got lost on you in the midst of trying to trivialize my result.

Thank you for your comment.

 

[Robert1986]

Robert1986, micromass, and gb7nash, those were my thoughts, too.

If you start with your quadratic as ax² = bx + c
then b and c will have the opposite signs as in the usual form of the Quadratic Formula, so the revised form becomes
\frac{b \pm \sqrt{b^2 + 4ac}}{2a}

The -b of the original becomes b, and the -4ac of the original becomes + 4ac. IMO, not that big of an improvement.

And the improvement only exists in writing this new quadratic formula, and it is only if you have some hatred of minus signs. In actual practice, it works out the exact same way. I'm glad I'm not the only one who thinks that this "trick" is a complete waste of time.

 

[Mark44]

That's the point. It's not a different way of thinking. It's the exact same formula and the old way of thinking. There's nothing novel going on here.

So you dismiss it?

The old way of thinking is 'you have to set it = 0 to find the roots'

The new way of thinking is 'the old way of thinking is a myth'

This point of mine got lost on you in the midst of trying to trivialize my result.

I agree with micromass. All that you are doing is changing from the standard form for a quadratic equation, ax² + bx + c = 0 to a different form, ax² = dx + f, where b = -d and f = -c. This is not a 'new way of thinking.' It's just a slightly different way of saying the same thing.

 

[pwsnafu]

The old way of thinking is 'you have to set it = 0 to find the roots'

Except we define "http://en.wikipedia.org/wiki/Root_of_a_function" " to be this.
That's why we teach it this way.

 

[Robert1986]

Thank you for the response, please keep the criticism comming.

Here is a difference you missed.

Look at the example x^2 + 3x + 2 you changed this to x^2 - 3x - 2 = 0 and used the OLD formula that has un-necessary minus signs. You also made a mistake somewhere cause your solutions don't work.

x = 3 +- sqrt(b^2 - 4(1)(c))/2a. Does Not give -1 and -2 for many reasons, many mistakes.

Where to begin, where to begin, where to begin.

First, -1 and -2 are solutions to x^2 + 3x +2.

(-1)^2 + 3(-1) + 2 = 1 -3 + 2 = 0 so -1 is a solution

(-2)^2 + 3(-2) +2 = 4 - 6 + 2 = 0 so -2 is a solution.


or to show it another way:

x^2 + 3x +2 = (x+1)(x+2) = 0, which means -1 and -2 are roots.


But, you didn't even do the quadratic formula correctly. You have to do (-3 +- sqrt(3^2-4(a)(2)))/2. This isn't what you did, which is why you didn't get the right answer.


Now, where are the "many mistakes"?

But your biggest mistake is the following subtle point...

No, my biggest mistake was getting roped into this...

The original was not set equal to anything but you transformed it into something else you thought i would do and set it equal to zero, which i would NEVER do.

I have no idea how you propose to find solutions to something that is not "set equal" to something. When you find roots of a quadratic, you are, in fact, setting it equal to 0 and finding what x's make the equation true.

It is not easy to attempt a different way of thinking when everything you have learned screams against the attempt.


I don't know how many of you can really follow a derivation cause if you think it is not correct then you should point out the flaw in the derivation instead of mixing up stuff and coming up with your own versions that do not follow from my derivation.

Hmmm. You see, I agree with your derivation. I find nothing wrong with what you did, mathematically. What I am saying is that what you did was silly, pointless and trivial.

For the example x^2 + 3x + 2 = 0

I propose to solve x^2 = -3x - 2 by isolating ax^2 term and using my method NOT x^2 - 3x - 2 = 0 as you claim i would.

Your method is to take the equation x^2 = -3x -2 and use your "revised" quadratic formula find the values of x for which this equation is true. Fine, I see nothing wrong with that, my only point is that it does not simplify anything in the slightest. I never said that you would solve it by solving x^2 - 3x - 2 =0. I said you would take the quadratic equation:

ax^2 + bx + c = 0 as ax^2 = bx +c then use a "different" (different in a superficial, fifth grader way) quadratic formula to find x's that make the equation true. This is EXACTLY what you described.


The equation x^2 = -3x -2 is the SAME FREAKING THING as the equation x^2 + 3x +2 = 0. And your quadratic formula is the SAME FREAKING thing as mine. Go ahead, plug the numbers in yours and in mine and, before you make any calculation you will see that they are IDENTICAL. So why the flip even bother moving stuff around?


You seem to not be able to grasp multiplication by minus 1.

 

[Robert1986]

So you dismiss it?

The old way of thinking is 'you have to set it = 0 to find the roots'

The new way of thinking is 'the old way of thinking is a myth'

This point of mine got lost on you in the midst of trying to trivialize my result.

Thank you for your comment.

Ok, solve this equation:

bx + c = 0.

You can do it multiple ways:

a) solve -bx - c = 0

b) solve bx = -c

c) solve x = -c/b

d) solve -bx -c + 100 = 100

and I could go on. The point isn't that what you have done is wrong, the point is that a)anyone with a middle-school level of education can figure this out b)you aren't doing anything new, in any sense of the word.

 

[Robert1986]

I agree with micromass. All that you are doing is changing from the standard form for a quadratic equation, ax² + bx + c = 0 to a different form, ax² = dx + f, where b = -d and f = -c. This is not a 'new way of thinking.' It's just a slightly different way of saying the same thing.

It's no wonder his math profs initially said "you're wrong" and then changed it to "you're right." I'm guessing that this is how the conversation went:

AGENTREDLUM: Prof., I have a new way to solve a quadratic. x = \frac{b +- \sqrt{b^2 + 4ac}{2a}.

PROF: thinking agentredlum means he proposes to solve the equation ax^2 + bx + c = 0, which is what anyone would think No freaking way.

AGENTREDLUM: long explanation of his easy arithmetic

PROF: Oh yeah. Well you're certainly correct. My only question is why it took prof half an hour to be convinced.

 

[agentredlum]

I agree with micromass. All that you are doing is changing from the standard form for a quadratic equation, ax² + bx + c = 0 to a different form, ax² = dx + f, where b = -d and f = -c.

Thank you for the response. That's not what I'm doing in the derivation and if you try to do this for a particular example it would fail to give the right answer using my formula. You see if you replace b by -d and f by -c as you claim then you get back ax^2 + bx + c = 0 so this method hasn't DONE anything. The subsequent derivation will yield the textbook quadratic formula.

This is what i derived

If ax^2 = bx + c then x = (b +-sqrt(b^2 + 4ac))/(2a)

You can't set it = 0 and use this version... you must isolate ax^2 term in equation given to you.

I fear my shortcut post had at least 2 undesirable effects

1) It caused confusion

2) It made the result look trivial.

When using shortcuts one must use them wisely because there are underlying intricacies involved.

 

[micromass]

2) It made the result look trivial.

That's because it is.

 

[micromass]

"Oh, let's try to solve the equation ax+b=0. But let's not set it equal to zero. Let's look at the equation ax=b instead. We can see that the solution is x=\frac{b}{a}.

This is basically what you did in your post.

No, I won't be impressed until you find a solution for a quintic polynomial.

 

[agentredlum]

"Oh, let's try to solve the equation ax+b=0. But let's not set it equal to zero. Let's look at the equation ax=b instead. We can see that the solution is x=\frac{b}{a}.

This is basically what you did in your post.

No, I won't be impressed until you find a solution for a quintic polynomial.

Let's look at ax = b

If you had been taught that you MUST set it equal to zero in order to find the root, what would you think? Would you think that rule is a myth?

So you only get impressed when someone performs an impossible task? It doesn't take much to impress me, i am thankful for any progress made, no matter how small, especially in a subject such as this which has been considered 'dead' for centuries.

I'm not asking for anyone to be impressed, this is not a cure for a disease. I am asking for some of the posters to be a bit more civilized and check their calculations before showering me with derision.

 

[pwsnafu]

So you only get impressed when someone performs an impossible task?

To be pedantic, micromass never asked for a solution using radicals. I don't remember if Bring radicals solve all quintics, but I think elliptic integrals can be used.

 

[agentredlum]

Now, where are the "many mistakes"? .

Your many mistakes are on this line below, originally posted by you.

x = 3 +- sqrt(b^2 - 4(1)(c))/2a

1) You don't have parenthesis on the immediate left of 3

2) you have 3, should be -3. If you use 3 you can't get the answers -1 and -2 that's why i said your answers were incorrect.

3) 1 parenthesis is missing to the immediate right of c

4) 2a should be enclosed in parenthesis

You also didn't substitute for all a, b, c but let that slide

 

[agentredlum]

To be pedantic, micromass never asked for a solution using radicals. I don't remember if Bring radicals solve all quintics, but I think elliptic integrals can be used.

Yes, I know that, i was trying to avoid flying off on a tangent by posting something like x^5 - 1 = 0

You don't understand me at all. The main idea of that post is that any progress, no matter how trivial is perceived by others, actually is benficial, especially in a subject that has been considered 'dead'

Where does pedantic stop? Sooner or later you will resort to correcting my punctuation in order to make your point?

 

[pwsnafu]

Where does pedantic stop? Will you sooner or later resort to correcting my punctuation in order to make your point?

Couldn't resist.

 

[agentredlum]

Couldn't resist.

We're cool.

Bring brings solutions to many quintics using Bring-Jerrard radicals but Bring cannot bring all solutions to quintics brought by able Abel-Ruffini using Bring-Jerrard radicals.

http://en.m.wikipedia.org/wiki/Quintic_function

 

[Robert1986]

Your many mistakes are on this line below, originally posted by you.

x = 3 +- sqrt(b^2 - 4(1)(c))/2a

1) You don't have parenthesis on the immediate left of 3

2) you have 3, should be -3. If you use 3 you can't get the answers -1 and -2 that's why i said your answers were incorrect.

3) 1 parenthesis is missing to the immediate right of c

4) 2a should be enclosed in parenthesis

You also didn't substitute for all a, b, c but let that slide

Yeah, I couldn't get tex working correctly so I screwed up writting the quadratic formula. Ya got me there. But this doesn't change the fact that the solutions are -1 and -2 and you get those solutions using any of the methods I described or "your method" (which is really just using the quadratic formula everyone else uses.)

 

[agentredlum]

Yeah, I couldn't get tex working correctly so I screwed up writting the quadratic formula. Ya got me there. But this doesn't change the fact that the solutions are -1 and -2 and you get those solutions using any of the methods I described or "your method" (which is really just using the quadratic formula everyone else uses.)

We're cool.

I'll try again to explain the 'bare bones mechanics'

My definition...

If ax^2 = bx + c

Then x = (b +-sqrt(b^2 + 4ac))/(2a)

Textbook definition...

If ax^2 + bx + c = 0

Then x = (-b +-sqrt(b^2 - 4ac))/(2a)

From your post I understand you are not disputing this. You are disputing it's usefulness. That is an opinion and it's fine by me.

What I am asking you (and others) is...

'Why do you claim it is the same formula as the textbook version when clearly one can see, just by looking at it, that it isn't. All you have to do is write them down side by side and compare'

This is not an opinion but a question that can be tested by observation, experimentation, and mathematical arguments.

Then I point out 2 MAJOR differences.

1) The idea that you must set the equation to zero is a myth because (as you agree) my version gives the correct answers and it is not set equal to zero.

2) My formula uses less symbols, in the definition and in the formula itself.So my derivation answers the following question...

'Do you have to set it equal to zero to find the roots?'

The old way of thinking says 'You have to'

My derivation says 'You don't have to'

It brings into question the long standing belief that setting the equation to zero is an unquestionable procedure.

This is not an opinion. One can put the 2 definitions side by side, textbook version, my version.

To bring into question a long standing belief, by providing evidence in the form of a derivation as I have done,
is not a trivial matter.

Let me be absolutely clear. I do not question the validity of the textbook version. I question the statement 'you must set it equal to zero' because of the word 'must'

 

[Robert1986]

We're cool.

I'll try again to explain the 'bare bones mechanics'

My definition...

If ax^2 = bx + c

Then x = (b +-sqrt(b^2 + 4ac))/(2a)

Textbook definition...

If ax^2 + bx + c = 0

Then x = (-b +-sqrt(b^2 - 4ac))/(2a)

From your post I understand you are not disputing this. You are disputing it's usefulness. That is an opinion and it's fine by me.

What I am asking you (and others) is...

'Why do you claim it is the same formula as the textbook version when clearly one can see, just by looking at it, that it isn't. All you have to do is write them down side by side and compare'

Yes, if you look at the two formulas, they are different. But you are starting with -b instead of b and -c instead of c. Your formula, then, is no surprise. It is a mind-numbingly simple arithmetic "trick" that, in practice, does nothing.

Since you have -b instead of b, then it is natural that -(-b) = b, which is what you have in your formula.

This is not an opinion but a question that can be tested by observation, experimentation, and mathematical arguments.

Then I point out 2 MAJOR differences.

1) The idea that you must set the equation to zero is a myth because (as you agree) my version gives the correct answers and it is not set equal to zero.
[\QUOTE]

But you ARE setting the quadratic to 0, then just shuffling stuff to the other side of the "=". There is no deep insight here (there is no shallow insight, either). This is 4th grade math.

2) My formula uses less symbols, in the definition and in the formula itself.

Yeah, but to get to your equation, you have to multiply b and c by -1 in the general quadratic equation, so what's the point?

So my derivation answers the following question...

'Do you have to set it equal to zero to find the roots?'

The old way of thinking says 'You have to'

My derivation says 'You don't have to'

But your derivation DOES set it equal to 0, then you just shuffle stuff to the right of the "=". Again, this is 4th grade stuff.

It brings into question the long standing belief that setting the equation to zero is an unquestionable procedure.

This is not an opinion. One can put the 2 definitions side by side, textbook version, my version.

To bring into question a long standing belief, by providing evidence in the form of a derivation as I have done,
is not a trivial matter.

Let me be absolutely clear. I do not question the validity of the textbook version. I question the statement 'you must set it equal to zero' because of the word 'must'

Oh boy; yes, moving stuff to the right side of an equation is real ground-breaking stuff.

 

[Robert1986]

You want to solve:

ax^2 + bx + c = 0 by doing this:

ax^2 = bx + c and then using this formula:

x = \frac{b +- \sqrt{b^2 + 4ac}}{2a}[\tex]<br /> <br /> right?<br /> <br /> OK, well, this equation:<br /> <br /> ax^2 = bx + c is just this equation:<br /> <br /> ax^2 + bx + c = 0 with the b and c multiplied by -1. And your quadratic formula is just the text-book quadratic formula with the b and c multiplied by -1. <br /> <br /> As I have been trying to explain, your method does nothing. Once you plug the actual numbers into the two quadratic formulas, they are the SAME THING. Go ahead, try it with symbols. This is why I say it is useless and nothing new or deep.

 

[agentredlum]

Yes, if you look at the two formulas, they are different. But you are starting with -b instead of b and -c instead of c. Your formula, then, is no surprise. It is a mind-numbingly simple arithmetic "trick" that, in practice, does nothing.

Since you have -b instead of b, then it is natural that -(-b) = b, which is what you have in your formula.

This is not an opinion but a question that can be tested by observation, experimentation, and mathematical arguments.

Then I point out 2 MAJOR differences.

1) The idea that you must set the equation to zero is a myth because (as you agree) my version gives the correct answers and it is not set equal to zero.
[\QUOTE]

But you ARE setting the quadratic to 0, then just shuffling stuff to the other side of the "=". There is no deep insight here (there is no shallow insight, either). This is 4th grade math.

Yeah, but to get to your equation, you have to multiply b and c by -1 in the general quadratic equation, so what's the point?
But your derivation DOES set it equal to 0, then you just shuffle stuff to the right of the "=". Again, this is 4th grade stuff.

Oh boy; yes, moving stuff to the right side of an equation is real ground-breaking stuff.

Did you look at post #1? You are only concentrating on the shortcut. In post #1 i never set it equal to zero, which invalidates all of your counter reasoning above.

In the 'shortcut' post i used the well known form that is ALREADY set equal to zero so of course one can say that is NOT groundbreaking.

But in post #1 i set it equal to bx + c, that is not trivial as you believe.

You don't understand. Because of the way in which the subject is learned, it is not trivial for someone to proceed in this direction, however, the derivation in post #1 is correct and shows you do not have to set it equal to zero in order to derive a working quadratic formula.

Let's put arguments to the side. I have 1 question for you.

Do you believe the equation must be set equal to zero in order to find the roots?

 

[micromass]

Do you believe the equation must be set equal to zero in order to find the roots?

Yes. And I'll explain by a counter-question.

What is the definition of a root?

 

[agentredlum]

You want to solve:

ax^2 + bx + c = 0 by doing this:

ax^2 = bx + c and then using this formula:

x = \frac{b +- \sqrt{b^2 + 4ac}}{2a}[\tex]<br /> <br /> right?<br /> <br /> OK, well, this equation:<br /> <br /> ax^2 = bx + c is just this equation:<br /> <br /> ax^2 + bx + c = 0 with the b and c multiplied by -1. And your quadratic formula is just the text-book quadratic formula with the b and c multiplied by -1. <br /> <br /> As I have been trying to explain, your method does nothing. Once you plug the actual numbers into the two quadratic formulas, they are the SAME THING. Go ahead, try it with symbols. This is why I say it is useless and nothing new or deep.

<br /> <br /> 0.0) I don&#039;t set it equal to zero, you do.<br /> <br /> 0.1) You have to keep track of an extra symbol, zero, I don&#039;t so you&#039;re moving around more objects than I am. As a matter of fact, even if the equation given to me is set equal to zero, I get rid of zero, by a clever &#039;kindergarten&#039; argument you don&#039;t seem to understand<br /> <br /> 0.2) Your formula has an extra minus sign in front of b which you need to keep track of. This minus sign has nothing to do with a particular equation that has numerical values for a, b, c. The minus sign is a consequence of the definition set equal to zero, and the subsequent derivation by completing the square.<br /> <br /> etc.,etc.,etc.

 

[agentredlum]

Yes. And I'll explain by a counter-question.

What is the definition of a root?

It's trivial

 

[micromass]

It's trivial

So I'm right, then...

 

[agentredlum]

So I'm right, then...

Yes but it's sad to me because you do not want to see beyond definitions.

Now, I have become the supreme onion, the saddener of all worlds...

 

[agentredlum]

Some posters have pointed out that my equation can be set equal to zero. This is a trivial statement. Any equation can be set equal to zero. Some posters have argued adamantly that this is 'reeealy' what I'm doing. This is a misunderstanding, I am not doing that. I am doing the exact opposite, in my derivations using completing the square, Tschirnhaus transformation and the numerical examples presented. Even in the 25% chance that an equation is given already set equal to zero, I get rid of zero and still achieve the correct answers.

Forget about the shortcut, it is not a derivation that can stand on it's own because it uses THE OTHER definition. I would have been impressed if someone pointed THAT out, but no one did. Instead some posters used it to diminish the worth of my result and others agreed with them. I was aware of the shortcomings of the short cut, pun intended, that's why I didn't use it in post #1

Nevertheless, the shortcut is an interesting curiosity and I'm glad I posted it even though it has stirred a lot of controversy. I say again, the shortcut cannot stand very well on its own.

Many thanks for all the responses, let's keep it civilized, o-k?

 

[Robert1986]

OK, perhaps I have misunderstood something, is this what you do to solve a quadratic:

Given this quadratic: ax^2 + bx + c = 0 you write this equation:

ax^2 = bx + c

Am I right so far (forget about where I set it equal to zero, just ignore that if you want, it isn't important; I only wrote it because it isn't an equation without it.)

Then you use this quadratic formula:

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

So do I still understand what you are doing?

What I am saying is that you haven't done anything. You have taken the original equation:

ax^2 + bx + c = 0 then you multiplied b and c by -1 to get:

ax^2 - bx -x = 0 (which is 100% identical to what you wrote, which you correctly call "trivial") and solved it using the normal text-book quadratic formula, AFTER you multiplied b and c by -1 in the text-book quadratic formula. Can't you see that your quadratic formula and the text-book quadratic formula only differ in that the b and c have been multiplied by -1 in yours, which accounts for the fact that you have multiplied b and c by -1 in your quadratic equation (you never said you were doing this, but you have, even if you don't realize it.)

It doesn't matter if you start with:

ax^2 = bx + c

and then solve for x from there by completing the square and doing other stuff. It doesn't matter becuase without knowing what a,b and c are, you can multiply them by anything (other than 0) and as long as you make the appropriate adjustments in the quadratic formula (which you did) you get the right answers. This isn't rocket science.

 

[Mark44]

Thank you for the response. That's not what I'm doing in the derivation and if you try to do this for a particular example it would fail to give the right answer using my formula.

No, it would not.

Here is a simple example, solved in the usual way and solved in your way.
1. x² + 3x + 2 = 0
I would normally just factor the quadratic expression on the left, but for this example I will use the Quadratic Formula.

x = \frac{-3 \pm \sqrt{3^2 - 4*1*2}}{2*1} = \frac{-3 \pm 1}{2}
So x = -2 or x = -1

2. x² = -3x - 2
This equation is equivalent to the equation in #1.
Here you are calling the coefficient of x, b and the constant term c, so b = -3 and c = -2. By your formula
x = \frac{(-3) \pm \sqrt{3^2 + 4*1*(-2)}}{2*1} = \frac{-3 \pm 1}{2}
So x = -2 or x = -1, exactly the same as before.

The reason I used d and f was to lessen the possible confusion that results in identifying different numbers with the same variable.

You see if you replace b by -d and f by -c as you claim then you get back ax^2 + bx + c = 0 so this method hasn't DONE anything. The subsequent derivation will yield the textbook quadratic formula.

 

[Mark44]

Then I point out 2 MAJOR differences.

1) The idea that you must set the equation to zero is a myth because (as you agree) my version gives the correct answers and it is not set equal to zero.

No mathematician will say that you must set any equation to zero. You can set some expression to zero, but an equation is not something that has a numerical value.

Here is a simple equation: 2x = 4
If I set this equation to zero (whatever that means), is this what I get?
2x = 4 = 0

Clearly that doesn't make any sense.

As others and I have pointed out, all that you are doing is working with a slightly different (but equivalent) equation. Equivalent equations have exactly the same solutions.

Starting from x² + 3x + 2 = 0, I can quickly find that the solutions are x = -2 and x = -1.

If you start with x² = 3x + 2, then you are working with a different equation, so you will get completely different solutions. However, if you start with x² = -3x - 2 and use your revised quadratic formula, you will get the same solutions that I showed above, because the two quadratic equations are equivalent.

2) My formula uses less symbols, in the definition and in the formula itself.

This is true, but of marginal importance IMO.

To summarize, your technique has ONE relatively minor difference, not TWO MAJOR differences.

So my derivation answers the following question...

'Do you have to set it equal to zero to find the roots?'

The old way of thinking says 'You have to'

My derivation says 'You don't have to'

It brings into question the long standing belief that setting the equation to zero is an unquestionable procedure.

Again, setting any equation to zero is neither a longstanding belief nor unquestionable procedure. No mathematician would tell you to set an equation to zero.

 

[sankalpmittal]

OK, perhaps I have misunderstood something, is this what you do to solve a quadratic:

Given this quadratic: ax^2 + bx + c = 0 you write this equation:

ax^2 = bx + c

Am I right so far (forget about where I set it equal to zero, just ignore that if you want, it isn't important; I only wrote it because it isn't an equation without it.)

Then you use this quadratic formula:

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

So do I still understand what you are doing?

What I am saying is that you haven't done anything. You have taken the original equation:

ax^2 + bx + c = 0 then you multiplied b and c by -1 to get:

ax^2 - bx -x = 0 (which is 100% identical to what you wrote, which you correctly call "trivial") and solved it using the normal text-book quadratic formula, AFTER you multiplied b and c by -1 in the text-book quadratic formula. Can't you see that your quadratic formula and the text-book quadratic formula only differ in that the b and c have been multiplied by -1 in yours, which accounts for the fact that you have multiplied b and c by -1 in your quadratic equation (you never said you were doing this, but you have, even if you don't realize it.)

It doesn't matter if you start with:

ax^2 = bx + c

and then solve for x from there by completing the square and doing other stuff. It doesn't matter becuase without knowing what a,b and c are, you can multiply them by anything (other than 0) and as long as you make the appropriate adjustments in the quadratic formula (which you did) you get the right answers. This isn't rocket science.

Exactly , your explanation is concurable .
agentredlum's formula is nothing new but I praise his innovation .
His formula evaluates the same thing as the general textbook formula .

What he did is
ax²+bx+c=0
He replaced b with -b and c with -c.
Thus ,
ax²-bx-c=0

Now in general textbook quadratic formula :
x=-b+-sqrt(b²-4ac)/2a

He replaced it with his changes ,
Hence ,
x=-(-b)+-sqrt(-b²-4(a)(-c))/2a
or₁
x=b+-sqrt(b²+4ac)/2a


What if I say replace a with -a and c with -c !
ax²+bx+c=0
-ax²+bx-c=0!
x=-b+-sqrt(b²-4ac)/2a

If I replaced it with my changes ,
Hence ,
x=-b+-sqrt(b²-4(-a)(-c))/2(-a)
or₁
x=-b+-sqrt(b²-4ac)/2(-a)

x²+3x+2=0

hence with my changes ,
-x²+3x-2=0


x=-b+-sqrt(b²-4(-a)(-c))/2(-a)
x=-3+-sqrt(3²-4(-1)(-2))/2(-1)
x=-3+-1/-2
x=2 or x=1

which is the wrong answer .
Hence we see there is loss of generality . Thus this condition is only true for replacing b with -b and c with -c without any change in a .


Well done ,
agentredlum

 

[Robert1986]

Exactly , your explanation is concurable .
agentredlum's formula is nothing new but I praise his innovation .

But this ISN'T an "innovation." I don't think many people would call multiplying two numbers by -1 an innovation.

His formula evaluates the same thing as the general textbook formula .

What he did is
ax²+bx+c=0
He replaced b with -b and c with -c.
Thus ,
ax²-bx-c=0

Now in general textbook quadratic formula :
x=-b+-sqrt(b²-4ac)/2a

He replaced it with his changes ,
Hence ,
x=-(-b)+-sqrt(-b²-4(a)(-c))/2a
or₁
x=b+-sqrt(b²+4ac)/2a


What if I say replace a with -a and c with -c !
ax²+bx+c=0
-ax²+bx-c=0!
x=-b+-sqrt(b²-4ac)/2a

If I replaced it with my changes ,
Hence ,
x=-b+-sqrt(b²-4(-a)(-c))/2(-a)
or₁
x=-b+-sqrt(b²-4ac)/2(-a)

x²+3x+2=0

hence with my changes ,
-x²+3x-2=0


x=-b+-sqrt(b²-4(-a)(-c))/2(-a)
x=-3+-sqrt(3²-4(-1)(-2))/2(-1)
x=-3+-1/-2
x=2 or x=1

which is the wrong answer .
Hence we see there is loss of generality . Thus this condition is only true for replacing b with -b and c with -c without any change in a .

Yeah, I'm well aware of what he is doing, and I have explained it several times: he is multiplying b and c by -1 in the original quadratic and in the text-book (ie normal) quadratic formula.

I don't see how this can even be of any practical value, either.

 

[agentredlum]

OK, perhaps I have misunderstood something, is this what you do to solve a quadratic:

Given this quadratic: ax^2 + bx + c = 0 you write this equation:

ax^2 = bx + c

Am I right so far (forget about where I set it equal to zero, just ignore that if you want, it isn't important; I only wrote it because it isn't an equation without it.)

Then you use this quadratic formula:

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

So do I still understand what you are doing?

What I am saying is that you haven't done anything. You have taken the original equation:

ax^2 + bx + c = 0 then you multiplied b and c by -1 to get:

ax^2 - bx -x = 0 (which is 100% identical to what you wrote, which you correctly call "trivial") and solved it using the normal text-book quadratic formula, AFTER you multiplied b and c by -1 in the text-book quadratic formula. Can't you see that your quadratic formula and the text-book quadratic formula only differ in that the b and c have been multiplied by -1 in yours, which accounts for the fact that you have multiplied b and c by -1 in your quadratic equation (you never said you were doing this, but you have, even if you don't realize it.)

It doesn't matter if you start with:

ax^2 = bx + c

and then solve for x from there by completing the square and doing other stuff. It doesn't matter becuase without knowing what a,b and c are, you can multiply them by anything (other than 0) and as long as you make the appropriate adjustments in the quadratic formula (which you did) you get the right answers. This isn't rocket science.

I will go step by step and with respect to help everyone understand using this post because it is a good one to dispel the misconceptions. I am not 'picking' on you robert and I understand the 'No freaking way!' Idea.

I would like to adress the first 4 lines of your quote above.

I CANNOT take ax^2 + bx + c = 0 as a given, because it relies on the OTHER definition. Do you understand? You ask me to forget about it, but i cannot because it is the first step in the derivation and the most important.

So, where do i get my equation? I explained it in a previous post, I'll explain it again. I consider equivalent forms. Careful here, EQUIVALENT, not EQUAL.

I consider 4 equivalent forms for my derivation. Each form, by itself, gives all possible 2nd degree equations applicable to completing the square. Each form is detached from the other 3 because no 2 are identically equal.

The 4 forms are

1) ax^2 + bx + c = 0

2) ax^2 + bx = c

3) ax^2 + c = bx

4) ax^2 = bx + c

I picked #4, completed the square and derived a different quadratic formula.

Then I made a NEW definition.

If ax^2 = bx + c

Then x = (b +-sqrt(b^2 + 4ac))/(2a)

That's it. Now posters continue to say I used #1. NO I DID NOT. I did NOT multiply by -1 anywhere in the derivation, explicitly or implicitly, I did not multiply by -1 anywhere in the forms above, implicitly, or explicitly. I did not multiply by -1 in my Quadratic formula. My derivation stands alone and does not rely on ANY of the other 3 forms in ANY way. I hope it's clear to everyone now.

I NEVER said take b, c, and multiply them by -1. I said b is INVARIANT to -b and c is INVARIANT to -c. It's not the same thing! b is NOT equal to -b, c is NOT equal to -c (except for zero) but any real number can be represented by b or -b equaly well without loss of generality, same goes for c and -c. This is the way I used the idea of invariance. CAREFUL, 1 is NOT invariant to -1. The idea of invariance only works when considering generalities, NOT when you pick 2 different members of the set of real numbers.

What I said is 'If you want to find the roots, isolate the ax^2 term first, THEN identify a, b, c, and use this new formula that i derived for you.'

Thank you for the responce.

 

[agentredlum]

No, it would not.

Here is a simple example, solved in the usual way and solved in your way.
1. x² + 3x + 2 = 0
I would normally just factor the quadratic expression on the left, but for this example I will use the Quadratic Formula.

x = \frac{-3 \pm \sqrt{3^2 - 4*1*2}}{2*1} = \frac{-3 \pm 1}{2}
So x = -2 or x = -1

2. x² = -3x - 2
This equation is equivalent to the equation in #1.
Here you are calling the coefficient of x, b and the constant term c, so b = -3 and c = -2. By your formula
x = \frac{(-3) \pm \sqrt{3^2 + 4*1*(-2)}}{2*1} = \frac{-3 \pm 1}{2}
So x = -2 or x = -1, exactly the same as before.

The reason I used d and f was to lessen the possible confusion that results in identifying different numbers with the same variable.

Right, you isolated the ax^2 term, identified a, b, c, used my formula and got the right answers. Well done!

I misunderstood what you were saying in your previous post.

I am calling the coefficient of x, b and the constant term c AFTER i isolate the ax^2 term, because i DEFINED the equation that way and then completed the square, not because I'm multiplying by -1 and certainly not before i isolate ax^2. b means nothing to me until i isolate ax^2. The same goes for the textbook definition, we can't use b until we isolate 0.

Thanks for the response.

 

[Mark44]

I CANNOT take ax^2 + bx + c = 0 as a given, because it relies on the OTHER definition. Do you understand? You ask me to forget about it, but i cannot because it is the first step in the derivation and the most important.

So, where do i get my equation? I explained it in a previous post, I'll explain it again. I consider equivalent forms. Careful here, EQUIVALENT, not EQUAL.

Yes, and this was the point I made before where you were talking about setting equations equal to zero. I was very careful to talk about equivalent equations.

I consider 4 equivalent forms for my derivation.

What do you mean by "equivalent forms"? You couldn't possibly mean "equivalent equations", because none of the equations is equivalent to any of the others.

It would be better to say "4 forms".

Each form, by itself, gives all possible 2nd degree equations applicable to completing the square. Each form is detached from the other 3 because no 2 are identically equal.

Identically equal to each other? That doesn't make any sense because equations aren't equal (identically or otherwise) to other equations. Two equations can be equivalent, but no two of the equations below are equivalent because of how you are defining b and c.

The 4 forms are

1) ax^2 + bx + c = 0

2) ax^2 + bx = c

3) ax^2 + c = bx

4) ax^2 = bx + c

I picked #4, completed the square and derived a different quadratic formula.

Then I made a NEW definition.

If ax^2 = bx + c

Then x = (b +-sqrt(b^2 + 4ac))/(2a)

We get it already. You are defining a new standard form for quadratic equations, which necessarily causes a couple of changes in your revised quadratic formula. IMO, not that big a deal.

I NEVER said take b, c, and multiply them by -1. I said b is INVARIANT to -b and c is INVARIANT to -c. It's not the same thing! b is NOT equal to -b, c is NOT equal to -c (except for zero) but any real number can be represented by b or -b equaly well without loss of generality, same goes for c and -c. This is the way I used the idea of invariance. CAREFUL, 1 is NOT invariant to -1. The idea of invariance only works when considering generalities, NOT when you pick 2 different members of the set of real numbers.

This whole paragraph makes no sense to me. Invariance is already defined in mathematics, and you seem to be using your own definition. Please define this term for us as you mean it.

 

[agentredlum]

No mathematician will say that you must set any equation to zero. You can set some expression to zero, but an equation is not something that has a numerical value.

Here is a simple equation: 2x = 4
If I set this equation to zero (whatever that means), is this what I get?
2x = 4 = 0

Clearly that doesn't make any sense.

As others and I have pointed out, all that you are doing is working with a slightly different (but equivalent) equation. Equivalent equations have exactly the same solutions.

Starting from x² + 3x + 2 = 0, I can quickly find that the solutions are x = -2 and x = -1.

If you start with x² = 3x + 2, then you are working with a different equation, so you will get completely different solutions. However, if you start with x² = -3x - 2 and use your revised quadratic formula, you will get the same solutions that I showed above, because the two quadratic equations are equivalent.

This is true, but of marginal importance IMO.

To summarize, your technique has ONE relatively minor difference, not TWO MAJOR differences.

Again, setting any equation to zero is neither a longstanding belief nor unquestionable procedure. No mathematician would tell you to set an equation to zero.

No, that's not what i mean. If you set 2x = 4 to zero you get 2x - 4 =0

That is the sense in which I am talking about it here and I have heard it in this sense from many mathematicians and scientists. So i don't understand why you make this point twice. From now on i'll just use isolate zero instead of set equal to zero.

If you ask a math professor 'how do i solve 3x^2- 2x = 17 using the quadratic formula?' the first thing he/she will say is set it equal to zero, or collect everything on one side of the equation, or any words you feel are more appropriate. The point is the professor tells you to put it in a form where zero is on one side, everything else on the other.

Thank you for your response.

 

[pwsnafu]

I NEVER said take b, c, and multiply them by -1. I said b is INVARIANT to -b and c is INVARIANT to -c. It's not the same thing! b is NOT equal to -b, c is NOT equal to -c (except for zero) but any real number can be represented by b or -b equaly well without loss of generality, same goes for c and -c. This is the way I used the idea of invariance. CAREFUL, 1 is NOT invariant to -1. The idea of invariance only works when considering generalities, NOT when you pick 2 different members of the set of real numbers.

Invariance is always "with respect to an operation". You say that a real number b is invariant to -b. The only operation which takes b to -b is "multiplication by -1".

Using a different notation (or denoting something with a different symbol) is not what invariance means.

 

[agentredlum]

Yes, and this was the point I made before where you were talking about setting equations equal to zero. I was very careful to talk about equivalent equations.What do you mean by "equivalent forms"? You couldn't possibly mean "equivalent equations", because none of the equations is equivalent to any of the others.

This whole paragraph makes no sense to me. Invariance is already defined in mathematics, and you seem to be using your own definition. Please define this term for us as you mean it.

I want to say...Using one form does not depend on using another. What is a better way to say that?

I want to say all real numbers can be represented by c or -c. How should i say that?

 

[agentredlum]

Invariance is always "with respect to an operation". You say that a real number b is invariant to -b. The only operation which takes b to -b is "multiplication by -1".

Using a different notation (or denoting something with a different symbol) is not what invariance means.

No that's not what I mean. -b can cycle through all real numbers just as well as b. I explained this in a previous post a few days ago. If you are going to be pedantic, I'm going to have to write a book on every post. A little latitude please?