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

[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?

 

[agentredlum]

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)/2aWhat 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=0x=-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

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

The solutions you got satisfy the equation -x^2 + 3x - 2 = 0 which is on the line after you say 'hence my changes' . Do you know why?

There is more to this than other posters think there is. I will derive a quadratic formula where a is replaced by -a and c is replaced by -c And post it.

In the meantime, think about what happened in your counter-example, concidence? or something deeper.

I want to stress again. The shortcut cannot stand on it's own.

Thank you again.

[edit] Are you sure you didn't make a mistake? Please check your work again.

 

[Mark44]

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

No, you're not setting 2x = 4 to zero. You are rewriting the equation as an equivalent equation.

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.

I don't believe that a mathematician would say that you are setting an equation to zero.

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.

"Collect all the terms on one side" is accurate; "set it equal to zero" is not.

My point in being pedantic about this is that you were claiming that one of the important advantages in your technique is that you don't have to set the equation to zero, which is a meaningless thing to do.

 

[pwsnafu]

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

You just did.

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?

This isn't about being pedantic. You used a mathematical term incorrectly, and people have criticized you for it.

 

[Mark44]

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.

If you are going to misuse commonly understood definitions, we have to be pedantic.

A little latitude please?

This is the way mathematics works. If you propose a theorem, you need to be able to support it by way of definitions and other theorems. If you are using different definitions than the rest of us, you're going to be called on them.

 

[agentredlum]

No, you're not setting 2x = 4 to zero. You are rewriting the equation as an equivalent equation.
I don't believe that a mathematician would say that you are setting an equation to zero.
"Collect all the terms on one side" is accurate; "set it equal to zero" is not.

My point in being pedantic about this is that you were claiming that one of the important advantages in your technique is that you don't have to set the equation to zero, which is a meaningless thing to do.

So what do you want from me now?

 

[agentredlum]

You just did.
This isn't about being pedantic. You used a mathematical term incorrectly, and people have criticized you for it.

What do you want me to say?

I already gave you a definition, 2 derivations and a formula. WHAT MORE DO YOU WANT?

 

[agentredlum]

Set the equation to zero is common usage where i come from.

I explained my use of the word invariance in many post's. I never claimed to use some textbook definition of invariance.

You are focussing on minutia to prove your opinion that a particular result is worthless and has no application. Thats just an opinion. You can't know about every possible application.

But here is what you ARE doing. You are giving the result a bad reputation and misleading others about it.

80% of arguments made by 'opposers' are flawed. 90% of responders have no clue about what I did and attempt to explain it away as a 'trick' AFTER I POSTED THE TRICK!

Where were you all for the first 2 months and 10000+ views?

Now you show up and 'generalize' the result?

Why don't you come up with something original instead of using my trick for arguments against ME.

Does anyone have any tricks or curiousities to post?

 

[Robert1986]

Careful here, EQUIVALENT, not EQUAL.

Huh?

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.
[\QUOTE]

Bless your heart, you did, in fact multiply b and c by -1 implicitly. Your "fourth form"

ax^2 = bx +c is equivalent to ax^2 - bx - c = 0 which is just ax^2 +bx + c = 0 after you multiply b and c by -1.

And your quadratic formula is the text-book quadratic formula but with b and c multiplied by -1. You never said to multiply by -1 but it is the EXACT SAME thing as what you did. You took a long way to do it and did some interesting things along the way, but we come to the same result your way or "my way" (which is much easier to come t0, BTW).

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.
[\QUOTE]

I'm not sure what you mean by "invariant" here. The solutions to ax^2 = bx +c are not the same as the solutions to ax^2 + bx + c = 0 (which is the definition of roots of a polynomial). For example, the roots of x^2 + 3x + 2 are found by solving: x^2 + 3x + 2 = 0 which gives x=-1,-2. However, these are NOT roots of the polynomial x^2 - 3x -2 which means they are NOT solutions to x^2 -3x -2 = 0 which means they are NOT solutions to x^2 = 3x + 2. This means, BY DEFINITION, b and c are not invariant under multiplication by -1. Otherwise x^2 + 3x + 2 = 0 and x^2 = 3x + 2 which differ only in the fact that b and c are off by a fact of -1, would have the same exact solution set, but they don't.

You do understand that ax^2 = bx +c and x^2 + bx + c = 0 DO NOT have the same solution set, right?

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.

But why the heck would we even do that? It isn't even practical (and it doesn't really make sense). You take a quadratic (that everyone above 7th grade knows how to find the roots of) and write something else, namely ax^2 = bx + c, and then find values of x that don't satisfy the new equation but instead satisfy the old. I cannot see how this can possibly be any easier. Can you please explain, with an EXAMPLE how this makes life any easier?

 

[Robert1986]

Set the equation to zero is common usage where i come from.

I explained my use of the word invariance in many post's. I never claimed to use some textbook definition of invariance.

You are focussing on minutia to prove your opinion that a particular result is worthless and has no application. Thats just an opinion. You can't know about every possible application.

But here is what you ARE doing. You are giving the result a bad reputation and misleading others about it.

80% of arguments made by 'opposers' are flawed. 90% of responders have no clue about what I did and attempt to explain it away as a 'trick' AFTER I POSTED THE TRICK!

Where were you all for the first 2 months and 10000+ views?

Now you show up and 'generalize' the result?

Why don't you come up with something original instead of using my trick for arguments against ME.

Does anyone have any tricks or curiousities to post?

Yes, you are a modern-day Galileo. No one understand what you are doing but you are correct.


What you don't understand is that none of us (that i know of) question that your method finds roots to ax^2 + bx + c. We (well, I, and I'm guessing most people) question is a)how does this make life easier and b)why you don't realize that all you are doing is multiplying b and c by -1 in the original equation and in the quadratic formula.

 

[agentredlum]

Yes, you are a modern-day Galileo. No one understand what you are doing but you are correct.What you don't understand is that none of us (that i know of) question that your method finds roots to ax^2 + bx + c. We (well, I, and I'm guessing most people) question is a)how does this make life easier and b)why you don't realize that all you are doing is multiplying b and c by -1 in the original equation and in the quadratic formula.

Find the error in post #243 that causes the malfunction, and I will be happy to continue this debate.

Can you please do this 1 favor for me?

Thank you for your response.

 

[sankalpmittal]

Genuinely posted by Robert1986

Bless your heart, you did, in fact multiply b and c by -1 implicitly. Your "fourth form"

ax^2 = bx +c is equivalent to ax^2 - bx - c = 0 which is just ax^2 +bx + c = 0 after you multiply b and c by -1.

And your quadratic formula is the text-book quadratic formula but with b and c multiplied by -1. You never said to multiply by -1 but it is the EXACT SAME thing as what you did. You took a long way to do it and did some interesting things along the way, but we come to the same result your way or "my way" (which is much easier to come t0, BTW).

I'm not sure what you mean by "invariant" here. The solutions to ax^2 = bx +c are not the same as the solutions to ax^2 + bx + c = 0 (which is the definition of roots of a polynomial). For example, the roots of x^2 + 3x + 2 are found by solving: x^2 + 3x + 2 = 0 which gives x=-1,-2. However, these are NOT roots of the polynomial x^2 - 3x -2 which means they are NOT solutions to x^2 -3x -2 = 0 which means they are NOT solutions to x^2 = 3x + 2. This means, BY DEFINITION, b and c are not invariant under multiplication by -1. Otherwise x^2 + 3x + 2 = 0 and x^2 = 3x + 2 which differ only in the fact that b and c are off by a fact of -1, would have the same exact solution set, but they don't.

You do understand that ax^2 = bx +c and x^2 + bx + c = 0 DO NOT have the same solution set, right?But why the heck would we even do that? It isn't even practical (and it doesn't really make sense). You take a quadratic (that everyone above 7th grade knows how to find the roots of) and write something else, namely ax^2 = bx + c, and then find values of x that don't satisfy the new equation but instead satisfy the old. I cannot see how this can possibly be any easier. Can you please explain, with an EXAMPLE how this makes life any easier?

Of course yes .
ax²-bx-c=0 is not equal to ax²+bx+c=0
If we use same textbook formula viz. x=-b+-sqrt(b²-4ac)/2a
for above two representations , then answer will not come same .

But what he did is he changed general quadratic formula according to his replacement somewhat like this ,
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

Now taking a , -b and -c he got same answer in his quadratic formula .

It is nothing new , just a numb meaningless derivation according to you but it has meaning !
Strange thing is that this condition is only applicable when we replace b with -b and c with -c !
Try replacing :
1. a with -a
2. b with -b
3. a with -a and b with -b
4. c with -c
5. a with -a and c with -c.

In all above 5 conditions above the answer will not be correct . This will only work for replacement of b with -b and c with -c . Hence the real number cycle is independent of them !

Eg . 5th condition

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=0x=-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 .
This is what is meaningful and astonishing in his derivation !However , I totally agree with you in one aspect : his formula has no practical value in solving quadratic equation as it evaluates the same result as the general textbook quadratic formula .
[Edit] There is no malfunction in post 243 . I checked it . It is cent percent correct .

 

[agentredlum]

Of course yes .
ax²-bx-c=0 is not equal to ax²+bx+c=0
If we use same textbook formula viz. x=-b+-sqrt(b²-4ac)/2a
for above two representations , then answer will not come same .

But what he did is he changed general quadratic formula according to his replacement somewhat like this ,
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

Now taking a , -b and -c he got same answer in his quadratic formula .

It is nothing new , just a numb meaningless derivation according to you but it has meaning !
Strange thing is that this condition is only applicable when we replace b with -b and c with -c !
Try replacing :
1. a with -a
2. b with -b
3. a with -a and b with -b
4. c with -c
5. a with -a and c with -c.

In all above 5 conditions above the answer will not be correct . This will only work for replacement of b with -b and c with -c . Hence the real number cycle is independent of them !

Eg . 5th condition

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=0x=-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 .
This is what is meaningful and astonishing in his derivation !However , I totally agree with you in one aspect : his formula has no practical value in solving quadratic equation as it evaluates the same result as the general textbook quadratic formula .
[Edit] There is no malfunction in post 243 . I checked it . It is cent percent correct .

Sorry to bring bad news but there are a few algebraic mistakes in post # 243 and you have reproduced them in post #262

One algebraic error is particularly fatal to your counter-example. Let's see who finds the fatal error first.

Thank you for your response.

 

[sankalpmittal]

Sorry to bring bad news but there are a few algebraic mistakes in post # 3^5 and you have reproduced them in post # 2^8 + 6

One algebraic error is particularly fatal to your counter-example. Let's see who finds the fatal error first.

Thank you for your response.

There are no mistakes !
[Note added] There is general quadratic formula which works perfectly . Then why are you making it more complicated .

For example , to write 5 :
We can also write it as
5+0
4+1
5/5 x 5
1/5/1/25

This is exactly what you are doing . You are making the formula more complicated . It just evaluates same result .
[BTW Can't you just write post 243 and 262 ? I'm not overwhelmed by your mathematical refutation . You got enough praise . Please read my post 262 again. ]

 

[agentredlum]

You cannot get form4 from form1 because if you multiply by -1 you have to do it to every term on both sides of the equation otherwise the expression is not mathematically sound.

form4 stands alone, independant of the other 3.

Again I caution about the shortcut, I had good reason to replace b with -b and c with -c and my reason DID NOT involve multiplication by -1 in any way.

 

[sankalpmittal]

Argentridlum says :
You cannot get form4 from form1 because if you multiply by -1 you have to do it to every term on both sides of the equation otherwise the expression is not mathematically sound.

form4 stands alone, independant of the other 3.

Again I caution about the shortcut, I had good reason to replace b with -b and c with -c and my reason DID NOT involve multiplication by -1 in any way.

Read my post 262 again . I am replacing not multiplying , ok . To multiply with -1 , I have to do it on both sides viz. RHS and LHS
.

Read posts :
243 , 262 and 264

[btw , you still have to tell the mistake]

 

[agentredlum]

Argentridlum says :
You cannot get form4 from form1 because if you multiply by -1 you have to do it to every term on both sides of the equation otherwise the expression is not mathematically sound.

form4 stands alone, independant of the other 3.

Again I caution about the shortcut, I had good reason to replace b with -b and c with -c and my reason DID NOT involve multiplication by -1 in any way.

Read my post 262 again . I am replacing not multiplying , ok . To multiply with -1 , I have to do it on both sides viz. RHS and LHS
.

Read posts :
243 , 262 and 264

[btw , you still have to tell the mistake]

Sorry, the post was not meant for anyone who agrees multiplying by -1 is not a good idea.

 

[sankalpmittal]

Sorry, the post was not meant for anyone who agrees multiplying by -1 is not a good idea.

Ok Ok .

Your private message has been replied and forwarded to you . My substitution , in post 262 and 243 is correct . Value of a is -1 .

Do me a favour . Go through the posts 262,243 and 262 again .

 

[Robert1986]

Find the error in post #243 that causes the malfunction, and I will be happy to continue this debate.

Can you please do this 1 favor for me?

Thank you for your response.

Did I say there was an error? This is part of what was said in post 243:

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

"He replaced b with -b and c with -c." Is the SAME EXACT THING as saying "multiply b and c by -1."

and
"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"

is the SAME EXACT THING as saying "multiply b and c by -1 in the text-book quadratic formula"

both of which I have long insisted are what you are doing. Now, it seems we agree: all you are doing is multiplying b and c by -1, even if you are not doing it explicitly.

 

[sankalpmittal]

Did I say there was an error? This is part of what was said in post 243:



"He replaced b with -b and c with -c." Is the SAME EXACT THING as saying "multiply b and c by -1."

and
"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"

is the SAME EXACT THING as saying "multiply b and c by -1 in the text-book quadratic formula"

both of which I have long insisted are what you are doing. Now, it seems we agree: all you are doing is multiplying b and c by -1, even if you are not doing it explicitly.

Hiii Robert ,

Please go through my posts 262 , 264 and 243 .

 

[Robert1986]

Can I just get two questions answered, so that I understand a little more:

a) Do you think that the following equations have the same solution set:

ax^2 + bx + c = 0

and

ax^2 = bx + c

b)please post an example which uses this trick to make my life easier. I still do not see how finding the roots of ax^2 + bx + c is made easier by writting ax^2 = bx +c and then using your quadratic formula which does not give solutions to ax^2 = bx + c but, in fact, gives solutions to ax^2 + bx + c = 0. How do the added steps make it easier than just using the original quadratic formula from the get-go?