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

[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?

 

[Robert1986]

IMO, about 80% of the confusion in this thread has been caused by your inability to use standard terminology to explain what you mean. If you don't mind, I would like to re-state what you have said, but in a way that math people might better understand:

Given the quadratic ax^2 + bx +c, we all know how to find the roots with the quadratic formula. The following method also works: let b' = -b and c' = -c and use this formula:

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

Proof:

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a] = x = [-b +- sqrt(b^2 -4ac)]/[2a] which is the quadratic formula associated with ax^2 + bx + c. QED.


If you had said that, I would have understood exactly what you mean much sooner. Now, you might not have used the same method to derive your formula, but that doesn't matter. Usually, when giving a proof, you give the most efficient and easily understood proof, which is what I did above. For example, Gauss would never really explain why something was true. He would just state something and then prove it. He felt that all written theorems should appear as though they just came from the brow of the author of the proof.


Now, let's move on to this thing about not having to set and equation equal to 0 (in your terms) to find the roots. When someone says to find the roots of a polynomial p(x) it means to find the value of x such that p(x) = 0. This is by definition. It confuses us when you write down a completley new polynomial, say p'(x), and then describe a method that does not find the roots of p'(x) but of the original p(x).


The polynomials p(x) and p'(x) don't have the same solutions so when you start with this:

ax^2 + bx + c = 0

then

ax^2 = bx + c

doesn't make sense. Saying that these two equations are equivilant (which, BTW, is no difference than saying they are equal) is simply not true.


Now, please explain how your method makes life easier.

 

[Robert1986]

I should also point out that in ax^2 + bx + c, b and c are not variables, they are constants so they are not "cycling" through anything.

 

[agentredlum]

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?

If a, b, c are particular values, example a = 1 b = 2 c = 3 then the equations above DO NOT have the same solution set. I never claimed the 2 forms above have the same solution set for PARTICULAR values.

If you consider a, b, c, abstractly, as running through all real numbers then the 2 forms above have the same solution set. TAKEN AS A WHOLE SET. I emphasize this last part because it is very important for the definition and derivation that I did.

To post 1 example would not be enough. I have to post 7, to show the way signs effect both formulas, a total of 14 calculations. I have been dreading this but i will do it. Give me some time.

My argument was the textbook definition and formula has more symbols than my version so that makes my version simpler.

 

[Robert1986]

If a, b, c are particular values, example a = 1 b = 2 c = 3 then the equations above DO NOT have the same solution set. I never claimed the 2 forms above have the same solution set for PARTICULAR values.

If you consider a, b, c, abstractly, as running through all real numbers then the 2 forms above have the same solution set. TAKEN AS A WHOLE SET. I emphasize this last part because it is very important for the definition and derivation that I did.

I don't even know what you mean by this. If by "solution set" you mean that the text-book quadratic formula associated with ax^2 + bx + c = 0 and your quadratic formula associated with ax^2 = bx +c give the same answers, then yes. But this is not what people mean by solution set.

And I have no idea what gave you the impression that I wasn't look at the whole solution set.

 

[agentredlum]

I will post an example that supports my case without question.

Problem: Solve for x, x^2 = 2x + 5

Teextbook version solution...

Step1) Move all terms to one side of the equation.

x^2 - 2x - 5 = 0

Step2) Identify a, b, c, a = 1 b = -2 c = -5

Step3) Plug into x = (-b +-sqrt(b^2 - 4ac))/(2a)

x = (-(-2) +-sqrt((-2)^2 - 4(1)(-5))/(2(1))

Step4) Simplify

x = (2 +-sqrt(4 + 20))/2

Let's stop here because the rest is the same for both formulas.

My version solution...

Step1) Isolate ax^2 term

No manipulations necessary

Step2) Identify a, b, c, a =1 b = 2 c = 5

Step3) Plug into x = (b +-sqrt(b^2 + 4ac))/(2a)

x = ((2) +-sqrt((2)^2 + 4(1)(5))/(2(1))

Step4) Simplify

x = (2 +-sqrt(4 + 20))/2

Let's stop here and compare the number of calculations at each step.

In step one textbook version you performed 2 subtractions and added an extra symbol, zero.

I did nothing in my version.

In step 2 you identified a = 1 b = -2 c = -5

In step 2 i identified a = 1 b = 2 c = 5

you now have 2 extra minus signs that you have to carry over to step 3. I don't have that problem.

In step 3 We both substituted I had positive numbers, you had a few negatives, your formula also contains an extra negative.

Step 4 you had to compute -(-2) i had to compute (2) you have 2 extra minus signs

you had to compute (-2)^2 so you had to square the minus sign i had to compute (2)^2

you had to compute 4(1)(-5) and then subtract from b^2 I had to compute 4(1)(5) and then add to b^2 again you had that pesky minus sign and subsequent subtraction forces you to do addition. I didn't have that problem.

So you did 3 extra things in step1, 2 extra in step 2, and at least 4 extra in step 4. I won't count step 3 cause i want to be fair to you.

Satisfied now?

 

[Robert1986]

OK, now I see where you are comming from.

IF you are given an equation in this form:

ax^2 = bx + c

then in some sense your way is easier (though, marginally so, anyone who does a lot of math wouldn't actually move the stuff around as you suggested, they would just know what a, b and c are by looking at the equation) but it requires one to memorize another quadratic formula when the text-book one works just fine given one has enough experience to make simple algebraic manipulations in one's head.


But, I assume you would still use the text-book quadratic formula if given this:

ax^2 + bx + c = 0, right?


Also, did you look what I posted about explaining this more clearly? Did what I write make sense to you, I mean, did my way of getting this result make sense to you?

For me, I don't really see a benefit. If you do, that's fine; I don't think you're insane if it helps you. For me, the extra negatives really don't pose much of a problem.

 

[agentredlum]

OK, now I see where you are comming from.

IF you are given an equation in this form:

ax^2 = bx + c

then in some sense your way is easier (though, marginally so, anyone who does a lot of math wouldn't actually move the stuff around as you suggested, they would just know what a, b and c are by looking at the equation) but it requires one to memorize another quadratic formula when the text-book one works just fine given one has enough experience to make simple algebraic manipulations in one's head. But, I assume you would still use the text-book quadratic formula if given this:

ax^2 + bx + c = 0, right?Also, did you look what I posted about explaining this more clearly? Did what I write make sense to you, I mean, did my way of getting this result make sense to you?

For me, I don't really see a benefit. If you do, that's fine; I don't think you're insane if it helps you. For me, the extra negatives really don't pose much of a problem.

Finally, but it took much more than half an hour LOL. That's because we're not in the same room talking.

I looked at what you wrote and I don't see a problem with it initially, except you put a b^2 where it didn't belong, but that doesn't matter to me because it did not affect your final result.

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a]

The rest of it seems fine to me but I must caution you, I am not as rigorous as others. To me, it seems a tiny bit like circular reasoning but I am no expert in logic. Let others more qualified make their assessment of your proof.

I wanted to avoid any connection to the textbook version, that's why i considered the 4 forms, made my arguments, definitions, and derivations.

If i had an equation to solve in the form ax^2 + bx + c = 0 Then I would use the textbook version.

If you gave me 10 equations to solve in the form ax^2 = bx + c then I would feel silly using the textbook version because i know a better version for this form. My version. Imagine... sitting there with pencil and paper, collecting terms on one side, carrying around un-necessary minus signs, transforming subtractions to additions, squaring negatives, on and on, for 10 equations. Or better yet, ask a professional mathematician, who doesn't know, or won't accept my version because it's 'trivial', to do 10 equations by hand, would you chuckle as you watch him struggle?

If you were throwing random forms at me, I wouldn't know what i was going to get next out of possible

form1 ax^2 + bx + c = 0

form2 ax^2 + bx = c

form3 ax^2 + c = bx

form4 ax^2 = bx + c

And you asked me to solve 100 random forms or 1000 or 10^6 random 2nd degree equations, Then I would use my version because I would expect 75% of the forms thrown at me would not be form1

Don't forget that the textbook version has an extra minus sign and subtraction instead of addition. Those 2 differences are going to cause more problems. Does that make sense?

Thank you for your reply, thank you for your comments.

 

[agentredlum]

I have 2 questions about -b and - 4ac in the formula...

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

If you are given 2nd degree equations at random (that can be solved using above formula)

Question1: What percent of the equations given would you expect to compute a double negative for -b?

Question2: What percent of the equations given would you expect to compute a triple negative for - 4ac?



For instance if b is -5 then -(-5) must be evaluated, how often would you expect to see 2 negatives?

If a is -2 and c is -3 then - 4(-2)(-3) must be evaluated, how often would you expect to see 3 negatives?

 

[Char. Limit]

Wait... so all of this is to avoid minus signs? Are people really that scared of a -4ac?

 

[micromass]

Wait... so all of this is to avoid minus signs? Are people really that scared of a -4ac?

Minusphobia...

 

[Robert1986]

Wait... so all of this is to avoid minus signs?

Yeah, and here's the funny thing: they don't actually go away in practice.

Are people really that scared of a -4ac?

People in general? No.

Agenredlum? Apparently very much so.

 

[Robert1986]

Finally, but it took much more than half an hour LOL. That's because we're not in the same room talking.

I looked at what you wrote and I don't see a problem with it initially, except you put a b^2 where it didn't belong, but that doesn't matter to me because it did not affect your final result.

x = [b' +- sqrt(b'^2 + 4ac')]/[2a] = [(-b)^2 +- sqrt((-b)^2 + 4a(-c))]/[2a]

My mistake.

The rest of it seems fine to me but I must caution you, I am not as rigorous as others.

Yes, I have observed.

To me, it seems a tiny bit like circular reasoning but I am no expert in logic. Let othe

It only seems that way because the result is trivial.

Don't forget that the textbook version has an extra minus sign and subtraction instead of addition. Those 2 differences are going to cause more problems. Does that make sense?

No, that doesn't make sense. I would say that writting down a completely new equation, and then using that to solve the original one will cause problems. I'm a math guy, a minus sign doesn't scare me this much.

 

[Hurkyl]

Okay, you've spent long enough trying to repeal the use of negative numbers to recognize that all of the various forms of a quadratic equation differ only superficially. Time to wrap things up.