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

[agentredlum]

You mean http://en.wikipedia.org/wiki/Elliptic_curve#The_group_law" I thought everyone was using it. Just fired up Wolfram Alpha. All I did was start with (3,5) and doubled it, then double the answer.

Come to think of it, I haven't calculated (3,5) + (129/1000, -383/1000). So that's another solution.

The group law is beyond my comprehension at this time. What do you mean you started with (3,5)?
Please post the calculation step by step i would be greatfull.

My method was to plug in every fraction in the sequence 3/2, 5/2, 5/3, 7/2, 7/3, 7/4, 7/5, 8/3, 8/5, 9/4, 9/5, 9/7, 10/3, 10/7, 11/2, 11/3, 11/4, 11/5, 11/6, 11/7, 11/8, 12/5, 12/7, 13/2, 13/3, 13/4, 13/5, 13/6, 13/7, 13/8, 13/9, 13/10,... until i found a fraction that worked or got bored.

To my credit i only picked fractions greater than the cube root of 2, and other fractions in this sequence are missing either because they have been used already, example 6/4, 10/4, 10/6 are skipped, or they reduce to integer also skipped so i am not a total moron.

I guess i am MORON - dx*(MORON) and as dx goes to zero, i am total moron.

Particularly annoying in this method are the prime numbers. If i nad the courage to reach 127, using my method, that number alone would have required me to test 99 fractions.

 

[agentredlum]

Spock: "Computer, calculate to the last digit, the decimal expansion of pi"

Computer: "Working..."

Spock: "Captain, the computer will now devote all of its electronic circuits to the completion of this impossible task"

 

[agentredlum]

So what is the connection between pythagorean triples, x^3 + y^2 = 2, rational angle measure, rational sine and rationl cosine?

 

[agentredlum]

The above should be x^3 - y^2 = 2

micromass...are you there?

Do you have a way to generate rational angle measure that has rational sine and cosine simultaneously?

 

[micromass]

All the solutions look good. Congratulations everybody.

 

[pwsnafu]

Do you have a way to generate rational angle measure that has rational sine and cosine simultaneously?

Lambert's proof of the irrationality of pi relies on the lemma that for nonzero rational x, tan x is irrational. He does this by finding the http://en.wikipedia.org/wiki/Proof_that_π_is_irrational#Lambert.27s_proof".

There is also http://someclassicalmaths.wordpress.com/2009/07/17/nivens-proof-that-the-trigonometric-and-inverse-trigonometric-functions-are-irrational-for-rational-non-zero-arguments/" for irrationality of cos.

 

[agentredlum]

Lambert's proof of the irrationality of pi relies on the lemma that for nonzero rational x, tan x is irrational. He does this by finding the http://en.wikipedia.org/wiki/Proof_that_π_is_irrational#Lambert.27s_proof".

There is also http://someclassicalmaths.wordpress.com/2009/07/17/nivens-proof-that-the-trigonometric-and-inverse-trigonometric-functions-are-irrational-for-rational-non-zero-arguments/" for irrationality of cos.

Thank you very much. C'MON PEOPLE POST SOME TRICKS!

 

[sankalpmittal]

I would like to have this thread dedicated to showing math tricks from all areas of mathematics. Hopefully the title has aroused your interest and you have an interesting trick you would like to share with everyone. Let me start by showing one of my favorite tricks, perhaps something that has not occurred to many of you?

Start with a general quadratic, do not set it equal to zero, set it equal to bx+c

ax^2 = bx + c

multiply everything by 4a

4(ax)^2 = 4abx + 4ac

subtract 4abx from both sides

4(ax)^2 - 4abx = 4ac

add b^2 to both sides

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

factor the left hand side

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

take square roots of both sides

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

add b to both sides

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

divide by 2a, a NOT zero

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

This quadratic formula works perfectly fine for quadratic equations, just make sure you isolate the ax^2 term BEFORE you identify a, b, and c

1) Notice that this version has 2 less minus signs than the more popular version
2) The division in the derivation is done AT THE LAST STEP instead of at the first step in the more popular derivation, avoiding 'messy' fractions.
3) In this derivation there was no need to split numerator and denominator into separate radicals
4) Writing a program using this version, instead of the more popular version, requires less memory since there are less 'objects' the program needs to keep track of. (Zero is absent, 2 less minus symbols)

I hope you find this interesting and i look forward to seeing your tricks.

The method of completing the square... multiplying by 4a and adding b^2 i learned from NIVEN AND ZUCKERMAN in their book ELEMENTARY NUMBER THEORY however it was an example they used on a congruence, they did not apply it to the quadratic formula.

In post #1 you posted a trick right ? It is known as Sriadhacharya's method in India . He derived quadratic formula by that method only . We can also use that method for perfect squaring .

Your last step is wrong .
It will be x= [-b +- sqrt(b^2 + 4ac)]/(2a)
See :
ax^2+bx+c=0
Multiplying with 4a on both sides and transporting c to rhs :
4a^2x^2+4abx = -4ac
To make perfect square add b^2 on both sides :
4a^2x^2+4abx+b^2 = b^2-4ac
Now ,
(2ax+b)^2 = b^2-4ac
So
2ax+b = +- sqrt(b^2 + 4ac)
therefore
x = [-b +- sqrt(b^2 + 4ac)]/2a

 

[agentredlum]

In post #1 you posted a trick right ? It is known as Sriadhacharya's method in India . He derived quadratic formula by that method only . We can also use that method for perfect squaring .

Your last step is wrong .
It will be x= [-b +- sqrt(b^2 + 4ac)]/(2a)
See :
ax^2+bx+c=0
Multiplying with 4a on both sides and transporting c to rhs :
4a^2x^2+4abx = -4ac
To make perfect square add b^2 on both sides :
4a^2x^2+4abx+b^2 = b^2-4ac
Now ,
(2ax+b)^2 = b^2-4ac
So
2ax+b = +- sqrt(b^2 + 4ac)
therefore
x = [-b +- sqrt(b^2 + 4ac)]/2a

If it is wrong then it should not work, right? Please show me a counter-example.

Also please check your work on the 3rd line from the bottom.

The idea that a quadratic equation must be set to zero in order to find the roots is a MYTH, yet everyone is taught that it is an unquestionable truth.

You don't have to set it equal to zero to find the roots, as shown by my derivation.

And, as a curiosity in this case, if you DON'T do as you have been taught, you get a better formula.

 

[kdbnlin78]

Agentredlum.

In your "proof" of the solutions to the quadratic equation you state:

"Start with a general quadratic, do not set it equal to zero, set it equal to bx+c

ax^2 = bx + c"

The line above is not representative of what you intimated you would do.

It should read "Start with a general quadratic, do not set it equal to zero, set it equal to ex+f"

Then:

Most general quadratic = ex + f

ax^2 + bx + c = ex + f.

Your "trick" is to suggest you have a general quadratic on the left hand side. You do not.

All you have done is to take the basic quadratic equation:

ax^2 + bx + c = 0

and subtracted bx + c (re-written to absorb the negative signs)

Really you have ax^2 = -bx - c.

 

[agentredlum]

Agentredlum.

In your "proof" of the solutions to the quadratic equation you state:

"Start with a general quadratic, do not set it equal to zero, set it equal to bx+c

ax^2 = bx + c"

The line above is not representative of what you intimated you would do.

It should read "Start with a general quadratic, do not set it equal to zero, set it equal to ex+f"

Then:

Most general quadratic = ex + f

ax^2 + bx + c = ex + f.

Your "trick" is to suggest you have a general quadratic on the left hand side. You do not.

All you have done is to take the basic quadratic equation:

ax^2 + bx + c = 0

and subtracted bx + c (re-written to absorb the negative signs)

Really you have ax^2 = -bx - c.

That is a good point, and allows me to bring up a subtle point not mentioned in my derivation.

Since b, c can cycle through all real numbers...-b, -c also cycles through all real numbers so -bx -c can be replaced by bx + c without loss of generality.

What I'm saying is the following...

ax^2 + bx + c = 0 cycles through all possible numerical combinations of a, b, c with 'a' not equal to zero.

so does ax^2 = -bx - c

so does ax^2 = bx + c

You can use any of these representation to derive a working quadratic formula and other representations are possible. I picked ax^2 = bx + c because it has some interesting properties.

Thank you very much for the response, I hope I have alleviated your concern

 

[agentredlum]

I have a shortcut to my derivation which I was saving for a special occasion. Time has passed and not too many people have shown an interest so I might as well post it now.

This shortcut uses the well known result...

If ax^2 + bx + c = 0 then x = (-b +-sqrt(b^2 - 4ac))/(2a) for 'a' not equal 0

This is the standard textbook definition

My shortcut (trick) uses the idea of invariance. What I mean by that is b can be replaced by - b and c can be replaced by -c without loss of generality. Do this in the standard textbook definition above. If you have a hard time accepting this then I fear my arguments will not convince you.

If ax^2 - bx - c = 0 then x = (-(-b) +- sqrt((-b)^2 - 4a(-c))/(2a)

now simplify IN A CLEVER WAY to get

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

Which is what I got by my original derivation. I also have 1 more way of deriving my result which is a completely different derivation

 

[sankalpmittal]

I have a shortcut to my derivation which I was saving for a special occasion. Time has passed and not too many people have shown an interest so I might as well post it now.

This shortcut uses the well known result...

If ax^2 + bx + c = 0 then x = (-b +-sqrt(b^2 - 4ac))/(2a) for 'a' not equal 0

This is the standard textbook definition

My shortcut (trick) uses the idea of invariance. What I mean by that is b can be replaced by - b and c can be replaced by -c without loss of generality. Do this in the standard textbook definition above. If you have a hard time accepting this then I fear my arguments will not convince you.

If ax^2 - bx - c = 0 then x = (-(-b) +- sqrt((-b)^2 - 4a(-c))/(2a)

now simplify IN A CLEVER WAY to get

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

Which is what I got by my original derivation. I also have 1 more way of deriving my result which is a completely different derivation

Huh ??
ax^2 - bx - c = 0 then x = (-(-b) +- sqrt((-b)^2 - 4a(-c))/(2a)

What !? What are you talking about exactly ? The notification of textbook is correct, that precisely :
ax²+bx+c=0 , then x = (-b +-sqrt(b^2 - 4ac))/(2a) for 'a' not equal 0

Exactly this is the quadratic formula , yours is some crap and I have even proved it wrong .

You say : "My shortcut (trick) uses the idea of invariance. What I mean by that is b can be replaced by - b and c can be replaced by -c without loss of generality. Do this in the standard textbook definition above. If you have a hard time accepting this then I fear my arguments will not convince you.

If ax^2 - bx - c = 0 then x = (-(-b) +- sqrt((-b)^2 - 4a(-c))/(2a)

now simplify IN A CLEVER WAY to get

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

Which is what I got by my original derivation. I also have 1 more way of deriving my result which is a completely different derivation
"

Not exactly , you just cannot replace b with -b and c with -c without loss of generality .

ax²+bx+c=0
right ?
Then to replace with -b and -c your equation will become :
-ax²-bx-c=0
You are ignoring the signs selectively , that's your first mistake .
then
x=(-b +- sqrt(b^2 - 4ac)/(2a)

Thats it !

"If ax^2 - bx - c = 0 then x = (-(-b) +- sqrt((-b)^2 - 4a(-c))/(2a)

now simplify IN A CLEVER WAY to get

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

suppose you have an equation :
x²+3x+2=0
then we say :
x²=-(3x+2)
Where a=1, b=3 , c=2.
then
x= -b+-sqrt(b^2 - 4ac)/(2a)
x=-3+-sqrt(3x3-4x1x2)/2x1
x=-2 or x=-1
Lets try it by your method :
x = (b +- sqrt(b^2 + 4ac)/(2a)
x=-3+- sqrt(-3^2 + 4x1x-2)/2
x=3+-sqrt(-17)/2
Absurd !
We know by factor theorem that function f(x) of this equation has factor +-1 and +-2 of equation but on replacing x with -2 or -1 ie f(-2) or f(-1) then equation results in 0 .

YOUR EQUATION HAS BEEN CONTRADICTED .

How can you assume discriminant of your equation to be b²+4ac ?

It will entirely change the nature of roots !

Oh you acquire one more method to derive your result ? Please show it to me so that I can be convinced . And prove that by experimentation .

 

[agentredlum]

Your counter-example x^2 + 3x + 2 = 0. You didnt follow the instructions which say isolate the x^2 term.

I'll do it for you so that you can see how it works.

x^2 = -3x - 2

a = 1 b = -3 c = -2

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

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

x = (-3 +-sqrt(9 - 8))/2

x = (-3 +-sqrt(1))/2

x = (-3 +-1)/2

x = -2/2 = -1

x = -4/2 = -2

Which are the correct answers. You did not understand the method and inserted your own 'element' when you decided to write

x^2 = -(bx + c)

I never derived it that way. This you threw in because of a misunderstanding... no surprise it didn't work.

Thank you for your response. I will be happy to answer any questions but please don't be so confrontational in the future.

It is hard for you to accept that b and c can be replaced by -b and -c without loss of generality, but it is true.

I have shown the derivation to 10 math professors. All of them, as a first impression said it was wrong.

I spent at least half an hour with each professor going over the steps, after that, all of them admitted it was correct.

My third derivation I will save for someone who asks nicely.

 

[sankalpmittal]

Your counter-example x^2 + 3x + 2 = 0. You didnt follow the instructions which say isolate the x^2 term.

I'll do it for you so that you can see how it works.

x^2 = -3x - 2

a = 1 b = -3 c = -2

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

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

x = (-3 +-sqrt(9 - 8))/2

x = (-3 +-sqrt(1))/2

x = (-3 +-1)/2

x = -2/2 = -1

x = -4/2 = -2

Which are the correct answers. You did not understand the method and inserted your own 'element' when you decided to write

x^2 = -(bx + c)

I never derived it that way. This you threw in because of a misunderstanding... no surprise it didn't work.

Thank you for your response. I will be happy to answer any questions but please don't be so confrontational in the future.

It is hard for you to accept that b and c can be replaced by -b and -c without loss of generality, but it is true.

I have shown the derivation to 10 math professors. All of them, as a first impression said it was wrong.

I spent at least half an hour with each professor going over the steps, after that, all of them admitted it was correct.

My third derivation I will save for someone who asks nicely.

It is hard for me to believe that your method worked . I made a mistake of typing -9-8 instead of 9-8 .
But how can you consider the discriminant to be b²+4ac ?

Ok , a simple problem .

x²+3x+2=0
Tell me the nature of this root by your formula .
Please show me your 3rd derivation also .

b²-4ac =0 gives equal and real roots .
b²-4ac >0 gives real and distinct roots .
b²-4ac <0 gives imaginary roots namely i .

Can you really contradict above three laws by your discriminant b² +4ac ?

Answer above questions and I may be totally convinced that your derivation is correct .

(I am in 10th class , 14 years)

I wasn't confrontational , just typed it that way .

In earlier posts you also mentioned that your formula depicts invariance ? In what way is it special ? What can it do that normal textbook quadratic formula can't ? (Not going against but just clearing my doubts .)

 

[agentredlum]

It is hard for me to believe that your method worked . I made a mistake of typing -9-8 instead of 9-8 .
But how can you consider the discriminant to be b²+4ac ?

Ok , a simple problem .

x²+3x+2=0
Tell me the nature of this root by your formula .
Please show me your 3rd derivation also .

b²-4ac =0 gives equal and real roots .
b²-4ac >0 gives real and distinct roots .
b²-4ac <0 gives imaginary roots namely i .

Can you really contradict above three laws by your discriminant b² +4ac ?

Answer above questions and I may me totally convinced that your derivation is correct .

(I am in 10th class , 14 years)

I wasn't confrontational , just typed it that way .

All right, I am happy to explain further. You must isolate the x^2 term, that is essential, then you can identify a, b, c and use

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

THEN

If b^2 + 4ac = 0 you get 1 real double root

If b^2 + 4ac > 0 you get 2 distinct real roots

If b^2 + 4ac < 0 you get 2 distinct imaginary roots

There is no major difference here from the textbook explanation about nature of roots, except for the plus sign, but the nature of the roots is governed by whatever is under the radical.

Now using my formula i compute the nature of the roots to your example x^2 + 3x + 2 = 0

Step 1: Isolate the ax^2 term

x^2 = -3x - 2

Step2: Identify a, b, c

a = 1 b = -3 c = -2

Step3: compute b^2 + 4ac

(-3)^2 + 4(1)(-2)

9 - 8 = 1

This is greater than 0 so there will be 2 real distinct roots.

Did you follow my derivation in post #1? I used a method of completing the square that i learned from Niven and Zuckerman in their book 'Elementary Theory of Numbers' they completed the square on a congruence, they DID NOT apply it to the quadratic formula.

If you had no preconceived notions about what a quadratic formula should look like, AND you start with

ax^2 = bx + c

Then completing the square by any method you like will lead you to

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

The reason the textxbooks don't get this answer is because they insist on setting the equation to 0

 

[sankalpmittal]

All right, I am happy to explain further. You must isolate the x^2 term, that is essential, then you can identify a, b, c and use

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

THEN

If b^2 + 4ac = 0 you get 1 real double root

If b^2 + 4ac > 0 you get 2 distinct real roots

If b^2 + 4ac < 0 you get 2 distinct imaginary roots

There is no major difference here from the textbook explanation about nature of roots, except for the plus sign, but the nature of the roots is governed by whatever is under the radical.

Now using my formula i compute the nature of the roots to your example x^2 + 3x + 2 = 0

Step 1: Isolate the ax^2 term

x^2 = -3x - 2

Step2: Identify a, b, c

a = 1 b = -3 c = -2

Step3: compute b^2 + 4ac

(-3)^2 + 4(1)(-2)

9 - 8 = 1

This is greater than 0 so there will be 2 real distinct roots.

Did you follow my derivation in post #1? I used a method of completing the square that i learned from Niven and Zuckerman in their book 'Elementary Theory of Numbers' they completed the square on a congruence, they DID NOT apply it to the quadratic formula.

If you had no preconceived notions about what a quadratic formula should look like, AND you start with

ax^2 = bx + c

Then completing the square by any method you like will lead you to

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

The reason the textxbooks don't get this answer is because they insist on setting the equation to 0

You astounded me by proving the myth and allegory correct ! However just the last question :
In earlier posts you also mentioned that your formula depicts invariance ? In what way is it special ? What can it do that normal textbook quadratic formula can't ? (Not going against but just clearing my doubts .)

And the last request :
Please show your third derivation .

I am in class 10th 14 years .

Yes I read your first post in this topic .

 

[agentredlum]

My formula doesn't do anything new, or anything that the old formula cannot do. It is a different way to get the same results. The formula itself does NOT require you to set it equal to zero, (major difference from the textbook definition), My formula has 2 fewer minus signs, (major difference from the textbook definition).

Do me a favor and write both formulas down side by side, clearly.

Now, clear your mind of many years of preconceived notions and look at them just as formulas.

Since they both give the same answers, as a 10th grade student, which one would you prefer to use?

My third derivation involves a technique invented by Ehrenfried Walther Von Tschirnhaus and will require me a few hours to reproduce on paper and then post the simplified version.

The way I used invariance is like this...

ax^2 + bx + c describes all possible numerical combinations of a, b, c with a not equal to zero for purposes regarding the derivation

BUT SO DOES ax^2 - bx - c (!)

To put it another way...

As far as second degree equations are concerned ax^2 - bx - c and ax^2 + bx + c are invariant because they describe the same set of infinite second degree equations of this form.

To put it another way...

Algebra doesn't care if you use plus or minus in an ABSTRACT derivation, just as long as your 'use' describes the SAME phenomenon, and you follow the rules of algebra carefully.

In this sense b is invariant to -b and c is invariant to -c

Now, one may say big deal, so what? But in this case, i have shown that this seemingly trivial notion of invariance leads to a nice and elegant result.

 

[pwsnafu]

Not exactly , you just cannot replace b with -b and c with -c without loss of generality .

Why not? The quadratic formula holds if a is non-zero. There are no constraints on b or c.

ax²+bx+c=0
right ?
Then to replace with -b and -c your equation will become :
-ax²-bx-c=0

Why did you place a negative sign in front of a? He never said change a.

Look, here's the quadratic formula. For a non-zero
ax^2+bx+c=0 \implies x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
So replacing b with -b, and c with -c we obtain
ax^2-bx-c=0 \implies x = \frac{b\pm\sqrt{(-b)^2+4a(-c)}}{2a}
And we simplify the equations to
ax^2 = bx+c \implies x= \frac{b\pm\sqrt{b^2+4ac}}{2a}
Which is exactly the formula he wrote in https://www.physicsforums.com/showpost.php?p=3377731&postcount=1".

It really is that trivial.

It is hard for me to believe that your method worked . I made a mistake of typing -9-8 instead of 9-8 .
But how can you consider the discriminant to be b²+4ac ?

It's not. The original post makes no mention about the discriminant. The discriminant is statement of a quadratic that is written in the form ax^2+bx+c.

x²+3x+2=0
Tell me the nature of this root by your formula .

What does that have to do with the discussion? Or you know, just evaluate it!

Can you really contradict above three laws by your discriminant b² +4ac ?

I'll say this again. agentredlum never said the discriminant was b² +4ac.

 

[agentredlum]

Why not? The quadratic formula holds if a is non-zero. There are no constraints on b or c.Why did you place a negative sign in front of a? He never said change a.

Look, here's the quadratic formula. For a non-zero
ax^2+bx+c=0 \implies x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}
So replacing b with -b, and c with -c we obtain
ax^2-bx-c=0 \implies x = \frac{b\pm\sqrt{(-b)^2+4a(-c)}}{2a}
And we simplify the equations to
ax^2 = bx+c \implies x= \frac{b\pm\sqrt{b^2+4ac}}{2a}
Which is exactly the formula he wrote in https://www.physicsforums.com/showpost.php?p=3377731&postcount=1".

It really is that trivial.It's not. The original post makes no mention about the discriminant. The discriminant is statement of a quadratic that is written in the form ax^2+bx+c. What does that have to do with the discussion? Or you know, just evaluate it!I'll say this again. agentredlum never said the discriminant was b² +4ac.

Thank you for your response. and your support is much appreciated by me in this matter.

It's very hard even for trained mathematicians to disregard years of training.

pwsnafu, from your post i understand that you did not get it that the discriminant is also invariant if you isolate the ax^2 term.

To put it another way, in the DERIVATION...

If you set equal to 0 discriminant is b^2-4ac

If you set equal to bx + c the discriminant is b^2 + 4ac

For any arbitrary quadratic applicable to the quadratic formula, these 2 discriminants GIVE THE SAME VALUE.

The choice is yours whether you want to set it equal to 0 or set it equal to bx + c

if you set it equal to 0 you must use b^2 - 4ac

if you set it equal to bx + c (Isolate ax^2 term) then you must use b^2 + 4ac

It is interesting to me that for centuries people have been isolating 0

Look at ax^2 + bx + c = 0

I see many things, i also see that 0 has been isolated.

 

[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.