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

[agentredlum]

Read http://en.wikipedia.org/wiki/Charles_II_of_Spain" and I think you'll cheer up. With heroes like him, and his horizontal family tree, the ancestor paradox is easily explained.

LOL this is so funny i have to quote some of it.

'He is noted for his extensive physical, intellectual, and emotional disabilities along with his consequent ineffectual rule as well as his role in the developments preceding the War of the Spanish Succession'

'Empress Maria Anna was simultaneously his aunt and grandmother and Margarita of Austria was both his grandmother and great-grandmother.'

'The indolence of the young Charles was indulged to such an extent that at times he was not expected to be clean. When his half-brother Don John of Austria, a natural son of Philip IV, obtained power by exiling the queen mother from court, he covered his nose and insisted that the king at least brush his hair.[3]The only vigorous activity in which Charles is known to have participated was shooting. He occasionally indulged in the sport in the preserves of the Escorial.'

Well, you did it Guffel, you made me happy for a while.

I've heard of Don Juan, but Don John?!?

The history of humanity is one of trial and error, unfortunately, mostly error.

 

[math4math]

Cool tricks to do math quick

Thanks

 

[Anamitra]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
{[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

 

[agentredlum]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
{[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+4!+5!+6!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

I got #3 pretty quick the answer is 3. The only factorials that contribute to the last digit are 1, 2, 3, 4 since 5! ends in zero all other factorials con tribute zero to the units position.

It is not too hard to find the next to last digit.

 

[agentredlum]

Posted by Anamitra

'4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12'

The last 2 digits of this sum are 13 so when divided by 4 the remainder is 1.
This gaurantees that the above expression must be of the form 12n + 1 or 12n + 5 or 12n + 9
When these are divided by 3 they leave remainder 1, 2, 0
So the best i can do for now is say the sum leaves remainder 1, or 5, or 9
Still working on #4

 

[agentredlum]

Anamitra posted

'2. Prove that the following:{[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]'

The expression in parenthesis must be of the form 5w, or 5w + 1, or 5w + 2, or 5w + 3, or 5w + 4

The exponent 35*36*37 is even

Carefull consideration of all 5 cases will reveal allowable forms.

(5w)^even is clearly of the form 5n

(5w + 1)^even is clearly of the form 5n + 1 since all terms in the binomial expansion will be multiples of 5 except the last term which will be 1^even or simply 1

(5w + 2)^even is clearly of the form 5n - 1 or 5n + 1 since 2^even ends in 4 or 6 and anything that ends in 4 is of the form 5n - 1, anything that ends in 6 is of the form 5n + 1. Again all terms in the binomial expansion will be multiples of 5 except the last term.

(5w + 3)^even By now it should be obvious that we only need to consider 3^even. This always ends in 9 or 1 and again this is of the form 5n -1 or 5n + 1

(5w + 4)^even Consider 4^even always ends in 6. This is of the form 5n + 1

This completes the proof

 

[pwsnafu]

Posted by Anamitra

'4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12'

The last 2 digits of this sum are 13 so when divided by 4 the remainder is 1.
This gaurantees that the above expression must be of the form 12n + 1 or 12n + 5 or 12n + 9
When these are divided by 3 they leave remainder 1, 2, 0
So the best i can do for now is say the sum leaves remainder 1, or 5, or 9
Still working on #4

All you need to know is that 4! = 24 and any larger factorial is a multiple of 4! (e.g. 6! = 6 times 5 times 4!). So only the first three terms are relevant.


While we are on the topic of divisibility: take any three digit number, repeat it to get a six digit number. This number is divisible by 7. Tends to surprise people who haven't seen it before.

 

[agentredlum]

All you need to know is that 4! = 24 and any larger factorial is a multiple of 4! (e.g. 6! = 6 times 5 times 4!). So only the first three terms are relevant. While we are on the topic of divisibility: take any three digit number, repeat it to get a six digit number. This number is divisible by 7. Tends to surprise people who haven't seen it before.

Oh WOW! why didn't i think of that? So the answer is 9, Thanx pwsnafu!

A*10^5 + B*10^4 + C*10^3 + A*10^2 + B*10 +C=A*10^2(10^3 + 1) + B*10(10^3 + 1) + C(10^3 +1)

(10^3 + 1)(100A +10 B + C) = 1001(100A +10 B + C) = 7*143(100A + 10B + C)

7*11*13(100A +10B + C) So this proves that your example is divisible by 7, 11, 13, 77, 91, and 143 in one proof.

Which completes the proof of your VERY COOL observation.

 

[Anamitra]

What pwsnafu has said is correct.In fact agentredlum has been shown diligent efforts with this one[the fourth sum] .All factorials onwards from 4! are divisible by 12. Agentredlum did the third one quite well and I was expecting him and many others, to do the fourth one by the far simpler method which pwsnafu has mentioned just now.Incidentally the second sum[in the set I have given] has an interesting alternative[solution technique].

 

[Anamitra]

If you take a number N=abcabc ,where a,b c are digits,you may write

N=abc*1000+abc
=abc[1001]
1001 is divisible by 7.

 

[agentredlum]

If you take a number N=abcabc ,where a,b c are digits,you may write

N=abc*1000+abc
=abc[1001]
1001 is divisible by 7.

Oh WOW! Your solution is more elegant.

 

[Anamitra]

Regarding sum 2 of post 153:

The square of any number[integral number] can be expressed as 5n or5n+1 or 5n-1

If you square a number the possible last digits[units place] are 0,1,4,6,5 and 9
If the last digit is one or nine the squared number is at a unit's distance from a number ending with zero.
If the squared number ends with 6, it is again at a unit's distance from a number ending with 5.
If the squared number ends with zero or five it is of the form 5n

This result,alternatively, may be established by using Fermat's Theorem
{N}^{p-1}{-}{1} is divisible by p[a prime] if N does not contain p as a factor.

Therefore,
{N}^{4}{-}{1} is a multiple of five[if N does not contain 5]
But
{N}^{4}{-}{1}{=}{(}{N}^{2}{+}{1}{)}{(}{N}^{2}{-}{1}{)}
Hence the result.

 

[Anamitra]

Problem 1 of Post #153:Solution

The last digit[unit's digit] of any number[natural number] can occur in 10 ways from 0 through 9.We don't have eleven choices for it.
So, if we have eleven or more numbers[say 12 numbers] of we can always locate a pair which have the same value for the last digit.
Take their difference--you have a zero in the unit's place. So it must be divisible by five.

 

[agentredlum]

Problem 1 of Post #153:Solution

The last digit[unit's digit] of any number[natural number] can occur in 10 ways from 0 through 9.We don't have eleven choices for it.
So, if we have eleven or more numbers[say 12 numbers] of we can always locate a pair which have the same value for the last digit.
Take their difference--you have a zero in the unit's place. So it must be divisible by five.

Very nice sir, PLEASE POST MORE!

 

[Anamitra]

Here is simple tip for framing your own problems.

You raise a natural number to the fifth power. The digit in the unit's place does not change.
n and n to the power 5 have the same digit in the unit's place.
Therefore expressions like,
{n}^{5}{-}{n} are multiples of 5[and 2 of course]

You can always substitute n by some frightening formula that represents a natural number.

 

[micromass]

Question: find me 10 values for t\in [0,2\pi[ such that both cos(t) as sin(t) are rational?
Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

If you did that question, then this should become not so hard anymore:
Find 10 values for t such that \sqrt{x^3-2} is rational.

 

[agentredlum]

Question: find me 10 values for t\in [0,2\pi[ such that both cos(t) as sin(t) are rational?
Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

If you did that question, then this should become not so hard anymore:
Find 10 values for t such that \sqrt{x^3-2} is rational.

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

 

[cubzar]

post #166
1. cos(t) and \sqrt{}1-cos²(t) are rational if cos(t)=a/c, with c=largest member of a pythagorean triple and a=one of the other members.
t=0, \pi, cos^-1(3/5), cos^-1(4/5), cos^-1(5/13), cos^-1(12/13), cos^-1(8/17), cos^-1(15/17), cos^-1(12/20), cos^-1(7/25), cos^-1(24/25)

 

[SteveL27]

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

Micromass must want x rational. Otherwise it's too easy: Any x such that x^3 is two more than a square works. For example x = cube root of {2, 3, 6, 11, 18, ...}.

 

[dimension10]

Four Simple Problems:
1. You have a set of twelve natural numbers[arbitrary natural numbers].Show that a pair can be always located from this set so that their difference is divisible by 5

2. Prove that the following:
{[}{{n}_{1}}^{35}{+}{{n}_{2}}^{36}{+}{{n}_{3}}^{37}{]}^{35*36*37}
where n1,n2 and n3 are natural numbers can be expressed in the form 5n or 5n+1 or 5n-1, where n is a natural number.[It many not be true the other way round]

3. What is the last digit[unit's place] in the sum:

1!+2!+3!+....99!

4. What remainder do you have if 1!+2!+3!+...+2011! is divided by 12

1.

If the numbers are a, b, c d, e, f, g, h, i, j, k and l, then

a \equiv m (mod 5)
b \equiv n (mod 5)
c \equiv o (mod 5)
...
l \equiv x (mod 5)

Then the question just requires that the difference between a certain 2 of the set {m, n, o,...,x} is divisible by 5. And we know that the {m, n, o,...,x} are either 0, 1, 2, 3 or 4. To fulfil the premises, two elements of the set {m, n, o,...,x} should be equal. This has to be as we have 12 numbers which is greater than 5.

3. 1!=1 (mod 10)
2!=2 (mod 10)
3!=6 (mod 10)
4!=4 (mod 10)
5!=0 (mod 10)
6!=0 (mod 10)
(I know I should put the congruence sign but I can't find that one on the keyboard :) )

Then all the factorials would have a remainder of 0. 1+2+6+4=13. The last digit of 13 is 3.

4.

1!=1 (mod 12)
2!=2 (mod 12)
3!=6 (mod 12)
4!=0 (mod 12)
5!=0 (mod 12)
...
1+2+6=9

The remainder is 9.

 

[agentredlum]

What is a rational angle? Please post any expression that represents a rational angle

 

[micromass]

All pythagorean triples will solve The first half of the first question.

For question 3 do you mean 10 values of x such that sqrt(x^3-2) is rational?

Do you want x rational?

Great questions!

Yes, I want x rational
You solved the first problem by using pythagorean triples, but that wasn't what I had in mind. What I had in mind is to find a formula that generates the pythagorean triples. The method of finding such a formula is useful in finding the rational values x such that \sqrt{x^3-2} is rational.

 

[pwsnafu]

Further question, find me 10 rational values for t such that both cos(t) as sin(t) are rationals.

Not many choices! What an understatement.

Even if you replace "both" with "either" you still won't have many choices!

 

[Anamitra]

If {\theta}{,}{cos}{(}{\theta}{)}{,}{sin}{(}{\theta}{)} are rational simultaneously,
{n}\theta{,}{cos}{(}{n}\theta{)}{,}{,}{sin}{(}{n}{\theta}{)}
are also rationals provided n is integral.This may not be true for the sub-multiple angles[I have assumed the angle to be in radians]

We can use Pythagorean triplets to generate the rationals sin[theta] and cos[theta] but the problem is to get a rational theta[not equal to zero] corresponding to the ratios of sin and cos.If we get one such set we can get a huge number of valid sets.
If {tan}{(}\frac{\alpha}{2}{)}{=}\frac{m}{n}
Where m/n is a rational[m, n are integers, n not equal to zero]
{sin}{(}\alpha{)}{=}\frac{2nm}{{m}^{2}{+}{n}^{2}}
{Cos}{(}\alpha{)}{=}\frac{{n}^{2}{-}{m}^{2}}{{m}^{2}{+}{n}^{2}}

Here Sin and cos of the angle are rational--but the angle itself may be rational or irrational.
[This method fails to locate a rational theta for rational values of sin and cos]
A relevant link:
http://www.yaroslavvb.com/papers/olmsted-rational.pdf

The link discusses the situation in relation to degrees. May be things are more favorable in terms of radians.

 

[Anamitra]

From the last post:
If {\theta}{,}{cos}{(}{\theta}{)}{,}{sin}{(}{\theta}{)} are rational simultaneously,
{n}\theta{,}{cos}{(}{n}\theta{)}{,}{,}{sin}{(}{n}{\theta}{)}
are also rationals provided n is integral.This may not be true for the sub-multiple angles[I have assumed the angle to be in radians]Now, if theta=a/b where a and b are integers and n=b we have an integral value a for which
Sin(a) and Cos(a) are rational

 

[cubzar]

x\equiv1(mod2)
x\equiv3(mod4)
x\equiv3(mod8)
These are the first 20 solutions.
3,
7483515,
10688219,
11862731,
12268371,
13866203,
14157699,
15058035,
18056043,
18507683,
18786627,
19403251,
20014659,
20412827,
21074339,
21953027,
22411283,
22475523,
23136619,
23207451,

 

[agentredlum]

x\equiv1(mod2)
x\equiv3(mod4)
x\equiv3(mod8)
These are the first 20 solutions.
3,
7483515,
10688219,
11862731,
12268371,
13866203,
14157699,
15058035,
18056043,
18507683,
18786627,
19403251,
20014659,
20412827,
21074339,
21953027,
22411283,
22475523,
23136619,
23207451,

3 is the only positive integer solution to x^3 - y^2 = 2

I don't know if micromass is asking for modular solutions.

Great work on generating all those large number solutions!

My gut tells me there should be rational solutions between integers but my gut has been wrong before...

 

[pwsnafu]

My gut tells me there should be rational solutions between integers but my gut has been wrong before...

Well 129/100 and 2340922881/58675600 are both solutions (approximately 1 and 40 respectively). If you are interested they evaluate to 383/1000 and 113259337279/449455096000 respectively.

 

[agentredlum]

Well 129/100 and 2340922881/58675600 are both solutions (approximately 1 and 40 respectively). If you are interested they evaluate to 383/1000 and 113259337279/449455096000 respectively.

THAT IS FANTASTIC!

I verified 129/100 using a simple calculator (wish i had TI 92)

Did you use a formula? A program? to generate these 2 numbers, the second fraction could not have been easy.

The method i was using would have taken me 30 days to find 129/100, a programmable calculator would do it in 3 hours, maple in 3 minutes, without mistakes.

To find the second fraction you posted would take me 300 billion years. LOL

 

[pwsnafu]

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.

Edit: Turns out (3,5) + (129/100, -383/1000) = (164323/29241, -66234835/5000211). There you go.

 

[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.