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

[agentredlum]

Anyway, what is Z in your equation...sqrt(D)

nice try throwing smoke bomb but please, you have to do better than that.

agentredlum-passed out...

 

[micromass]

Anyway, what is Z in your equation...sqrt(D)

nice try throwing smoke bomb but please, you have to do better than that.

agentredlum-passed out...

Like I said, writing $\sqrt{D}=Z$ is not allowed since D is complex. But writing $Z^2=D$ is allowed here. I'm not doing anything new or fishy here I'm just arguing on when to use which symbols...

 

[agentredlum]

Like I said, writing $\sqrt{D}=Z$ is not allowed since D is complex. But writing $Z^2=D$ is allowed here. I'm not doing anything new or fishy here I'm just arguing on when to use which symbols...

Yes but your formula doesn't give any answers until you take the square root, a square roooooot you do not allow yourself to take.

 

[agentredlum]

How many hours you been without sleep?

 

[micromass]

Yes but your formula doesn't give any answers until you take the square root, a square roooooot you do not allow yourself to take, who's silly now?

I don't need to use the square root symbol anywhere. Where do you think I need it??

And I don't appreciate being called silly. If that's your attitude, then I don't think this discussion will go on for much longer.

How many hours you been without sleep?

I just woke up.

 

[agentredlum]

I don't need to use the square root symbol anywhere. Where do you think I need it??

And I don't appreciate being called silly. If that's your attitude, then I don't think this discussion will go on for much longer.
I just woke up.

I apologize, forget i said it.

pwsnafu is still trying to figure out how i got the answer.

 

[micromass]

I apologize, forget i said it.

pwsnafu is still trying to figure out how i got the answer.

I think pwsnafy know very well how you got the answer. It's the first thing you should try...

 

[agentredlum]

I edited the post removing the remark.

 

[agentredlum]

I think pwsnafy know very well how you got the answer. It's the first thing you should try...

And what is that?

 

[pwsnafu]

I apologize, forget i said it.

pwsnafu is still trying to figure out how i got the answer.

I think pwsnafy know very well how you got the answer. It's the first thing you should try...

And what is that?

Wait what? I didn't know I supposed to work something out.

If we are talking about the 1/x+1/y problem, then yes, I know how to get the answer easily that's the whole point. Too often students don't analyze the question. If all you have is the quadratic formula, then everything is a quadratic equation. Especially if you're a high school student.

If we are not talking about that, what question?

 

[agentredlum]

Wait what? I didn't know I supposed to work something out.

If we are talking about the 1/x+1/y problem, then yes, I know how to get the answer easily that's the whole point. Too often students don't analyze the question. If all you have is the quadratic formula, then everything is a quadratic equation. Especially if you're a high school student.

If we are not talking about that, what question?

I know what i was talking about, i was talking about that.

I don't know what micromass was talking about, i can't read minds. It appears to me that micromass wants to explain the radical in the quadratic formula is not necessary to numerically determine the roots using the quadratic formula.

 

[dodo]

(We interrupt this program for a short message...)

According to this PS3 manual,
http://manuals.playstation.net/document/en/ps3/current/browser/menub.html

In your "Network" menu, under "Tools" you can find an option to turn Javascript "on" or "off". If Javascript is "off", then that is the reason why you're seeing formulas as gibberish. Give it a try!

 

[agentredlum]

(We interrupt this program for a short message...)

According to this PS3 manual,
http://manuals.playstation.net/document/en/ps3/current/browser/menub.html

In your "Network" menu, under "Tools" you can find an option to turn Javascript "on" or "off". If Javascript is "off", then that is the reason why you're seeing formulas as gibberish. Give it a try!

I tried it just now, it didn't work. Thank you anyway.

If I delete cache will i lose my bookmarks? I have 1000 bookmarks many about math and science so I'm afraid of losing them. If you can assist i would be greatful.

 

[dimension10]

For trignometric equalitites, https://www.physicsforums.com/showthread.php?t=514065.

 

[Guffel]

I tried it just now, it didn't work. Thank you anyway.

If I delete cache will i lose my bookmarks? I have 1000 bookmarks many about math and science so I'm afraid of losing them. If you can assist i would be greatful.

No, you won't lose your bookmarks. Automatic logins on forums etc may disappear, but as long as you remember your user name and password, there should be no problem.

 

[Anamitra]

Well, how would you solve $x^2=2+i$?? By saying $x=\sqrt{2+i}$?? How helpful...

The square root symbol has nothing to do with finding roots of polynomials. I can easily find the roots of $x^n=-1$:

$$\cos(\pi k /n)+i\sin(\pi k /n),~k\in \{0,...,n-1\}$$

but by your method, you would only find $x=\sqrt[n]{x}$ (even if we would allow the square root on complex numbers). That's not really what we want, is it??

A Direct Application of the Square-Root Operator
[ By using the Binomial Theorem]
$${(}{2}{+}{i}{)}^{1/2}$$ ------------ [Expression 1]
$${=}{2}^{1/2}{(}{1}{+}{i/2}{)}^{1/2}$$
$${=}{2}^{1/2}{(}{1}{+}{\Sigma}{(}{-1}{)}^{r-1}\frac{{(}{2r-2}{)}{!}}{{2}^{3r-1}{r}{!}{(}{r-1}{)}{!}}{i}^{r}{)}$$ ----------- [Expression 2]
[In the above expression r=1,2,3,4... to infinity]
=[1+1/32-5/2048+21/65536-429/8388608 ……………]+i [1/4-1/128+7/8192-33/262144+...]
=1.4553 + 0.3435 i
The convergence of the two series may be established by considering[for each series] the absolute values of the terms and then by applying D’Alembert’s ratio test or some other suitable test of convergence. The general term given in the third step(relation(2)) of the calculation proves useful in this respect. [One may consider the ratio between alternate terms[their absolute values] in the summation in Expression 2 $${\mid}\frac{{u}_{r+2}}{{u}_{r}}{\mid}$$ as r tends to infinity]

Suppose we are to find:
$${{(}2+i{)}}^{1/5}$$
We find the first root by using the binomial theorem . Let the root be a+ib
$${y}^{5}{=}{2+i}$$
$${y}^{4}{=}\frac{2+i}{a+ib}$$
After rationalizing the RHS we take the fourth root by applying the binomial theorem again.
The process is repeated---- we do not get more than five roots by this method.

De Moivre’s method is much more convenient for these evaluations. But the application of the square root and other roots [their validity]can be seen from the above calculations.

 

[agentredlum]

No, you won't lose your bookmarks. Automatic logins on forums etc may disappear, but as long as you remember your user name and password, there should be no problem.

I tried it, still didn't work. Thank you very much for your concern.

 

[Char. Limit]

Here's something interesting. Consider the equation:

$$a^2 + b^2 + c^2 + d^2 = e$$

Now, let e be any natural number (here, referring to any positive integer OR zero). Then a, b, c, and d have solutions in the natural numbers. In other words, any natural number can be rewritten as the sum of four square numbers.

This result is known as Four-Square Theorem[/url].

 

[agentredlum]

Here's something interesting. Consider the equation:

$$a^2 + b^2 + c^2 + d^2 = e$$

Now, let e be any natural number (here, referring to any positive integer OR zero). Then a, b, c, and d have solutions in the natural numbers. In other words, any natural number can be rewritten as the sum of four square numbers.

This result is known as Four-Square Theorem[/url].

Thanx for the post, PLEASE POST MORE!

I find this interesting as well, Lagrange was a mathematical and science genius.

 

[agentredlum]

No one is posting tricks *sniff* let me ask some questions that I believe have fabulous Answers. I won't give the answers right away so you can think about it.

1) Using all digits 0~9 only once how many different numbers can you make?

Example: The number 1234567890 is one possibility, the number 1023456789 is another possibility. No digit is to be repeated anywhere in the number and all digits must be used.

2) How many of these numbers asked for above are divisible by 9?

3) How many zeroes are at the end of 1000!

Example: 7! = 5040 so there is one zero at the end of 7!

Good Luck

 

[Anamitra]

Answers:
1. $${10}{!}{-}{9}{!}$$
=3265920
2. Sum of digits=0+1+2+...+9
=45
45 is divisible by 9. Therefore all numbers according to the given prescription should be divisible by 9
3. Theorem to be used: The highest power of a prime,p, in n! is given by:
{ [n/p]+[n/(p^2)]+[n/(p^3)]-----------}
You may include as many terms[possible] as you can in the above series.
[x] denotes the greatest integer less than or equal to x

The highest power of 5 in 1000! is 249
The highest power of 2 is even greater.
Therefore we have 249 zeros.

 

[agentredlum]

Answers:
1. $${10}{!}{-}{9}{!}$$
=3265920
2. Sum of digits=0+1+2+...+9
=45
45 is divisible by 9. Therefore all numbers according to the given prescription should be divisible by 9
3. Theorem to be used: The highest power of a prime in n! is given by:
{ [n/p]+[n/(p^2)]+[n/(p^3)-----------}
You may include as many terms[possible] as you can in the above series.
[x] denotes the greatest integer less than or equal to x

The highest power of 5 in 1000! is 249
The highest power of 2 is even greater.
Therefore we have 249 zeros.

YES! VERY NICE SIR!

I'snt it amazing that ALL numbers constructed in this way are divisible by 9?

Just a humble observation...you don't need ! on n in your formula #3

by the way, the highest power of 2 in the prime factorization of 1000! is 994 but you only use 249 of them to produce factors of ten cause there are 249 5's

Do you know any problems like this?

 

[Anamitra]

Just a humble observation...you don't need ! on n in your formula #3:

Thanks for pointing it out. It was just due to inadvertence!
[I have edited the thing: It was looking extremely odd]

 

[agentredlum]

Thanks for pointing it out. It was just due to inadvertence!
[I have edited the thing: It was looking extremely odd]

LOL

I would pay good money to know the exact value of 1000!/5

If you make cubes with side Planck Length and you packed the entire known universe wth these cubes, i don't think you would come even remotely close to 1000! cubes.

I edited it out of the quote by you in my post

 

[agentredlum]

I have a question about ancestors that seems to be related to number theory. Let's consider my total ancestors. 1 generation before me there is my father and mother so that's 2 ancestors. My father had parents and so did my mother, so 2 generations before me is total 6 ancestors for me. My grandparents also had parents so 3 generations before me is total 14 ancestors for me. You get the idea, 4 generations before me is total 30 ancestors, 5 generations, 62 ancestors, n generations, 2^(n + 1) - 2 ancestors.

Assuming 30 years for every generation and assuming 100020 years of humanity on this planet, then by the above argument I am supposed to have about 2^3335 - 2 ancestors.

This has to be wrong but I can't figure out where the mistake in the logic is. The only way I know to reduce this humungous number is to assume that at many places along the line, many parents were brother and sister, or father and daughter, or mother and son, or grandfather and granddaughter, or grandmother and grandson.

This is very disturbing to me...HELP!

 

[Guffel]

Google "ancestor paradox" and I hope you'll get happier.

 

[agentredlum]

Google "ancestor paradox" and I hope you'll get happier.

WOW! I can't believe all these years it never occurred to me to google it. Thanx Guffel, or should i call you cousin?

 

[agentredlum]

IMHO I accept this explanation

http://www.genetic-genealogy.co.uk/supp/ancestor_paradox.html

The othrs seem to be by the same person and they are against the posted link in some respects. However i have only looked at 10 of 35000 hits.

Still not happy.

 

[Guffel]

Still not happy.

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.

 

[agentredlum]

Is it me or does google like to exaggerate their own importance? The first page claimed 35700 hits. By the third page it went down to 332. I only get 10 hits per page...