TRICKS you know which weren't taught by your professors

this might be way too easy for the average /sci/ user but something I learned recently which helps is:

x^(17/40) = 5
x = 5^(40/17)

you can find the fraction of a number through a decent calculator

>>8481467
This is primordial though.

I'm trying to understand how did the guy who made the quadratic formula thought. I don't know if there is a third grade polynomial formula, it probably is useless, though.

>>8481479
>quadratic formula
Just solve x from ax^2 + bx + c = 0
are you fucking serious?

this has to be a troll thread
fuck you

>>8481489
At least you posted.

Do you know some tricks which can help people to operate simple problems easier?

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>>8481489
>this has to be a troll thread
lel, 3rd grade

>>8481489
>Just solve x from ax^2 + bx + c = 0
Is this how you solve ax^3+bx^2+cx+d=0?

Pleb

>>8481454
I've seen partial fractions taught a lot of ways, but I've never seen a professor or TA solve by power.

One of the easiest ways to solve is by factoring your coefficients by powers of x and then using a matrix or systems of equations to solve.

>>8481524
You could take the first derivative then use the quadratic....

>>8481585
>derivative, limits, another quadratic formula
Is that a formula?

File: IMG_20161118_183039.jpg (3MB, 3264x2448px) Image search: [Google]

3MB, 3264x2448px

It's not that hard come con.

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If you imply a=1 everything gets easier.

>>8481454
If the form ax^3+bx^2+cx+d has 3 repeated roots (real or complex), then d = n^3.

You don't have to do all the fancy stuff you described. Only takes a second to check.

>>8481703
What's "n"

>>8481454
What's an easy way to see if a number is prime? I know there is a method that uses the square root of the number itself, but I forgot the logic.

Anyone know?

>>8482217
to check if a number is a prime, you have to know all primes smaller than half that number

>>8482222
>>8482217
Smaller than the square root of the number, not half. If none of the primes smaller than the square root are factors, then the number is prime. Any prime larger than the square root would necessarily have a prime factor smaller than the square root, so you don't need primes larger than the square root.

>>8482236
Whoops, I meant "any prime factor larger than the number's square root would necessarily imply the existence of a prime factor smaller than the square root." Not that the prime factor itself would have a prime factor smaller than the square root of the number.

>>8482217
The logic is essentially "divide the number by all primes smaller than the square root of the number. If none of them are prime factors, then the number is prime." Obviously that shit won't fly for very large prime numbers, but it's a quick and dirty way to check the primeness of smaller numbers.

>>8481467
I mean.. Thats just how it works... Its not a trick