Conquer Club

same thing as the other thread except how do you find complex roots.

ehat other thread

the analytical trig thread.

I use a calculator.

I dig with a shovel.

reverend_kyle wrote:same thing as the other thread except how do you find complex roots.

In ax^2+bx+c=0, <-What in this???

The discriminant not be a square or 0.

b^2-4ac cannot be 0,1,4,9,16,25,36,49 and so on...

Syzygy wrote:

reverend_kyle wrote:same thing as the other thread except how do you find complex roots.

In ax^2+bx+c=0, <-What in this???

The discriminant not be a square or 0.

b^2-4ac cannot be 0,1,4,9,16,25,36,49 and so on...

yah, it depends how you are given the problems. im actually revising the complex roots of unity at the mo, its not too bad, as long as you can visualise where the roots are just draw them out on an argand diagram and its easy enough.

reverend_kyle wrote:same thing as the other thread except how do you find complex roots.

I asked my Girlfriend, and she said that you should go to a profesional hairdresser and let them find them.

b^2-4ac>0 ==> 2 real solutions (or roots)
b^2-4ac=0 ==> 1 real ("repeated") solution
b^2-4ac<0 ==> 0 real solutions.

That how I know it.

alex_white101 wrote:

Syzygy wrote:

reverend_kyle wrote:same thing as the other thread except how do you find complex roots.

In ax^2+bx+c=0, <-What in this???

The discriminant not be a square or 0.

b^2-4ac cannot be 0,1,4,9,16,25,36,49 and so on...

yah, it depends how you are given the problems. im actually revising the complex roots of unity at the mo, its not too bad, as long as you can visualise where the roots are just draw them out on an argand diagram and its easy enough.

Ah crap, he said complex roots as in non-real roots, and NOT irrational roots like I said above....

In that case, let me rephrase... ax^2 + bx + c has discriminant of
D = b^2 − 4ac;

b^2 − 4ac<0 for the roots to be complex.

johnnyrotten wrote:b^2-4ac>0 ==> 2 real solutions (or roots)
b^2-4ac=0 ==> 1 real ("repeated") solution
b^2-4ac<0 ==> 0 real solutions.

That how I know it.

That's right.

When it's less than 0 you get no real solutions, but you do get 2 complex solutions.

Syzygy wrote:

alex_white101 wrote:

Syzygy wrote:

reverend_kyle wrote:same thing as the other thread except how do you find complex roots.

In ax^2+bx+c=0, <-What in this???

The discriminant not be a square or 0.

b^2-4ac cannot be 0,1,4,9,16,25,36,49 and so on...

yah, it depends how you are given the problems. im actually revising the complex roots of unity at the mo, its not too bad, as long as you can visualise where the roots are just draw them out on an argand diagram and its easy enough.

Ah crap, he said complex roots as in non-real roots, and NOT irrational roots like I said above....

In that case, let me rephrase... ax^2 + bx + c has discriminant of
D = b^2 − 4ac;

b^2 − 4ac<0 for the roots to be complex.

Syzygy wrote:

johnnyrotten wrote:b^2-4ac>0 ==> 2 real solutions (or roots)
b^2-4ac=0 ==> 1 real ("repeated") solution
b^2-4ac<0 ==> 0 real solutions.

That how I know it.

That's right.

When it's less than 0 you get no real solutions, but you do get 2 complex solutions.

Only if you assume you are working with a quadratic.

Finding complex roots for the general case is much, much harder.

use de moivre's theorem