Using quadratic formula, when do we have imaginary roots ?

Question

anonymous · Answer

when your discriminant is less than zero!!
  D= b^2 -4ac <0
  then if you try to find root ....in the course you have to do sqare root(D).. as D is negative..thats why imaginary

anonymous · Answer

$$   b^{2}-4ac < 0 ===>$$
there are 32 imaginary roots

anonymous · Answer

But I thought that when you square root a negative number, you get something undefined. Cause whatever that gets squared must be a positive number.

zarkon · Answer

$$\sqrt{-25}=5i$$

anonymous · Answer

I know that figure but how is it defined?

zarkon · Answer

$$(5i)^2=-25$$

anonymous · Answer

I mean the "i"

anonymous · Answer

^ tsk tsk

anonymous · Answer

when b^2-4ac is negative or less than 0

anonymous · Answer

whenever you have to square a negative variable like -x  in a equation ..
          must check your final  answer by putting them into equation whether they satisfy or not.... because some will not satisfy...omit them from your final answer

zarkon · Answer

http://en.wikipedia.org/wiki/Imaginary_number

anonymous · Answer

Nono. What we are finding is "Is there such a thing as imaginary roots" as we have real roots which are:
$$b^{2}-4ac>0$$$$b^{2}-4ac=0$$$$b^{2}-4ac\ge0$$
So we know that $$b^{2}-4ac<0$$means that it has no "real" roots.
I've discussed this with my friend and we came up with the 5i idea but we don't know what it is.

anonymous · Answer

Sorry I was replying to Prashant. Thanks Zarkon!

anonymous · Answer

imaginary root doesn't make any sense...although complex number is a great branch..it helps in solving a lot of problems
 
to  overcome the concept of sqroot of (-1).Euler decided to take i as squareroot(-1).....
  i^2 =-1         
but in real world there is no existence of i.

anonymous · Answer

Of course. It's IMAGINARY. lol