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Using quadratic formula, when do we have imaginary roots ?

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when your discriminant is less than zero!! D= b^2 -4ac <0 then if you try to find root the course you have to do sqare root(D).. as D is negative..thats why imaginary
\[ b^{2}-4ac < 0 ===>\] there are 32 imaginary roots
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.

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I know that figure but how is it defined?
I mean the "i"
^ tsk tsk
when b^2-4ac is negative or less than 0
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
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.
Sorry I was replying to Prashant. Thanks Zarkon!
imaginary root doesn't make any sense...although complex number is a great 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.
Of course. It's IMAGINARY. lol

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