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anonymous
 5 years ago
Using quadratic formula, when do we have imaginary roots ?
anonymous
 5 years ago
Using quadratic formula, when do we have imaginary roots ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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
 5 years ago
Best ResponseYou've already chosen the best response.0\[ b^{2}4ac < 0 ===>\] there are 32 imaginary roots

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But I thought that when you square root a negative number, you get something undefined. Cause whatever that gets squared must be a positive number.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know that figure but how is it defined?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when b^24ac is negative or less than 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whenever 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nono. 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
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry I was replying to Prashant. Thanks Zarkon!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0imaginary 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
 5 years ago
Best ResponseYou've already chosen the best response.0Of course. It's IMAGINARY. lol
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