anonymous
  • anonymous
Using quadratic formula, when do we have imaginary roots ?
Mathematics
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
  • anonymous
\[ b^{2}-4ac < 0 ===>\] there are 32 imaginary roots
anonymous
  • anonymous
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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Zarkon
  • Zarkon
\[\sqrt{-25}=5i\]
anonymous
  • anonymous
I know that figure but how is it defined?
Zarkon
  • Zarkon
\[(5i)^2=-25\]
anonymous
  • anonymous
I mean the "i"
anonymous
  • anonymous
^ tsk tsk
anonymous
  • anonymous
when b^2-4ac is negative or less than 0
anonymous
  • anonymous
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
  • Zarkon
http://en.wikipedia.org/wiki/Imaginary_number
anonymous
  • anonymous
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
  • anonymous
Sorry I was replying to Prashant. Thanks Zarkon!
anonymous
  • anonymous
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
  • anonymous
Of course. It's IMAGINARY. lol

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