Imaginary numbers entertwined with quadratics

  • Thread starter runicle
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  • #1
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Hi, I'm not sure if this is calculas based or algebra based so here's the question.

For each pair of roots find a quadratic equation. Express your equation in the form of ax^2+bx+c=0
(
(A) 2i, -2i

For this question i don't know what is being asked so i guess the pairs could be x...
I know this:
ax^2+bx+c, a(x-h)^2+k and a(x-s)(x-t)
So the problem is how can i use the things that i know to answer the question. If you can please use analogies and don't give me the answer i just need to know what to do.
 

Answers and Replies

  • #2
Hootenanny
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Think about the process you go through to solve a quadratic and apply it in reverse. Take note of the complex numbers though.

-Hoot:smile:

Note to mentors/admin suggest moving this thread to pre calc forum.
 
  • #3
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Im trying this.... I just used (a+bi)(c+di)
-(a+bi)(c+di)=0
-ac+adi+bci+bdi^2=0
-from previous question 2i, -2i I represented a = 0.... and so forth
-
 
  • #4
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Although i don't know what 2i represents.... Is it x b c?
 
  • #5
Hootenanny
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From what I can see the solution is pretty simple

[tex](x - (2i))(x - (-2i)) = 0[/tex]

or am I missing something?

Remember that [itex]i^2 = -1 [/itex]

[EDIT] I've edited my post to change my stupid mistake
 
Last edited:
  • #6
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I doubt that hootenay... I'm thinking of x = -b -+ Square root of b^2-4ac over 2a equation and deriving it so that something will occur... im confused...
 
  • #7
dav2008
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Like hoot implied(I think slightly incorrectly), it's always true that if r1 and r2 are the two roots to a quadratic then the characteristic quadratic equation is (x-r1)(x-r2)=0. Then you just expand that to get it into the standard form.

The same argument can be extended to higher degree polynomial equations.
 
  • #8
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My davey jones's locker i think ive got It!!
Correct me if i'm right
0=(x-(2x))(x-(-2x))
0=(x-2i)(x+2i)
0=(x^2+2ix-2ix-4i)
0=(x^2+4)
am i correct?
now if the pairs were 1+3i, 1-3i I would have sub the same way like
0=(x-(1+3i))(x-(1-3i))
0=(x-1+3i)(x+1-3i)
and solve from that correct?
 
  • #9
Integral
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Are you correct?

It is very easy to check your work. What are the roots of

[tex] x^2 + 4 =0 [/tex] ?
 
  • #10
VietDao29
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runicle said:
now if the pairs were 1+3i, 1-3i I would have sub the same way like
0=(x-(1+3i))(x-(1-3i))
0=(x-1+3i)(x+1-3i)
and solve from that correct?
You made a small mistake when going from the first line to the second line. Your second line should reads:
0 = (x - 1 - 3i) (x - 1 + 3i).
Now just expand all the terms out, simplify it, and it's all done.
Can you go from here? :)
 

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