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Real find both roots of the equation

  • Thread starter UnD
  • Start date
  • #1
UnD
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x^2 +6x +k=0 has one root (a) where Im (a) =2, If k is real find both roots of the equation and k

So i got b+ 2i is the root

(b+2i)^2 +6(x+2i) +k=0
and after expanding it out, i have no clue what to do. Please help. THanks
 

Answers and Replies

  • #2
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Try using the quadratic formula and thinking about what comes underneath the square root in relation to Im(a) = 2
 
  • #3
Tide
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You know that if the roots are complex then the two roots are complex conjugates of each other. The sum of the roots is -6 (negative ratio of linear coefficient to quadratic coefficient) so you should be able figure out what what the real part has to be. Once you have a root you can find k.
 
  • #4
HallsofIvy
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UnD said:
x^2 +6x +k=0 has one root (a) where Im (a) =2, If k is real find both roots of the equation and k
So i got b+ 2i is the root
(b+2i)^2 +6(x+2i) +k=0
and after expanding it out, i have no clue what to do. Please help. THanks
I wish you had shown us what you got by expanding it! Clearly that "6(x+ 2i)" should be "6(b+ 2i)" but I don't know whether that's a typo or you actually left the x in your calculation.
Expand it out and set it equal to 0. For a complex number to be equal to 0, both real and imaginary parts must be 0. That gives you two (simple) equations for the two (real) unknown numbers, b and k.
 
  • #5
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sorry for bumping this topic

but could anyone please explain in detail how this question is done?
 
  • #6
HallsofIvy
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First try doing it yourself! You said "after expanding it out, i have no clue what to do." and I asked you to show what you got after expanding. You should get a complex number depending on b and k. As I said before, set real and imaginary parts equal to 0 and you get two equations for b and k. Solve those equations.
 

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