Real find both roots of the equation

[UnD]

x^2 +6x +k=0 has one root (a) where I am (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

 

[sqrt(-1)]

Try using the quadratic formula and thinking about what comes underneath the square root in relation to Im(a) = 2

 

[Tide]

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.

 

[HallsofIvy]

x^2 +6x +k=0 has one root (a) where I am (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.

 

[aarubui]

sorry for bumping this topic

but could anyone please explain in detail how this question is done?

 

[HallsofIvy]

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.