If z=a+2i is a root of x^2+6x+k=0, were ..... (elementary algebra)

Question

If z=a+2i is a root of x^2+6x+k=0, were a,k∈R... Find as well as modules and principle argument of z. Which quadrant does z lie in?

Answer 1

If z=a+2i is a root then the other root is a-2i, because the quadratic has real coefficients.

(x-a+2i)(x-a-2i)=(x-a)^2+4=x^2+6x+k=0. x^2-2ax+a^2+4=x^2+6x+k=0. So -2a=6 and a=-3. a^2+4=k=13, and the quadratic is x^2+6x+13=0. z=-3+2i so the real component of z is -3 and the imaginary component is 2. The point (-3,2) is a representation of the complex z, lying in the second quadrant, where x is negative and y is positive.

Comments

What is the value of a??

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See solution. a=-3 and k=13. Can you see how they were calculated?