$$x^2 + 4x + 20 = 0$$ $$x=-2 \pm 4i$$

Roots of Quadtratic $z^2 + (4+i+qi)z + 20 = 0$ - Roots are $w$ and $w*$

(a) When $q$ is real, explain why it must be $-1$


Roots are conjugates therefore the coefficient must be real - and to get rid of the imaginary parts $q$ must be $-1$?

(b) Where $w=p+2i$ and $p$ is real, find the values of $q$.


Subbed $p+2i$ into equation to give $p=2$, where it is real. Substitute $2+2i$ into the equation again and get $q=3$, and $q=-8$? Very sure this is not correct - Should be a complex number?

Thank you.

  • $\begingroup$ i think your first argument is ok $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 13:09
  • $\begingroup$ if for b) the same equation valid? $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 13:12
  • $\begingroup$ This is all information given - I would assume so - I think possible roots of equations is required. $\endgroup$ – Cicada Apr 21 '17 at 13:17

Hint for part (b): given the root $p+2i\,$, the other root is $\,\cfrac{20}{p+2i}\,$ by Vieta's relations.

Then, again by Vieta's relations for the sum of the roots:

$$ p+2i + \cfrac{20}{p+2i} = -(4+i+qi) $$

The latter gives $q$ in terms of $p$. The solution is not unique, for each $p \in \mathbb{R}$ there exists a complex $q$ given by the equation above such that $p+2i$ is a root.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.