How can I find $p$ and $q$ in $x^2+px+q = 0$
Knowing only one complex root, for example $2-i$?
$\endgroup$ 12 Answers
$\begingroup$Assuming $p$ and $q$ are real, the two roots must be either real, or if they are not real, they must be complex conjugates (i.e., of the form $a+bi$ and $a-bi$).
Reasoning: If the two roots are $r_1$ and $r_2$, we have
$$ (x-r_1)(x-r_2) = 0 $$
$$ x^2 - (r_1+r_2)x + r_1r_2 = 0 $$
Since $p$ is real, the imaginary components of $r_1$ and $r_2$ must cancel each other out. It then remains to confirm that complex conjugates have a real product:
$$ (a+bi)(a-bi) = a^2+b^2 \in \mathbb{R} $$
In your example, we have $r_1 = 2+i$, hence $r_2 = 2-i$, $r_1+r_2 = 4$, $r_1r_2 = 5$, and you can reconstruct the quadratic equation as above.
$\endgroup$ 0 $\begingroup$Hint:
The other root can be obtained by conjugation if $p$ and $q$ are real.
Hence the other root is $2+i$.
$\endgroup$ 1
