Finding the coefficients of a quadratic equation with complex number roots

Question

$$p(z)=ax^2+bx+c$$

The two roots are $4-i$ and $-4$, and $p(i)=-4$

How can I determine the coefficients $a$, $b$ and $c$?

Answer 1

If the roots of $p$ are $4-i$ and $-4$, then $$ p(z) = \alpha(z-(4-i))(z+4) = \alpha(z^2 + iz -4(4-i)) $$ for some $\alpha\in\mathbb{C}$. You can use $p(i) = -4$ to solve for $\alpha$. The coefficients are then $$ a = \alpha,\quad b = i\alpha, \quad c = -4(4-i)\alpha $$

Answer 2

Well, we have that:

$$\text{a}x^2+\text{b}x+\text{c}=0\space\Longleftrightarrow\space x=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}\tag1$$

So, now we can set up a system of equations:

$$ \begin{cases} \text{a}\cdot i^2+\text{b}\cdot i+\text{c}=-4\\ \\ \frac{-\text{b}+\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}=4-i\\ \\ \frac{-\text{b}-\sqrt{\text{b}^2-4\text{a}\text{c}}}{2\text{a}}=-4 \end{cases}\tag2 $$

And then you will find that:

$$\text{a}=\frac{18+4i}{85},\text{b}=\frac{18i-4}{85},\text{c}=\frac{8i-304}{85}\tag3$$