Find coefficients to a quadratic equation knowing roots and a point...

Question

Given a standard quadratic equation: $$p(z) = az^2 + bz + c$$

We know that $-10$ and $10-i$ are roots. We know that $p(i)=-10$

What are $a$, $b$ and $c$?

Answer 1

$$p(z)=C(z+10)(z-10+i)\\p(z)=C(z^2-10z+zi+10z-100+10i)\\p(z) = C(z^2+zi+10i-100)$$ The condition is that $$p(i) = -10$$ This means that $$ C(i^2+i\cdot i+10i-100) =-10$$ $$C(-102+10i) =-10\\C=\frac{10}{102-10i}=\frac{5}{51-5i}$$ $$C=\frac{5}{51-5i}\cdot\frac{51+5i}{51+5i}$$ $$C = \frac{255+25i}{2626}$$ Now plug in $C$, and we can get $$p(z) = \frac{255+25i}{2626}(z^2+zi+10i-100)$$ $$p(z) = \frac{255+25i}{2626}z^2+\frac{255+25i}{2626}zi+(10i-100)\cdot \frac{255+25i}{2626}$$ $$p(z) = \frac{255+25i}{2626}z^2+\frac{-25+255i}{2626}z+(5i-50)\cdot \frac{255+25i}{1313}$$

$$p(z) = \frac{255+25i}{2626}z^2+\frac{-25+255i}{2626}z+\frac{-12875+25i}{1313}$$ Result: $$a = \frac{255+25i}{2626},b=\frac{-25+255i}{2626},c=\frac{-12875+25i}{1313}$$

Answer 2

Try $p(z)=a(z-i)(z-10+i)$

Multiply and find coefficients using the information at $z=i$