quadratic equation with complex coeeficient

Question

Given that p and q are real and that $1+2i$ is a root of the equation :

$$Z^2+(p+5i)z+q(2-i)=0$$

Determine The value of p and q. The other root of the equation.

Answer 1

$z^2+(p+5i)z+q(2-i)=0$

$z=1+2i$ is a root therefore

$(1+2i)^2+(p+5i)(1+2i)+q(2-i)=0$

$1+4i-4+p+2pi+5i-10+2q-qi=0$

$(p+2q-13)+i(2p-q+9)=0$

we have the system

$ \left\{\begin{matrix} p+2q-13&=0 \\ 2p-q+9&=0 \end{matrix}\right. $

multiply the second equation by $2$ and add

$ \left\{\begin{matrix} p+2q-13&=0 \\ 4p-2q+18&=0 \end{matrix}\right. $

$5p=-5\to p=-1$

$-1+2q-13=0\to q=7$

The other root is $z_2=-7i$

$z^2+(-1+5i)z+14-7i=0$

$$z=\frac{1-5i\pm\sqrt{(-1+5i)^2-4(14-7i)}}{2}=\frac{1-5i\pm\sqrt{1-10i-25-56+28i}}{2}=\\=\frac{1-5i\pm\sqrt{-80+18i}}{2}=\frac{1-5i\pm\sqrt{(1+9i)^2}}{2}=\frac{1-5i\pm(1+9i)}{2}$$ $z_1=\dfrac{1-5i-1-9i}{2}=-7i$

$z_2=\dfrac{1-5i+1+9i}{2}=1+2i$

Hope this helps

Answer 2

Let $\alpha = 1+2i$

Let $\beta = a+bi$ be the other root

Then $\alpha+\beta = -(p+5i)$

and $\alpha\times\beta = q(2-i)$

Now figure out p and q with two equations.

$(1+2i+a+bi) = -p -5i$

$a+1 +(2+b) = -p-5i$

$(1+2i)(a+bi) = q(2-i)$

$a+1 =-p$

$b+2=-5$

$b = -7$

$a-2b = 2q$

$2a+b = -q$

$4a+2b = -2q$

Adding the last two, we get

$a = 0$ and $q = -b = 7$

$p = -1$

$\beta = -7i$

Answer 3

\begin{align} z^2+(p+5i)z+q(2-i) &= (z-a-bi)(z-1-2i) \\ &= z^2 -[(a+1)+(b+2)i]z+[(a-2b)+(2a+b)i] \\ \hline p &= -a-1 \\ 5 &= -b-2 \\ 2q &= a-2b \\ -q &= 2a+b \\ \hline p &= -a-1 \\ \color{red}b &\color{red}= \color{red}{-7} \\ 2q &= a+14 \\ \color{red}{-2q} &\color{red}= \color{red}{4a-14} \\ \hline \color{red}p &\color{red}= \color{red}{-1} \\ \color{red}b &\color{red}= \color{red}{-7} \\ \color{red}a &\color{red}= \color{red}0 \\ \color{red}{q} &\color{red}= \color{red}7 \\ \hline \end{align}

So $p=-1$, $q=7$, and the other root is $-7i$.

check

\begin{align} (z-0+7i)(z-1-2i) &= z^2+(-1+5i)z + (14-7i) \\ &= z^2+(p+5i)z+q(2-i) \end{align}