The equation $3{{z}^{3}}-7{{z}^{2}}+17z+m=0$, where $m$ is a real constant, has a root $z=1+2\mathbf

Question

The equation $3{{z}^{3}}-7{{z}^{2}}+17z+m=0$, where $m$ is a real constant, has a root $z=1+2\mathbf{i}$. Find the value of $m$. Hence using an algebraic method, find all the roots of the equation $3{{z}^{3}}-7{{z}^{2}}+17z+m=0$. Show your working clearly. [4] Hence, solve the equation $\frac{3}{{{w}^{3}}}+\frac{7}{{{w}^{2}}}+\frac{17}{w}-m=0$, giving your answers in the form $a+b\mathbf{i}$, where $a$, $b\in \mathbb{R}$. [2]

Accepted Answer

(i) $m=-5$, $z=\frac13$, $1+2\mathrm{i}$, $1-2\mathrm{i}$ (ii) $w=-3$, $-\frac15(1-2\mathrm{i})$, $-\frac15(1+2\mathrm{i})$

2022 ASRJC P2 Q1

2022 ASRJC P2 Q1

Video Solution:

Solution:

Solution locked