Question

(i) If $-2+i\beta ,\beta \in R-\left\{0\right\}$ is a root of $x^3+63x+\lambda =0,\lambda \in R$ then find other roots of equation.
(ii) If $\frac{-1}{2}+i\beta$, is a root of $2x^3+b{x}^2+3x+1=0,b,\beta \in R-\left\{0\right\}$, then find the value(s) of $b$.

Answer 1

$1).$ if $-2+i\beta$ is a root then $-2-i\beta$ is also a root

Given $x^3+63x+\lambda =0$

Let the other root be $p$

Sum of roots is $0⟹-2+i\beta -2-i\beta +p=0⟹p=4$

sum of product of roots is $63⟹4+{\beta }^2-16=63⟹\beta =5\sqrt{3}$

$\therefore$ the roots are $-2\pm i5\sqrt{2},4$

$2).$ if $\frac{-1}{2}+i\beta$ is a root then $\frac{-1}{2}-i\beta$ is other root

Given $2x^3+b{x}^2+3x+1=0$

Let the other root be $p$

Product of roots is $\frac{-1}{2}⟹p(\frac{1}{4}+{\beta }^2)=\frac{-1}{2}⟹\frac{1}{4}+{\beta }^2=\frac{-1}{2p}$

Sum of product of roots is $\frac{3}{2}⟹(\frac{1}{4}+{\beta }^2)+\left(p\right)(-1)=\frac{3}{2}⟹-\frac{1}{2p}-p=\frac{3}{2}⟹p=-1or\frac{-1}{2}$

Sum of roots is $-1+p=-2or-\frac{3}{2}$

$-\frac{b}{2}=-2or-\frac{3}{2}$

$b=4or3$