Question

Solve the equation $x^4-4x^2+8x+35=0$, if one of its roots is $2+\sqrt{3}i$

Answer 1

Since $2+\sqrt{3}$ is a root, $2-i\sqrt{3}$ is also root
sum of the roots $=2+i\sqrt{3}+2-i\sqrt{3}=4$
Product of the roots $=(2+i\sqrt{3})(2-i\sqrt{3})$
$=2^2-i^2(\sqrt{3}{)}^2=4+3=7$
$\therefore$ The corresponding factor is $x^2-4x+7$.
$\therefore x^4-4x^2+8x+35=(x^2-4x+7)(x^2+\rho x+5)$
Equating $x$-terms, we get
$8=7\rho -20\Rightarrow 8+20=7\rho$
$\Rightarrow 28=7\rho$

$\Rightarrow \rho =\frac{28}{7}20=4$
$\therefore$ Other factor is $x^2+4x+5$
$x^2+4x+5=0$

$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ....$[a=1,b=4,c=5]$
$\Rightarrow x=\frac{-4\pm \sqrt{16-4\left(1\right)\left(5\right)}}{2\left(1\right)}$
$\Rightarrow x=-4\pm \frac{\sqrt{-4}}{2}=\frac{-4\pm 2i}{2}$
$\Rightarrow x=+2\frac{(-2\pm i)}{2}$
$\Rightarrow x=-2\pm i$
Thus, the roots are $2\pm i\sqrt{3}$, and $-2\pm i$.