If the bi-quadratic equation f ( x ) = x 4 + 2 x 3 - 16 x 2 - 22 x + 7 = 0 has 2 + 3 as one of its roots,…

Given, $f\left(x\right)=x^4+2x^3-16x^2-22x+7=0$

$\because 2+\sqrt{3}$ is one of the roots of $f\left(x\right)=0$

$\therefore 2-\sqrt{3}$ is also one of the roots of $f\left(x\right)=0$

Let us assume the other two roots of the $f\left(x\right)=0$ are $\alpha , \beta$

$\Rightarrow$ Sum of roots$=-2$

$\Rightarrow \alpha +\beta +2+\sqrt{3}+2-\sqrt{3}=-2$

$\Rightarrow \alpha +\beta =-6 \dots \left(\mathrm{i}\right)$

And product of roots$=7$

$\Rightarrow \alpha \beta \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=7$

$\Rightarrow \alpha \beta =7 \dots \left(\mathrm{ii}\right)$

Put $\beta =\frac{7}{\alpha }$ in equation $\left(\mathrm{i}\right)$, we get

$\Rightarrow {\alpha }^2+6\alpha +7=0$

$\Rightarrow \alpha =\frac{-6\pm \sqrt{36-28}}{2}=-3\pm \sqrt{2}$

Put value of $\alpha$ in equation $\left(\mathrm{ii}\right)$, we get

$\beta =-3\mp \sqrt{2}$

$\therefore$ Roots of the $f\left(x\right)=0$ are $2+\sqrt{3}, 2-\sqrt{3}, -3+\sqrt{2}, -3-\sqrt{2}$.