Question

Solve : $a^2+\frac{1}{a^2}-2-3a+\frac{3}{a}$

Answer 1

$a^2+\frac{1}{a^2}-2-3a+1$

$\frac{a^4+1-2a^2-3a^3+3a}{a^2}=0$

$a^4-3a^3-2a^2+3a+1=0$

$(a+1)\frac{a^4-3a^3-2a^2+3a+1}{(a+1)}=0$

$(a+1)(a^3-4a^2+2a+1)=0$

$(a+1)\left[\right(a-1\left)\right(a^2-3a-1\left)\right]=0$

Solving the quadratic equation $(a^2-3a-1)$ we get,

$a=\frac{3\pm \sqrt{13}}{2}$

$\therefore$ Roots of biquadratic equation $a^2+\frac{1}{a^2}-2-3a+1$ are,

$a=\pm 1,\frac{3\pm \sqrt{13}}{2}$