Question

x${}^4$ +3x${}^2$ =16x + 60.

Answer 1

Adding $x^2+4$ to both sides of the given equation, we get
$x^4+4x^2+4=x^2+16x+64$ or $(x^2+2{)}^2-(x+8{)}^2=0$

Apply $a^2-b^2=(a+b)(a-b)$

$\therefore (x^2+x+10)(x^2-x-6)=0$

$\therefore x=3,-2$ and $\frac{-1\pm \sqrt{-39}}{2}$