Solutions of $x^4+2tx^2+1=0$

Question

I am looking for all the solutions of $x^4+2tx^2+1=0$

Is there any "quick" way to figure out all the complex-solutions $\pi_i(t)$ where $i=1,...4$ ?

I got $\pi_1(t)=\sqrt{-t+\sqrt{t^2-1}}\sqrt{t-\sqrt{t^2-1}}\sqrt{-t+\sqrt{1-t^2}}$

but apparently I am too "stupid" to verify my calculation. I always end up making mistakes (hopefully) in my calculation.

Would appreciate any help

Answer 1

My favourite hidden quadratic. Note: $$x^4+2tx^2+1=(x^2)^2+2t(x^2)^1+1(x^2)^0$$ Then using quadratic formula for $x^2$ gives $$x^2=\frac{-2t\pm\sqrt{4t^2-4}}{2}$$ And so by simplification $$x=\pm\sqrt{-t\pm\sqrt{t^2-1}}$$

Answer 2

$$x^4+2tx^2+1=0$$ Substitute $y=x^2$ $$y^2+2ty+1=0$$ Now use the formula $$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ So, here $a=1,b=2t,c=1$ $$y=\frac{-t\pm \sqrt{4t^2-4(1)(1)}}{2(1)}$$ $$y=\frac{-t\pm \sqrt{4t^2-4}}{2}$$ $$y=-t\pm \sqrt{t^2-1}$$

Substitute back $y=x^2$ $$x^2=-t\pm \sqrt{t^2-1}$$ $$x=\pm\sqrt{-t\pm \sqrt{t^2-1}}$$