Solving an equation containing 4th power of variable.

Question

I know how to solve Quadratic equations. Recently i came across the equation of type $ax^4 + bx^2 + c = 0$ and i had to solve it. So what i did is that i supposed $x^2 = y$ so that the above equation becomes: $ay^2 + by + c = 0$ and i found y by quadratic formula and took the square root of y getting x.I.e: $x = \sqrt{-b \pm \frac{\sqrt{b^2 - 4ac}}{2a}}$ But i have to confirm whether this solution is correct or not. If not then what is the true solution.

Answer 1

Yes, quite correct, except that the outer square-root has a separate $\pm$. So $x=\pm\sqrt{-b\pm\sqrt{b^2-4ac}/2a}$