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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.

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    $\begingroup$ $x=\pm\sqrt{\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$.It is the correct solution since quadratic formulae must work but may be complex. $\endgroup$
    – kingW3
    Aug 29 '15 at 14:24
  • $\begingroup$ This method will work well on this type of equation. You can plug in your final answers into the original equation to check that they're solutions. However, since you will get four solutions out of this method (and all solutions will arise in this way), you are safe knowing that these are all of the solutions. $\endgroup$ Aug 29 '15 at 14:24
  • $\begingroup$ Side note: but when you have equation of the form $ax^4+bx^3+cx^2+dx+e$, you can use Ferrar's formula. Look here: en.wikipedia.org/wiki/Quartic_function. $\endgroup$ Aug 29 '15 at 14:26
  • $\begingroup$ It is right! (y) $\endgroup$ Aug 29 '15 at 14:32
  • $\begingroup$ Often, people just leave it as 'a solution of $ax^4+bx^3+cx^2+dx+e=0$'. There are other methods (for example a method called 'Newton's Method') that give the answer to as many decimal places as you want. $\endgroup$
    – Empy2
    Aug 29 '15 at 14:34
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Yes, quite correct, except that the outer square-root has a separate $\pm$. So $x=\pm\sqrt{-b\pm\sqrt{b^2-4ac}/2a}$

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