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For: $ax+b=0\;;\; x= \frac{-b}{a}\;\;\;\;\;$ and for:$$ Ax^2 +bx+c = 0\;; \;x = \frac{-b\pm\sqrt {b^2-4ac}}{2a}$$

And for $$ Ax^3+bx^2+cx+d =0$$ Cubic Formula

Is there a constant transformation from equations $x = \frac{-b}a\;\; -$ Linear; to: $x = \frac{-b\pm\sqrt {b^2-4ac}}{2a} \;\;-$ Quadratic; to: Cubic Formula$-\;$Cubic; ect...

In my head there must be a tacit relationship between the equations, I ask this because I want to program an application that can resolve any polynomial thrown at it. I've got linear and Quadratic working but if I had the relationship between the equations I could make one function to work for all. I don't know if this is more of a StackOverflow question but it is heavily math-based.

Harry Peter
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Tobi
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  • Two things; the LaTeX code for $\pm$ is \pm, and you also know that above degree 4 there is no general solution for a polynomial? Look up the unsolvability of the quintic. – walkar May 17 '16 at 15:42
  • Well that's burst my bubble – Tobi May 17 '16 at 15:57

1 Answers1

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There does not exist any general solution for polynomials of degree $\geq 5$ (In terms of radicals). Not that we haven't discovered one, but it was proven impossible. See here.

Eff
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