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I'm a high school student trying to get critical intuition when learning algebraic equation solving.

For $x$ any complex number and $c$ constant, simple polynomial such as $x^n -c=0$ are easily solvable for $x$. Then if we know how to solve polynomial $g(x)=k$ for any constant $k$ we can now solve more complex polynomial $[g(x)]^n -c=0$. Hence we say Quadratic polynomial is solved when we deform it into $a(x-p)^2 -q=0$ since linear polynomial is solvable.

The reason for doing this when solving polynomials is because our intuition can only do linear calculations and fails to do calulations of "higher complexity" directly.

(I cant find sufficient word to describe this)

e.g.

  • $ab=1$ implies $a=b^{-1}$ and

  • $a+b=0$ implies $a=-b$

but $a^2 + b^2 +ab=0$ is whole new stuff.

Further from polynomials, I think algebra is all about deforming equations into reasonably simpler form and deciding if such progress is doable.

Am I making things right?

Tianlalu
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Solvable Potato
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  • What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation... – Jean Marie Dec 15 '18 at 03:59
  • ... (and the nth degree general equation with $n \geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily. – Jean Marie Dec 15 '18 at 04:10

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