# Solution for equation of 2,3,4, degree

1. Mar 26, 2014

### Jhenrique

I noticed that is always possible to find the roots of polynomials of kind:
$A\lambda^2 + B$
$A\lambda^3 + B$
$A\lambda^4 + B$
$A\lambda^5 + B$

So I thought to transform the quadratic, the cubic, the quartic and the quintic function in the forms above, look:

$ax^3+bx^2+cx+d=0$

$(x=\lambda+u+v)$

$a(\lambda+u+v)^3+b(\lambda+u+v)^2+c(\lambda+u+v)+d=0$

$\\a \left ( \begin{matrix} \lambda^3 & +3\lambda^2u & +3\lambda u^2 & +u^3\\ +3\lambda^2v & +6\lambda u v & +3u^2 v & \\ +3\lambda v^2 & +3uv^2 & & \\ +v^3 & & & \\ \end{matrix} \right ) + b \left ( \begin{matrix} \lambda^2 & +2\lambda u & +u^2\\ +2\lambda v & +2 u v & \\ +v^2 & & \end{matrix} \right ) + c (\lambda + u + v) + d = 0$

$a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0$

So, comparing the equations:

$A\lambda^3 + B = a\lambda^3+(3a(u+v)+b)\lambda^2+(3a(u+v)^2 + 2b(u+v) + c)\lambda + (a(u+v)^3 + b(u+v)^2 + c(u+v) + d) = 0$

implies that:
$\begin{cases} A=a\\ 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v) + c)\\ B=(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)\\ \end{cases}$

So,
$A\lambda^3 + B = 0\;\;\;\Rightarrow\;\;\;\;\lambda = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x-u-v = \sqrt[3]{\frac{-B}{A}} \;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-B}{A}} + (u+v)\;\;\;\Rightarrow\;\;\;\; x = \sqrt[3]{\frac{-(a(u+v)^3 + b(u+v)^2 + c(u+v) + d)}{a}} + (u+v)$

Happens that
$0=3a(u+v)+b\;\;\;\Rightarrow\;\;\;\;u+v = \frac{-b}{3a}$

Conclusion:
$$x=\sqrt[3]{\frac{d}{a}-\frac{2b^3}{3^3a^3}-\frac{bc}{3a^2}}-\frac{b}{3a}$$

This ideia is correct?

2. Mar 26, 2014

### micromass

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