My Math Forum Solution for algebraic equations

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 March 26th, 2014, 08:34 PM #1 Senior Member   Joined: Nov 2013 Posts: 137 Thanks: 1 Solution for algebraic equations I noticed that it's 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 $\begin{cases} 0=(3a(u+v)+b)\\ 0=(3a(u+v)^2 + 2b(u+v) + c)\\ \end{cases}\;\;\;\Rightarrow\;\;\;\;u+v = \frac{\pm \sqrt{9a^2-12ac+4b^2} + 3a - 2b}{6a}$ 1) Is this idea correct? 2) If yes, is it possible to solve equations of the 5th, 6th, 7th... degree with this process?
 March 29th, 2014, 11:46 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory I haven't looked at it closely for errors, seems almost obvious that this is false. I'll just directly check your final result for $x^3 + x^2 + x + 1= 0$ Let's compute $t= u + v$. This is $t= \frac{\pm 1 + 1}{6}$ If 0, $x= 0$ and this case is outright ruled out. Otherwise, $t= 1/3$ in which case $x= \sqrt[3]{-\frac{40}{27}} + \frac13 \approx 0.9033 + 0.9872i$ And this is clearly not a zero of the polynomial given!

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