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Omar Khayym's geometric solution of a cubic equation

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"Omar Khayyám's geometric solution of a cubic equation, for the case $m = 2$, $n = 16$, giving the root $2$. The intersection of the vertical line on the $x$-axis at the center of the circle is happenstance of the example illustrated."


Voorbeeld

De vergelijking $x^3+4x=16$. Volgens Derive is de oplossing gelijk aan $x=2$. We zien $a=2$ en $b=16$.

Volgens Khayyam:

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$ \begin{array}{l} \left\{ \begin{array}{l} y = \frac{1}{2}x^2 \\ \left( {x - 2} \right)^2 + y^2 = 4 \\ \end{array} \right. \\ \left( {x - 2} \right)^2 + \left( {\frac{1}{2}x^2 } \right)^2 = 4 \\ x^2 - 4x + 4 + \frac{1}{4}x^4 = 4 \\ \frac{1}{4}x^4 + x^2 - 4x = 0 \\ x^4 + 4x^2 - 16x = 0 \\ x(x^3 + 4x - 16) = 0 \\ x = 0 \vee x^3 + 4x - 16 = 0 \\ x = 0 \vee x^3 + 4x = 16 \\ \end{array} $

Opdracht

Benader de oplossing van $x^3+16x=64$ m.b.v. de methode van Khayyam.

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