Nog een voorbeeld
Los op: $
x^4 + 2x^3 + 5x^2 + 6x + 9 = 0
$
Uitwerking
$
\eqalign{
& x^4 + 2x^3 + 5x^2 + 6x + 9 = 0 \cr
& x^2 + 2x + 5 + \frac{6}
{x} + \frac{9}
{{x^2 }} = 0 \cr
& x^2 + \frac{9}
{{x^2 }} + 6 + 2\left( {x + \frac{3}
{x}} \right) - 1 = 0 \cr
& \left( {x + \frac{3}
{x}} \right)^2 + 2\left( {x + \frac{3}
{x}} \right) - 1 = 0 \cr
& z^2 + 2z - 1 = 0 \cr
& z = - \sqrt 2 - 1 \vee z = \sqrt 2 - 1 \cr
& x + \frac{3}
{x} = - \sqrt 2 - 1 \vee x + \frac{3}
{x} = \sqrt 2 - 1 \cr
& x^2 + 3 = \left( { - \sqrt 2 - 1} \right)x \vee x^2 + 3 = \left( {\sqrt 2 - 1} \right)x \cr
& x^2 + \left( {\sqrt 2 + 1} \right)x + 3 = 0 \vee x^2 - \left( {\sqrt 2 - 1} \right)x + 3 = 0 \cr
& x = \frac{{ - \left( {\sqrt 2 + 1} \right) \pm \sqrt {\left( {\sqrt 2 + 1} \right)^2 - 4 \cdot 1 \cdot 3} }}
{{2 \cdot 1}} \vee x = \frac{{ - \left( {\sqrt 2 - 1} \right) \pm \sqrt {\left( {\sqrt 2 - 1} \right)^2 - 4 \cdot 1 \cdot 3} }}
{{2 \cdot 1}} \cr
& x = \frac{{ - \sqrt 2 - 1 \pm \sqrt {2\sqrt 2 + 3 - 12} }}
{2} \vee x = \frac{{ - \sqrt 2 + 1 \pm \sqrt { - 2\sqrt 2 + 3 - 12} }}
{2} \cr
& x = \frac{{ - \sqrt 2 - 1 \pm \sqrt {2\sqrt 2 - 9} }}
{2} \vee x = \frac{{ - \sqrt 2 + 1 \pm \sqrt { - 2\sqrt 2 - 9} }}
{2} \cr
& x = - \frac{1}
{2}\sqrt 2 - \frac{1}
{2} \pm \frac{1}
{2}\sqrt {2\sqrt 2 - 9} \vee x = - \frac{1}
{2}\sqrt 2 + \frac{1}
{2} \pm \frac{1}
{2}\sqrt { - 2\sqrt 2 - 9} \cr
& x = - \frac{1}
{2}\sqrt 2 - \frac{1}
{2} \pm \frac{1}
{2}i\sqrt {9 - 2\sqrt 2 } \vee x = - \frac{1}
{2}\sqrt 2 + \frac{1}
{2} \pm \frac{1}
{2}i\sqrt {9 + 2\sqrt 2 } \cr
& dus: \cr
& x = - \frac{1}
{2}\sqrt 2 - \frac{1}
{2} - \frac{1}
{2}i\sqrt {9 - 2\sqrt 2 } \vee \cr
& x = - \frac{1}
{2}\sqrt 2 + \frac{1}
{2} + \frac{1}
{2}i\sqrt {9 + 2\sqrt 2 } \vee \cr
& x = - \frac{1}
{2}\sqrt 2 - \frac{1}
{2} + \frac{1}
{2}i\sqrt {9 - 2\sqrt 2 } \vee \cr
& x = - \frac{1}
{2}\sqrt 2 + \frac{1}
{2} - \frac{1}
{2}i\sqrt {9 + 2\sqrt 2 } \cr}
$
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