Los op: $x^3-5x-2=0$
$ \eqalign{ & a = 1,\,\,b = 0,\,\,c = - 5\,\,en\,\,d = - 2 \cr & p = {c \over a} - {{b^2 } \over {3a^2 }} \to p = {{ - 5} \over 1} - {{ - 0^2 } \over {3 \cdot 1^2 }} = - 5 \cr & q = {{2b^3 } \over {27a^3 }} - {{bc} \over {3a^2 }} + {d \over a} \to q = {{2 \cdot 0^3 } \over {27 \cdot 1^3 }} - {{0 \cdot - 5} \over {3 \cdot 1^2 }} + {{ = 2} \over 1} = - 2 \cr & W = \sqrt {q^2 + {4 \over {27}}p^3 } \to W = \sqrt {\left( { - 2} \right)^2 + {4 \over {27}} \cdot \left( { - 5} \right)^3 } = 1{5 \over 9}i\sqrt 6 \cr & x = \root 3 \of {{{ - q + W} \over 2}} + \root 3 \of {{{ - q - W} \over 2}} - {b \over {3a}} \cr & x = \root 3 \of {{{ - - 2 + 1{5 \over 9}i\sqrt 6 } \over 2}} + \root 3 \of {{{ - - 2 - 1{5 \over 9}i\sqrt 6 } \over 2}} - {0 \over {3 \cdot 1}} \cr & x = \sqrt 2 + 1 \cr} $
Eén oplossing gevonden!
Complexe tussenstap
$
\sqrt {\left( { - 2} \right)^2 + {4 \over {27}} \cdot \left( { - 5} \right)^3 } = 1{5 \over 9}i\sqrt 6
$. Dat gaat zo:
$
\eqalign{
& \sqrt {\left( { - 2} \right)^2 + {4 \over {27}} \cdot \left( { - 5} \right)^3 } = \cr
& \sqrt { - {{392} \over {27}}} = \cr
& i\sqrt {{{392} \over {27}}} = \cr
& i \cdot {{14} \over 9}\sqrt 6 = \cr
& 1{5 \over 6}i\sqrt 6 \cr}
$
$
\eqalign{
& \root 3 \of {{{2 + 1{5 \over 9}i\sqrt 6 } \over 2}} + \root 3 \of {{{2 - 1{5 \over 9}i\sqrt 6 } \over 2}} = \cr
& \root 3 \of {1 + {7 \over 9}i\sqrt 6 } + \root 3 \of {1 - {7 \over 9}i\sqrt 6 } = \cr
& {1 \over 2}\sqrt 2 + {1 \over 2} + i\left( {{1 \over 2}\sqrt 3 - {1 \over 6}\sqrt 6 } \right) + {1 \over 2}\sqrt 2 + {1 \over 2} + i\left( { - {1 \over 2}\sqrt 3 + {1 \over 6}\sqrt 6 } \right) = \cr
& \sqrt 2 + 1 \cr}
$
