Solve-x-3-15x-92-0-

Question Number 110182 by udaythool last updated on 27/Aug/20

Solve x^{3} + 15 x - 92 = 0

Commented by mr W last updated on 27/Aug/20

l e t x = u + v

{(u + v)}^{3} + 15 (u + v) - 92 = 0

u^{3} + v^{3} + 3 (u v + 5) (u + v) - 92 = 0

l e t u v + 5 = 0 \Rightarrow u v = - 5 \Rightarrow u^{3} v^{3} = - 125

\Rightarrow u^{3} + v^{3} - 92 = 0 \Rightarrow u^{3} + v^{3} = 92

u^{3} a n d v^{3} a r e r o o t s o f z^{2} - 92 z - 125 = 0

\Rightarrow z = 46 \pm \sqrt{46^{2} + 125} = 46 \pm 3 \sqrt{249}

\Rightarrow u = \sqrt[3]{46 + 3 \sqrt{249}}

\Rightarrow v = \sqrt[3]{46 - 3 \sqrt{249}}

\Rightarrow x = \sqrt[3]{46 + 3 \sqrt{249}} + \sqrt[3]{46 - 3 \sqrt{249}} \approx 3.4339

Commented by udaythool last updated on 28/Aug/20

��THANKS��

Answered by 1549442205PVT last updated on 28/Aug/20

Put x = \sqrt{5} y \Rightarrow 5 \sqrt{5} y^{3} + 15 \sqrt{5} y - 92 = 0

\Leftrightarrow y^{3} + 3 y - \frac{92}{5 \sqrt{5}} = 0 (1)

Put y = z - \frac{1}{z} \Rightarrow y^{3} = z^{3} - \frac{1}{z^{3}} - 3 (z - \frac{1}{z})

Replace into (1) we get

z^{3} - \frac{1}{z^{3}} - \frac{92 \sqrt{5}}{25} = 0 \Leftrightarrow z^{6} - \frac{92 \sqrt{5}}{25} z^{3} - 1 = 0

Prime causes double exponent: use braces to clarify

Δ = {(\frac{92 \sqrt{5}}{25})}^{2} + 4 = \frac{8464}{125} + 4 = \frac{8964}{125}

\Rightarrow z^{3} = \frac{\frac{92 \sqrt{5}}{25} \pm \frac{6 \sqrt{1245}}{25}}{2} = \frac{92 \sqrt{5} \pm 6 \sqrt{1245}}{50}

i) For z^{3} = \frac{92 \sqrt{5} + 6 \sqrt{1245}}{50} .

Since (a - b) (a + b) = a^{2} - b^{2}, we have

\frac{92 \sqrt{5} + 6 \sqrt{1245}}{50} \times \frac{92 \sqrt{5} - 6 \sqrt{1245}}{50} = - 1 (*)

Hence, . By {Vieta}^{'} s theorem we get

z =^{3} \sqrt{\frac{92 \sqrt{5} + 6 \sqrt{1245}}{50}}, - \frac{1}{z} =^{3} \sqrt{\frac{92 \sqrt{5} - 6 \sqrt{1245}}{50}}

\Rightarrow x = \sqrt{5} y = \sqrt{5} (z - \frac{1}{z}) =

\sqrt{5} (^{3} \sqrt{\frac{92 \sqrt{5} + 6 \sqrt{1245}}{50}} +^{3} \sqrt{\frac{92 \sqrt{5} - 6 \sqrt{1245}}{50}})

=^{3} \sqrt{\frac{2300 + 150 \sqrt{249}}{50}} +^{3} \sqrt{\frac{2300 - 150 \sqrt{249}}{50}}

==^{3} \sqrt{46 + 3 \sqrt{249}} +^{3} \sqrt{46 - 3 \sqrt{249}}

ii) For z^{3} =^{3} \sqrt{\frac{92 \sqrt{5} - 6 \sqrt{1245}}{50}} . From (*)

we infer z - \frac{1}{z} is the same value

Thus, given equation x^{3} + 15 x - 92 = 0

has unique root

x =^{3} \sqrt{46 + 3 \sqrt{249}} +^{3} \sqrt{46 - 3 \sqrt{249}}

Commented by udaythool last updated on 28/Aug/20

Really I enjoyed this approach. Thanks ��