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August 27th, 2017, 09:52 PM   #1
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finding x in an equation

x^4+5x^3+50x^2-140x+168=0
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August 27th, 2017, 10:51 PM   #2
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August 28th, 2017, 01:38 AM   #3
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Newton-Raphson

Best solution is to use Newton-Raphson Numerical method to evaluate the given function.

Last edited by skipjack; August 28th, 2017 at 08:32 AM.
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August 28th, 2017, 03:56 AM   #4
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That equation has no real roots.
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August 28th, 2017, 04:05 AM   #5
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QUARTIC EQUATION CALCULATOR
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August 28th, 2017, 05:13 AM   #6
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One could use CAS Maxima.

Code:
algexact: true;
tex(algsys([x^4+5*x^3+50*x^2-140*x+168],[x]));

$\displaystyle \left[ \left[ x={{\sqrt{{{5985}\over{2\,\sqrt{{{26464-975\,\left({{
301400}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}} \right)^{{{1
}\over{3}}}+36\,\left({{301400}\over{27}}+ {{56\,\sqrt{782431}\,i
}\over{3}}\right)^{{{2}\over{3}}}}\over{\left({{30 1400}\over{27}}+{{
56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3 }}}}}}}}-\left({{
301400}\over{27}}+ {{56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1
}\over{3}}}-{{6616} \over{9\,\left({{301400}\over{27}}+{{56\,\sqrt{782 431}\,i}\over{3}}\right)^{{{1}\over{3}}}}}-{{325}\over{6}}}
}\over{2}}+{{\sqrt{{{26464-975\,\left({{301400}\over{27}}+{{56\,
\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3}}}+ 36\,\left({{301400
}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right )^{{{2}\over{3}}}
}\over{\left({{301400}\over{27}}+{{56\,\sqrt{78243 1}\,i}\over{3}}
\right)^{{{1}\over{3}}}}}}}\over{12}}-{{5}\over{4}} \right] ,
\left[ x=-{{\sqrt{{{5985}\over{2\,\sqrt{{{26464-975\,\left({{301400
}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right )^{{{1}\over{3}}}+
36\,\left({{301400}\over{27}}+{{56\,\sqrt{782431}\ ,i}\over{3}}
\right)^{{{2}\over{3}}}}\over{\left({{301400}\over {27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1 }\over{3}}}}}}}}-\left({{301400
}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right )^{{{1}\over{3}}}-
{{6616}\over{9\,\left({{301400}\over{27}}+ {{56\,\sqrt{782431}\,i
}\over{3}}\right)^{{{1}\over{3}}}}}-{{325}\over{6}}}}\over{2}}+{{
\sqrt{{{26464-975\,\left({{301400}\over{27}}+ {{56\,\sqrt{782431}\,i
}\over{3}}\right)^{{{1}\over{3}}}+36\,\left({{3014 00}\over{27}}+{{56
\,\sqrt{782431}\,i}\over{3}}\right)^{{{2}\over{3}} }}\over{\left({{
301400}\over{27}}+ {{56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1
}\over{3}}}}}}}\over{12}}-{{5}\over{4}} \right] , \left[ x={{\sqrt{
-{{5985}\over{2\,\sqrt{{{26464-975\,\left({{301400}\over{27}}+{{56\,
\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3}}}+ 36\,\left({{301400
}\over{27}}+ {{56\,\sqrt{782431}\,i}\over{3}}\right)^{{{2}\over {3}}}
}\over{\left({{301400}\over{27}}+{{56\,\sqrt{78243 1}\,i}\over{3}}
\right)^{{{1}\over{3}}}}}}}}-\left({{301400}\over{27}}+{{56\,\sqrt{782431}\,i}\ over{3}}\right)^{{{1}\over{3}}}-{{6616}\over{9\,\left({{
301400}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}} \right)^{{{1
}\over{3}}}}}-{{325}\over{6}}}}\over{2}}-{{\sqrt{{{26464-975\,\left(
{{301400}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3 }}\right)^{{{1
}\over{3}}}+36\,\left({{301400}\over{27}}+ {{56\,\sqrt{782431}\,i
}\over{3}}\right)^{{{2}\over{3}}}}\over{\left({{30 1400}\over{27}}+{{ 56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3 }}}}}}}\over{12}}-
{{5}\over{4}} \right] , \left[ x=-{{\sqrt{-{{5985}\over{2\,\sqrt{{{
26464-975\,\left({{301400}\over{27}}+{{56\,\sqrt{782431} \,i}\over{3
}}\right)^{{{1}\over{3}}}+36\,\left({{301400}\over {27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right)^{{{2 }\over{3}}}}\over{\left({{301400
}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right )^{{{1}\over{3}}}
}}}}}-\left({{301400}\over{27}}+{{56\,\sqrt{782431}\,i}\ over{3}}
\right)^{{{1}\over{3}}}-{{6616}\over{9\,\left({{301400}\over{27}}+{{
56\,\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3 }}}}}-{{325}\over{
6}}}}\over{2}}-{{\sqrt{{{26464-975\,\left({{301400}\over{27}}+{{56\,
\sqrt{782431}\,i}\over{3}}\right)^{{{1}\over{3}}}+ 36\,\left({{301400
}\over{27}}+{{56\,\sqrt{782431}\,i}\over{3}}\right )^{{{2}\over{3}}}
}\over{\left({{301400}\over{27}}+{{56\,\sqrt{78243 1}\,i}\over{3}}
\right)^{{{1}\over{3}}}}}}}\over{12}}-{{5}\over{4}} \right]
\right]$

Last edited by skipjack; August 28th, 2017 at 08:58 AM.
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August 29th, 2017, 05:17 AM   #7
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Holy Maxima, Victor!!

x^4 - 10x^3 + 35x^2 - 50x + 24 = 0

Try that one...
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August 29th, 2017, 05:24 AM   #8
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Code:
factor(x^4 - 10*x^3 + 35*x^2 - 50*x + 24);
$\displaystyle (x-4)(x-3)(x-2)(x-1)$
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August 29th, 2017, 05:35 AM   #9
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Aha! Thanks to me, your post this time is only 1 line

Btw, should be (x - 1)(x - 2)(x - 3)(x - 4)
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August 29th, 2017, 11:20 AM   #10
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Quote:
Originally Posted by Denis View Post
Holy Maxima...
Our queen will be happy!
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