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Solve 3x^3 - 21x^2 - 50x + 220 = 0

Guest Sep 23, 2014

Best Answer 

 #2
avatar+20549 
+10

Solve 3x^3 - 21x^2 - 50x + 220 = 0

$$\begin{array}{c|l}
\quad & \quad 3x^3-21x^2-50x+220=0 \quad |\; :3 \\
\quad & \quad \\
x^3+ax^2+bx+c=0 \quad & \quad x^3-7x^2-\frac{50}{3}x+\frac{220}{3}=0 \\
\quad & \quad \\
\quad & \quad a=-7; \; b=-\frac{50}{3}; \;c=\frac{220}{3} \\
\quad & \quad \\
B=\frac{3b-a^2}{3} \quad C=\frac{2a^3-9ab+27c}{27} \quad & \quad B=\frac{3(-\frac{50}{3})-(-7)^2}{3}=-\frac{99}{3}=-33 \quad \\
\quad & \quad \\
\quad & \quad C=\frac{2(-7)^3-9(-7)(-\frac{50}{3})+27(\frac{220}{3})}{27} =-\frac{-686-1050+1980}{27}=\frac{244}{27} \quad \\
\quad & \quad \\
d=\sqrt{-\frac{4}{3}B} \quad & \quad d=\sqrt{-\frac{4}{3}(-33)}=\sqrt{4*11}=2\sqrt{11}=6.633249581\\
\quad & \quad \\
q=\dfrac{C}{d^3}} \quad & \quad q=\frac{244}{27}\frac{1}{8(\sqrt{11})^3}}=\frac{61}{54}\frac{1}{(\sqrt{11})^3}=0.030963286\\
\quad & \quad \\
4q <1 \;! \small{\text{ 3 real solutions }}\quad & \quad 4q = 0.123853144 < 1\;!\\
\quad & \quad \\
\phi\ensurement{^{\circ}}= \sin^{-1}{(4q)} \quad & \quad \phi\ensurement{^{\circ}}= \sin^{-1}{(0.123853144)} \\
\quad & \quad \\
\quad & \quad \phi\ensurement{^{\circ}}=7.114531178\ensurement{^{\circ}}\\
\quad & \quad \\
x_1=d*sin{(\frac{1}{3}\phi\ensurement{^{\circ}})}-\frac{a}{3} \quad & \quad x_1=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*7.11451178\ensurement{^{\circ}} ) }=\frac{7}{3}+2\sqrt{11}*0.041378847 \\
\quad & \quad x_1=2.607809554\\
\quad & \quad \\
x_2=d*sin{(\frac{1}{3}
(\phi\ensurement{^{\circ}} +360\ensurement{^{\circ}} )
)}-\frac{a}{3} \quad & \quad x_2=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*
(7.11451178\ensurement{^{\circ}}+360\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*0.844594254 \\
\quad & \quad x_2=7.935737816\\
\quad & \quad \\
x_3=d*sin{(\frac{1}{3}
(\phi\ensurement{^{\circ}} +720\ensurement{^{\circ}} )
)}-\frac{a}{3} \quad & \quad x_3=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*
(7.11451178\ensurement{^{\circ}}+720\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*(-0.885973101) \\
\quad & \quad x_3=-3.543547371\\
\end{array}$$

heureka  Sep 24, 2014
 #1
avatar+92382 
+5

3x^3 - 21x^2 - 50x + 220 = 0

The easiest way to solve this is by graphing.......see it here: https://www.desmos.com/calculator/nklgal58ln

Notice that we have 3 "real" roots.......the only other possibility would have been 1 real root and 2 non-real roots, since this is a 3rd degree polynomial.

 

CPhill  Sep 23, 2014
 #2
avatar+20549 
+10
Best Answer

Solve 3x^3 - 21x^2 - 50x + 220 = 0

$$\begin{array}{c|l}
\quad & \quad 3x^3-21x^2-50x+220=0 \quad |\; :3 \\
\quad & \quad \\
x^3+ax^2+bx+c=0 \quad & \quad x^3-7x^2-\frac{50}{3}x+\frac{220}{3}=0 \\
\quad & \quad \\
\quad & \quad a=-7; \; b=-\frac{50}{3}; \;c=\frac{220}{3} \\
\quad & \quad \\
B=\frac{3b-a^2}{3} \quad C=\frac{2a^3-9ab+27c}{27} \quad & \quad B=\frac{3(-\frac{50}{3})-(-7)^2}{3}=-\frac{99}{3}=-33 \quad \\
\quad & \quad \\
\quad & \quad C=\frac{2(-7)^3-9(-7)(-\frac{50}{3})+27(\frac{220}{3})}{27} =-\frac{-686-1050+1980}{27}=\frac{244}{27} \quad \\
\quad & \quad \\
d=\sqrt{-\frac{4}{3}B} \quad & \quad d=\sqrt{-\frac{4}{3}(-33)}=\sqrt{4*11}=2\sqrt{11}=6.633249581\\
\quad & \quad \\
q=\dfrac{C}{d^3}} \quad & \quad q=\frac{244}{27}\frac{1}{8(\sqrt{11})^3}}=\frac{61}{54}\frac{1}{(\sqrt{11})^3}=0.030963286\\
\quad & \quad \\
4q <1 \;! \small{\text{ 3 real solutions }}\quad & \quad 4q = 0.123853144 < 1\;!\\
\quad & \quad \\
\phi\ensurement{^{\circ}}= \sin^{-1}{(4q)} \quad & \quad \phi\ensurement{^{\circ}}= \sin^{-1}{(0.123853144)} \\
\quad & \quad \\
\quad & \quad \phi\ensurement{^{\circ}}=7.114531178\ensurement{^{\circ}}\\
\quad & \quad \\
x_1=d*sin{(\frac{1}{3}\phi\ensurement{^{\circ}})}-\frac{a}{3} \quad & \quad x_1=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*7.11451178\ensurement{^{\circ}} ) }=\frac{7}{3}+2\sqrt{11}*0.041378847 \\
\quad & \quad x_1=2.607809554\\
\quad & \quad \\
x_2=d*sin{(\frac{1}{3}
(\phi\ensurement{^{\circ}} +360\ensurement{^{\circ}} )
)}-\frac{a}{3} \quad & \quad x_2=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*
(7.11451178\ensurement{^{\circ}}+360\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*0.844594254 \\
\quad & \quad x_2=7.935737816\\
\quad & \quad \\
x_3=d*sin{(\frac{1}{3}
(\phi\ensurement{^{\circ}} +720\ensurement{^{\circ}} )
)}-\frac{a}{3} \quad & \quad x_3=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*
(7.11451178\ensurement{^{\circ}}+720\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*(-0.885973101) \\
\quad & \quad x_3=-3.543547371\\
\end{array}$$

heureka  Sep 24, 2014

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