+0

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

0
357
2

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

Guest Sep 23, 2014

#2
+19207
+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
Sort:

#1
+85918
+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
+19207
+10
$$\begin{array}{c|l} \quad & \quad 3x^3-21x^2-50x+220=0 \quad |\; :3 \\ \quad & \quad \\ x^3+ax^2+bx+c=0 \quad & \quad x^3-7x^2-\frac{50}{3}x+\frac{220}{3}=0 \\ \quad & \quad \\ \quad & \quad a=-7; \; b=-\frac{50}{3}; \;c=\frac{220}{3} \\ \quad & \quad \\ B=\frac{3b-a^2}{3} \quad C=\frac{2a^3-9ab+27c}{27} \quad & \quad B=\frac{3(-\frac{50}{3})-(-7)^2}{3}=-\frac{99}{3}=-33 \quad \\ \quad & \quad \\ \quad & \quad C=\frac{2(-7)^3-9(-7)(-\frac{50}{3})+27(\frac{220}{3})}{27} =-\frac{-686-1050+1980}{27}=\frac{244}{27} \quad \\ \quad & \quad \\ d=\sqrt{-\frac{4}{3}B} \quad & \quad d=\sqrt{-\frac{4}{3}(-33)}=\sqrt{4*11}=2\sqrt{11}=6.633249581\\ \quad & \quad \\ q=\dfrac{C}{d^3}} \quad & \quad q=\frac{244}{27}\frac{1}{8(\sqrt{11})^3}}=\frac{61}{54}\frac{1}{(\sqrt{11})^3}=0.030963286\\ \quad & \quad \\ 4q <1 \;! \small{\text{ 3 real solutions }}\quad & \quad 4q = 0.123853144 < 1\;!\\ \quad & \quad \\ \phi\ensurement{^{\circ}}= \sin^{-1}{(4q)} \quad & \quad \phi\ensurement{^{\circ}}= \sin^{-1}{(0.123853144)} \\ \quad & \quad \\ \quad & \quad \phi\ensurement{^{\circ}}=7.114531178\ensurement{^{\circ}}\\ \quad & \quad \\ x_1=d*sin{(\frac{1}{3}\phi\ensurement{^{\circ}})}-\frac{a}{3} \quad & \quad x_1=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}*7.11451178\ensurement{^{\circ}} ) }=\frac{7}{3}+2\sqrt{11}*0.041378847 \\ \quad & \quad x_1=2.607809554\\ \quad & \quad \\ x_2=d*sin{(\frac{1}{3} (\phi\ensurement{^{\circ}} +360\ensurement{^{\circ}} ) )}-\frac{a}{3} \quad & \quad x_2=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}* (7.11451178\ensurement{^{\circ}}+360\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*0.844594254 \\ \quad & \quad x_2=7.935737816\\ \quad & \quad \\ x_3=d*sin{(\frac{1}{3} (\phi\ensurement{^{\circ}} +720\ensurement{^{\circ}} ) )}-\frac{a}{3} \quad & \quad x_3=\frac{7}{3}+2\sqrt{11}*\sin{ (\frac{1}{3}* (7.11451178\ensurement{^{\circ}}+720\ensurement{^{\circ}}) ) }=\frac{7}{3}+2\sqrt{11}*(-0.885973101) \\ \quad & \quad x_3=-3.543547371\\ \end{array}$$