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# Solve for x

0
175
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x^3 = 2x+1

Guest Jan 7, 2015

#2
+18827
+10

x^3 = 2x+1

$$x^3 -2x-1 = 0 \qquad \Rightarrow x_1=-1 \\ \\ (x^3 -2x-1) : (x+1) = x^2-x-1 \\\\ (x+1)\underbrace{(x^2-x-1)}_{=0} = 0\\\\ x^2-x-1 = 0\\ x_{2,3}=\frac{ 1\pm\sqrt{1-1*4*(-1)} } {2*1} = \frac{ 1\pm\sqrt{5 } } {2} \\ \\ x_2 = \frac{1+\sqrt{5}}{2} =1.61803398875 \\\\ x_3 = \frac{1-\sqrt{5}}{2} =-0.61803398875$$

heureka  Jan 7, 2015
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#1
+80941
+10

x^3 = 2x+1    subtract everything on the right side to make it 0....so we have....

x^3 - 2x - 1  =  0     and from the Facror Theorem, we have that -1 is a root (solution)

Therefore, using a little synthetic division, we have

-1  [  1    0   - 2    -1 ]

- 1     1     1

-----------------------

1     -1   -1     0

This tells us that the polynomial remaining after we divide  x^3 - 2x -1  by (x + 1)  = x^2 - x - 1

And setting this = 0, we have that x =

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ {\mathtt{x}} = {\frac{\left({\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{{\mathtt{2}}}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\ {\mathtt{x}} = {\mathtt{1.618\: \!033\: \!988\: \!749\: \!894\: \!8}}\\ \end{array} \right\}$$

So these are the 3 solutions......BTW.....the last 2 solutions are better known as  "-phi" and "Phi"......

CPhill  Jan 7, 2015
#2
+18827
+10
$$x^3 -2x-1 = 0 \qquad \Rightarrow x_1=-1 \\ \\ (x^3 -2x-1) : (x+1) = x^2-x-1 \\\\ (x+1)\underbrace{(x^2-x-1)}_{=0} = 0\\\\ x^2-x-1 = 0\\ x_{2,3}=\frac{ 1\pm\sqrt{1-1*4*(-1)} } {2*1} = \frac{ 1\pm\sqrt{5 } } {2} \\ \\ x_2 = \frac{1+\sqrt{5}}{2} =1.61803398875 \\\\ x_3 = \frac{1-\sqrt{5}}{2} =-0.61803398875$$