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x|x|=2x+1 

The lines are absolute value

Guest Aug 18, 2017
 #1
avatar+178 
+1

Input: x|x|=2x+1

Intepretation: \(x\left|x\right|=2x+1\)

Rewrite into alternative form:

\(x\sqrt{x^2}=2x+1\)

Simplify the square:

\(x^2=2x+1\)

Move the terms to the left:

\(x^2-2x-1=0\)

This function doesn't look like that it is factorizable, applying formulaic solution:

For \(ax^2+bx+c=0,\space x = {-b \pm \sqrt{b^2-4ac} \over 2a}\),

Plug \(a=1,\space b=-2,\space c=-1\)

\(x=\frac{2\pm \sqrt{4+4}}{2}\)

\(=\frac{2\pm \sqrt{8}}{2}\)

\(=\frac{2\pm 2\sqrt{2}}{2}\)

\(=1±\sqrt2\)

\(x=1+\sqrt{2}\ or\ 1-\sqrt{2}\)

Jeffes02  Aug 18, 2017
edited by Jeffes02  Aug 18, 2017
edited by Jeffes02  Aug 18, 2017
 #2
avatar+20001 
+3

x|x|=2x+1 

The lines are absolute value

 

\(\begin{array}{|rcll|} \hline x|x| &=& 2x+1 \quad & | \quad \text{square both sides} \\ x^2\cdot |x|^2 &=& (2x+1)^2 \quad & | \quad |x|^2 = x^2 \\ x^2\cdot x^2 &=& (2x+1)^2 \\ x^4 &=& 4x^2+4x+1 \\ \mathbf{x^4 - 4x^2+4x+1} & \mathbf{=} & \mathbf{0} \\ \hline \end{array} \)

 

solutions of \(x^4 - 4x^2+4x+1 = 0\)

see: http://www.wolframalpha.com/input/?i=x%5E4-4x%5E2-4x-1%3D0

\(\begin{array}{rcll} x_1 &=& -1 \\ x_2 &=& 1 - \sqrt{2} \\ x_3 &=& 1 + \sqrt{2} \\ \end{array}\)

 

Solutions of \(x|x|=2x+1\):

\(x_1 = -1 \\ \begin{array}{|rcll|} \hline (-1)\cdot |-1| & \overset{?}{=} & 2\cdot (-1)+1 \\ (-1)\cdot 1 & \overset{?}{=} & -2+1 \\ -1 & \overset{!}{=} & -1\ \checkmark \\ \hline \end{array} \)

x = -1 is a solution

 

\(x_2 = 1 - \sqrt{2} \\ \begin{array}{|rcll|} \hline (1 - \sqrt{2})\cdot |1 - \sqrt{2}| & \overset{?}{=} & 2\cdot (1 - \sqrt{2})+1 \\ (-0.41421356237)\cdot |-0.41421356237| & \overset{?}{=} & 2\cdot (-0.41421356237)+1 \\ (-0.41421356237)\cdot 0.41421356237 & \overset{?}{=} & -0.82842712475 + 1 \\ -0.17157287525 & \ne & 0.17157287525 \\ \hline \end{array}\)

\(\mathbf{x = 1-\sqrt{2} }\) is not a solution

 

\(x_2 = 1 + \sqrt{2} \\ \begin{array}{|rcll|} \hline (1 + \sqrt{2})\cdot |1 + \sqrt{2}| & \overset{?}{=} & 2\cdot (1 + \sqrt{2})+1 \\ (2.41421356237)\cdot |2.41421356237| & \overset{?}{=} & 2\cdot (2.41421356237)+1 \\ 2.41421356237\cdot 2.41421356237 & \overset{?}{=} & 4.82842712475 + 1 \\ 5.82842712475 & \overset{!}{=} & 5.82842712475\ \checkmark \\ \hline \end{array}\)

\(\mathbf{x = 1+\sqrt{2} }\) is a solution

 

laugh

heureka  Aug 18, 2017
 #3
avatar+93303 
+1

x|x|=2x+1 

The lines are absolute value

 

I'd do it in 2 parts.

 

If x is positive then

\(x^2=2x+1\\ x^2-2x-1=0\\ x=\frac{2\pm\sqrt{4--4}}{2}\\ x=\frac{2\pm\sqrt{8}}{2}\\ x=\frac{2\pm2\sqrt{2}}{2}\\ x=1\pm\sqrt{2}\\ \text{Since x is positive the only answer is }\\ x=1+\sqrt2\)

 

 

If x is negative then

\(-x^2=2x+1\\ -x^2-2x-1=0\\ x=\frac{2\pm\sqrt{4-4}}{2}\\ x=1\\ \text{but since x is negative there is no solution }\\ \)

 

\(\text{So the only solution is }\quad x=1+\sqrt{2}\)

Melody  Aug 18, 2017

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