x|x|=2x+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 $a=1,\space b=-2,\space c=-1$

$x=\frac{2\pm \sqrt{4+4}}{2}$

$=\frac{2\pm \sqrt{8}}{2}$

$=\frac{2\pm 2\sqrt{2}}{2}$

$=1±\sqrt2$

$x=1+\sqrt{2}\ or\ 1-\sqrt{2}$

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

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 }\\ $

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