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# x|x|=2x+1

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x|x|=2x+1

The lines are absolute value

Guest Aug 18, 2017
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#1
+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
+18564
+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 $$x^4 - 4x^2+4x+1 = 0$$

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

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

heureka  Aug 18, 2017
#3
+90192
+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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