+0  
 
0
63
3
avatar

x|x|=2x+1 

The lines are absolute value

Guest Aug 18, 2017
Sort: 

3+0 Answers

 #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+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 \(x^4 - 4x^2+4x+1 = 0\)

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+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

5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details