What are the solutions to this absolute value equation?

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Discussion Overview

The discussion revolves around solving the absolute value equation |1/2x + 1| = |x|. Participants explore different methods to find solutions and clarify potential misunderstandings regarding the expression's notation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially claims the solutions are x = 2 and x = -2/3.
  • Another participant presents a method involving two cases:
    • For the case where 1/2x + 1 = x, they derive the quadratic equation 2x² - 2x - 1 = 0, leading to solutions x = (1 ± √3)/2.
    • For the case where 1/2x + 1 = -x, they derive the quadratic equation 2x² + 2x + 1 = 0, resulting in complex solutions x = (-1 ± i)/2.
  • A later reply corrects the second equation's derivation, confirming the complex solutions.
  • Another participant emphasizes the importance of notation, clarifying that 1/2x is often interpreted as 1/(2x) rather than (1/2)x, suggesting the use of brackets for clarity.

Areas of Agreement / Disagreement

Participants generally agree on the need for clarity in notation and the validity of some proposed solutions, but multiple competing views on the correct interpretation and solutions remain unresolved.

Contextual Notes

There are potential ambiguities in the notation used for the equation, which may affect the interpretation of the solutions. The discussion includes both real and complex solutions, and the derivation steps contain corrections and refinements.

Alexstrasuz1
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I have trouble solving this equation
|1/2x+1|=|x|

My answers are x=2 and x=-2/3
 
Last edited:
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corrected mistake in (2) as pointed out by MarkFL in the successive post...

Alexstrasuz said:
I have trouble solving this equation
|1/2x+1|=|x|

My answers are x=2 and x=-2/3

An easy way is to find the solution of the equations...

$\displaystyle \frac{1}{2 x} + 1 = x \implies 2\ x^{2} - 2\ x - 1 =0 \implies x = \frac{1 \pm \sqrt{3}}{2}\ (1)$

$\displaystyle \frac{1}{2 x} + 1 = - x \implies 2\ x^{2} + 2\ x + 1 =0 \implies x = \frac{- 1 \pm i}{2}\ (2)$

Kind regards

$\chi$ $\sigma$

P.S. MarlFL has 'discovered' a mistake in (2) and I corrected it... sorry!...
 
Last edited:
chisigma said:
An easy way is to find the solution of the equations...

$\displaystyle \frac{1}{2 x} + 1 = x \implies 2\ x^{2} - 2\ x - 1 =0 \implies x = \frac{1 \pm \sqrt{3}}{2}\ (1)$

$\displaystyle \frac{1}{2 x} + 1 = - x \implies 2\ x^{2} + 2\ x - 1 =0 \implies x = \frac{- 1 \pm \sqrt{3}}{2}\ (2)$

Kind regards

$\chi$ $\sigma$

The second equation should be:

$$2x^2+2x+1=0\implies x=\frac{-1\pm i}{2}$$
 
Alexstrasuz said:
I have trouble solving this equation
|1/2x+1|=|x|

My answers are x=2 and x=-2/3

Yes, your answers are correct.
Other posts were solving...
|1/2/x+1|=|x|
 
RLBrown said:
Yes, your answers are correct.
Other posts were solving...
|1/2/x+1|=|x|

Just to be clear, I was solving:

$$\left|\frac{1}{2x}+1\right|=|x|$$

More often than not, when someone uses 1/2x, they mean 1/(2x) as opposed to (1/2)x.

This is why is is better to use bracketing symbols (or even better, use $\LaTeX$) to remove doubt. :D
 

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