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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

bob bundy
2012-10-27 22:43:17

Thanks.  In my head it said that, but, as you know, I'm getting old and doddery, so what my fingers type doesn't always square with what my brain thinks.  I have edited the post.

If you look carefully at the graph, you see the end point pixels are included.  smile


2012-10-27 22:05:47

True. Then the solution set should be {x: -1/2<=x<=1/2}

2012-10-27 22:03:58


- 1 / 2 is a solution, isn't it?

2012-10-27 22:02:37

Hi Bob

The solution is correct, but the interval is wrong. It should be an open interval: (-1/2,1/2)=]-1/2,1/2[.

bob bundy
2012-10-27 19:00:01

hi Sabbir,

Welcome to the forum.

For any 'linear' expression

The critical value for x is

So, for your equation you need to consider these cases:

(i) x < - 1/2

expression becomes

Although, strictly that value is outside the range that I'm checking, a numeric check shows it is a solution.

(ii) x = -1/2 Just done that.

(iii) -1/2 < x < 1/2

expression becomes

This is true for all values of x in the range.

(iv) x = +1/2

expression becomes

So this is a solution.


expression becomes

Putting (i) to (v) together, the solution set is {x : -1/2 ≤ x ≤ 1/2}  or [-1/2,1/2] as noelevans has already explained.

Hope that helps,


2012-10-27 16:35:08

Could you please show the solving?

2012-10-27 15:32:48


The general approach to absolute value equalities is to consider where the quantities inside the
absolute values are positive, zero, or negative.  Try the following:

Where both of 2x+1 and 2x-1 are positive replace the absolute values with these and solve.
Where both of 2x+1 and 2x-1 are negative replace the absolute values with the opposite of these
    and solve.
Where 2x+1 is positive and 2x-1 is negative replace |2x+1| with 2x+1 and |2x-1| with -(2x-1)
    and solve.
You will also find solutions where each of these quantities individually are zero.

You will find the solutions in [-1/2, 1/2]. smile

2012-10-27 14:16:16

How do I solve this equation

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