Question 1208938
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Solve |x + |3x - 2|| = 2
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<pre>
Starting equation is

    |x + |3x - 2|| = 2.    (1)


It means that 

    either  x + |3x - 2| =  2    (2)

    or      x + |3x - 2| = -2.   (3)


Next consider equations (2) and (3) separately.


       <U>Equation (2)</U>


Equation (2)  is equivalent to

    |3x-2| = 2-x.    (4)



In the domain 3x-2 >= 0,  equation (4) is equivalent to

    3x-2 = 2-x,  3x+x = 2+2,  4x = 4,  x= 4/4 = 1.

    For this value of x,  the expression 3x-2 = 3*1-2 = 3-2 = 1 is positive,
    so, the premise 3x-2 >= 0  is valid;  hence,  x= 1 is a valid solution to equation (4).



In the domain 3x-2 < 0,  equation (4) is equivalent to

    3x-2 = -(2-x),  3x-2 = -2+x,  3x - x = -2 + 2,  2x= 0,  x= 0.

    For this value of x,  the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative,
    so, the premise 3x-2 < 0  is valid;  hence,  x= 0 is a valid solution to equation (4).



       <U>Equation (3)</U>


Equation (3)  is equivalent to

    |3x-2| = -2-x.    (5)



In the domain 3x-2 >= 0,  equation (5) is equivalent to

    3x-2 = -2-x,  3x+x = -2+2,  4x = 0,  x= 0.

    For this value of x,  the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative,
    so, the premise 3x-2 >= 0  is NOT valid;  hence,  x= 0 is NOT a valid solution to equation (5).



In the domain 3x-2 < 0,  equation (5) is equivalent to

    3x-2 = -(-2-x),  3x-2 = 2+x,  3x - x = 2 + 2,  2x= 4,  x= 4/2 = 2.

    For this value of x,  the expression 3x-2 = 3*2-2 = 6-2 = 4 is positive,
    so, the premise 3x-2 < 0  is NOT valid;  hence,  x= 2 is NOT a valid solution to equation (5).


<U>ANSWER</U>.  After this analysis, we see that the only solutions for the given equation (1)  

         are x= 0  and  x= 1.
</pre>

Solved.


Notice that this solution follows the strict logic at every step, with the analysis of possible
domains, so it provides the true roots of the original equation without the necessity to check them at the end.


This method of solution and this logic do not create excessive erroneous solutions,
and therefore do not require checking the solutions at the end.


The possible excessive erroneous solutions are rejected (are excluded) in the course of analysis.


The plot using plotting tool www.desmos/calculator (free of charge for common use) does confirm the solution visually

<A HREF=https://www.desmos.com/calculator/siipdt7pnk>https://www.desmos.com/calculator/siipdt7pnk</A>