Question 1208939: Solve |3x - |2x + 1|| = 4
Found 3 solutions by Edwin McCravy, math_tutor2020, ikleyn:
Answer by Edwin McCravy(20056)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answers: x = -1 and x = 5
Explanation
The rule we need is |x| = k breaks down into x = k or x = -k where k is nonnegative.
For instance, |x| = 27 means x = 27 or x = -27.
Both -27 and 27 are the same distance from zero on the number line.
Using that rule we can break
|3x - |2x+1|| = 4
into
3x - |2x+1| = 4 and 3x - |2x+1| = -4
Let's solve the first piece.
3x - |2x+1| = 4
|2x+1| = 3x-4
2x+1 = 3x-4 or 2x+1 = -(3x-4)
2x+1 = 3x-4 or 2x+1 = -3x+4
2x-3x = -4-1 or 2x+3x = 4-1
-x = -5 or 5x = 3
x = 5 or x = 3/5
Those are two potential solutions so far.
The key term is "potential" since we need to verify each solution.
Now solve the 2nd piece found earlier.
3x - |2x+1| = -4
|2x+1| = 3x+4
2x+1 = 3x+4 or 2x+1 = -(3x+4)
2x+1 = 3x+4 or 2x+1 = -3x-4
2x-3x = 4-1 or 2x+3x = -4-1
-x = 3 or 5x = -5
x = -3 or x = -5/5 = -1
We found two more potential solutions.
---------------
The four potential solutions we need to check are
x = 5 or x = 3/5
x = -3 or x = -1
Let's return to the original equation and plug in x = 5
|3x - |2x+1|| = 4
|3*5 - |2*5+1|| = 4
|15 - |10+1|| = 4
|15 - |11|| = 4
|15 - 11| = 4
|4| = 4
4 = 4 ..... works
The solution x = 5 is confirmed.
Now test x = 3/5
|3x - |2x+1|| = 4
|3*3/5 - |2*3/5+1|| = 4
|9/5 - |6/5+1|| = 4
|9/5 - |6/5+5/5|| = 4
|9/5 - |11/5|| = 4
|9/5 - 11/5| = 4
|(9-11)/5| = 4
|-2/5| = 4
2/5 = 4 ..... this doesn't work out
We determine that x = 3/5 doesn't work, so it is extraneous.
We must cross it off the list.
Now onto x = -3
|3x - |2x+1|| = 4
|3*(-3) - |2*(-3)+1|| = 4
|-9 - |-6+1|| = 4
|-9 - |-5|| = 4
|-9 - 5| = 4
|-14| = 4
14 = 4 ...... this doesn't work
Remove x = -3 from the list.
Lastly we need to check x = -1
|3x - |2x+1|| = 4
|3*(-1) - |2*(-1)+1|| = 4
|-3 - |-2+1|| = 4
|-3 - |-1|| = 4
|-3 - 1| = 4
|-4| = 4
4 = 4 ...... this works
Therefore we confirm that only x = -1 and x = 5 are the answers.
Verification using a Desmos graph is shown here
https://www.desmos.com/calculator/lr9kd0nk5f
The two intersection points have x coordinates of x = -1 and x = 5.
Answer by ikleyn(52790)  (Show Source):
You can put this solution on YOUR website! Solve |3x - |2x + 1|| = 4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Looking into the solutions by other tutors, you may ask yourself, if there exist a method,
which allows to get true answers without creating excessive roots.
Yes, such a method does exist. It is shown below.
Starting equation is
|3x - |2x + 1|| = 4. (1)
It means that
either 3x - |2x + 1| = 4 (2)
or 3x - |2x + 1| = -4. (3)
Next consider equations (2) and (3) separately.
Equation (2)
Equation (2) is equivalent to
|2x+1| = 3x-4. (4)
In the domain 2x+1 >= 0, equation (4) is equivalent to
2x+1 = 3x-4, 1+4 = 3x - 2x, 5 = x, x= 5.
For this value of x, the expression 2x+1 = 2*5+1 = 11 is positive,
so, the premise 3x-2 >= 0 is valid; hence, x= 5 is a valid solution to equation (4).
In the domain 2x+1 < 0, equation (4) is equivalent to
2x+1 = -(3x-4), 2x+1 = -3x+4, 2x+3x = 4 - 1, 5x= 3, x= 3/5.
For this value of x, the expression 2x+1 = 3*(3/5)+1 = 9/5+1 is positive,
so, the premise 2x+1 < 0 is NOT valid; hence, x= 3/5 is NOT a valid solution to equation (4).
Equation (3)
Equation (3) is equivalent to
|2x+1| = 3x+4. (5)
In the domain 2x+1 >= 0, equation (5) is equivalent to
2x+1 = 3x+4, 1-4 = 3x-2x, -3 = x, x= -3.
For this value of x, the expression 2x+1 = 2*(-3)+1 = -6+1 = -5 is negative,
so, the premise 2x+1 >= 0 is NOT valid; hence, x= -3 is NOT a valid solution to equation (5).
In the domain 2x+1 < 0, equation (5) is equivalent to
2x+1 = -(3x+4), 2x+1 = -3x-4, 2x + 3x = -4 - 1, 5x= -5, x= -5/5 = -1.
For this value of x, the expression 2x+1 = 2*(-1)+1 = -2+1 = -1 is negative,
so, the premise 2x+1 < 0 is valid; hence, x= -1 is a valid solution to equation (5).
ANSWER. After this analysis, we see that the only solutions for the given equation (1)
are x= -1 and x= 5.
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
https://www.desmos.com/calculator/ajbgvspnhp
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