Equation involving modulus/absolute value

[James10492]

Hi again, here is another one that I cannot figure out:

[math]3x + 40 = 2|x+4| - 5[/math]
rearrange:

[math]\frac{(3x+45)}{2} = |x + 4|[/math]
so now I think there should be two solutions to the equation,

Solution 1:

[math]\frac{(3x+45)}{2} = x+4 \\ (3x+45) = 2x+8 \\ x = -37[/math]
Solution 2:

[math]\frac{(3x+45)}{2} = -x-4 \\ (3x+45) = -2x-8 \\ 5x = -53 \\ x = -\frac{53}{5}[/math]
Now it turns out that solution one is not actually a solution to the equation at all! Why is this?

 

[topsquark]

Hi again, here is another one that I cannot figure out:

[math]3x + 40 = 2|x+4| - 5[/math]
rearrange:

[math]\frac{(3x+45)}{2} = |x + 4|[/math]
so now I think there should be two solutions to the equation,

Solution 1:

[math]\frac{(3x+45)}{2} = x+4 \\ (3x+45) = 2x+8 \\ x = -37[/math]
Solution 2:

[math]\frac{(3x+45)}{2} = -x-4 \\ (3x+45) = -2x-8 \\ 5x = -53 \\ x = -\frac{53}{5}[/math]
Now it turns out that solution one is not actually a solution to the equation at all! Why is this?

Sometimes the original equation only has one solution, which is why you need to check your solutions after you have solved it.

The graph below is y = (3x + 4) - (2 |x + 4| -5). It clearly only has one root.

-Dan

 

[BigBeachBanana]

Hi again, here is another one that I cannot figure out:

[math]3x + 40 = 2|x+4| - 5[/math]
rearrange:

[math]\frac{(3x+45)}{2} = |x + 4|[/math]
so now I think there should be two solutions to the equation,

Solution 1:

[math]\frac{(3x+45)}{2} = x+4 \\ (3x+45) = 2x+8 \\ x = -37[/math]
Solution 2:

[math]\frac{(3x+45)}{2} = -x-4 \\ (3x+45) = -2x-8 \\ 5x = -53 \\ x = -\frac{53}{5}[/math]
Now it turns out that solution one is not actually a solution to the equation at all! Why is this?

For solution 1, you're implicitly assuming the domain of the function is [imath]x>-4[/imath]. But [imath]-37 \cancel{>}-4[/imath]

 

[Dr.Peterson]

Here is a graph of the two sides of the original equation:

​

The red is the left side, the green solid line is the right side, and the dotted lines are the extensions of the right side. You found the intersections of the two dotted lines with the red; only one of them is on the actual absolute value graph.

That is why you have to check each solution to see whether it is in the domain of that case in your solution.

 

[James10492]

Thanks for the great answers!