|x^2-2x| = |x^2+6x|

Question

solve the absolute value equation

 

Answer 1

x = 0 is obviously a solution.

If x ≠ 0, divide both sides of |x^2-2x| = |x^2+6x| by x,

|x-2| = |x+6|

Therefore are two ways to break the absolute signs:

x-2 = x+6, which has no solutions,

or

x-2 = -(x+6) => x = -2

Answer: x = 0, -2

 

 

Answer 2

Initially, there appears to be four possibilities from this:

x² - 2x = x² + 6x

x² - 2x = -(x² + 6x)

-(x² - 2x) = -(x² + 6x)

-(x² - 2x) = x² + 6x

After further examination, multiplying the first equation by -1 produces the third equation and multiplying the second equation by -1 produces the fourth equation.  We only need to solve two problems:

x² - 2x = x² + 6x        and          x² - 2x = -(x² + 6x)

For the first possibility:

x² - 2x = x² + 6x                Equation found when absolute value signs removed

x² - 2x - x² = x² + 6x - x²    Subtract x² from each side

-2x = 6x                           Simplify

-2x - 6x = 6x - 6x              Subtract 6x from each side

-8x = 0                            Simplify

-8x/(-8) = 0/(-8)               Divide each side by -8

x = 0                                Simplify

For the second possibility:

x² - 2x = -(x² + 6x)            Second equation found when removing absolute value signs

x² - 2x = -x² - 6x                Distribute the negative

x² - 2x + x² = -x² - 6x + x²   Add x² to both sides

2x² - 2x = -6x                     Simplify

2x² - 2x + 6x = -6x + 6x       Add 6x to both sides

2x² + 4x = 0                        Simplify

2x(x + 2) = 0                      Factor the left side

2x = 0          x + 2 = 0         Set each factor equal to 0

x = 0             x = -2            Solve each factor for x

The possible answers are 0 and -2.  x = {-2,0}

To check the problem, lets plug the answers into the originals and see if they are correct.

l(-2)² - 2(-2)l = l(-2)² + 6(-2)l

l(4 + 4)l = l4 - 12l

l8l = l-8l

8 = 8

-2 checked as an answer

l0² - 2*0l = l0² + 6*0l

l0-0l = l0 -0l

l0l = l0l

0 = 0

0 checked as an answer.