Absolute value example

Q:  Solve for x:

|x² – 2x| = 3

A:  You always get ride of the absolute value signs by breaking an absolute value problem into two equations (a plus version and a minus version):

So,

x² – 2x = 3  or x² – 2x = -3

Now, solve them one at a time:

x² – 2x = 3

x² – 2x – 3 = 0

(x – 3)(x + 1) = 0

x = 3 or -1

And, solve the second equation:

x² – 2x = -3

x² – 2x + 3 = 0

This problem needs to be solved using the quadratic formula (it cannot be factored):

So, x = (-b ± √(b² – 4ac))/(2a)

x =(-(-2) ± √((-2)² – 4*1*3))/(2*1)

x =(2 ± √(4 – 12))/2 = (2 ± √(-8))/2

Now, if you have not studied (or are not studying) imaginary numbers, you would say that there are no solutions at this point because there is a negative number under the square-root, and we cannot take the square-root of a negative number.

If you are studying imaginary numbers, you would continue as follows:

x = (2 ± √(-8))/2 = (2 ± 2√(2)i)/2 = 1 ± √(2)i

Chances are, imaginary numbers should not be included, so your answers are:

x = 3 or -1

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