alle1985 wrote:I have to solve: |x^2-3x|-2=0
i have to solve for |x^2-3x|=2 and for |x^2-3x|=-2 right?
for |x^2-3x|=-2 i have....
Can an absolute value ever
be negative? So can this equation have any
alle1985 wrote:but what can i do for |x^2-3x|=2?
The general process for solving absolute-value equations may be found here
. For this particular equation, locate the zeroes of x^2 - 3x (being x = 0 and x = 3), and determine the intervals of positivity and negativity (which may easily be discerned from the graph of y = x^2 - 3x). Then solve the derivative equations x^2 - 3x = 2 and -x^2 + 3x = 2 on the relevant intervals.
Remember to verify any solutions against the conditions. For instance, is the solution value within the relevant interval?