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
x^2-3x=-2
x^2-3x+2=0
(x-1)(x-2)=0
so x=1,2
but what can i do for |x^2-3x|=2?
Can an absolute value ever be negative? So can this equation have any valid solution?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....
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.but what can i do for |x^2-3x|=2?