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?

2 posts
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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?

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?

- alle1985
**Posts:**1**Joined:**Sun Nov 10, 2013 6:58 pm

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 valid solution?

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?

- nona.m.nona
**Posts:**252**Joined:**Sun Dec 14, 2008 11:07 pm

2 posts
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