Asked by Devon
x^3-2x^2-9x-2 is greater than or equal to -20. I got (-9+x^2)(x-2)(x-2). That would mean for the first grouping, the answer would be -3 or 3. My book listed them separately. Could I instead go +/- 3? I then got double root of two. But my book says the other answer is infinity. Where's that come from? Thanks!!!!!!!!!
Answers
Answered by
MathMate
The given inequality is
x^3-2x^2-9x-2 ≥ -20
or
f(x)=x^3-2x^2-9x-18 ≥ 0
F(x) can be factored into (x-3)*(x-2)*(x+3) so the roots of f(x) are +3, -3 and 2.
F(x) has a positive leading coefficient, so the graph goes to -∞ on the left, and to +∞ on the right.
We can therefore deduce that for x<-3, f(x) is negative, and for x>3, f(x) is positive.
Also, since the intermediate root is x=2, we can also deduce that f(x) is non-negative between x=-3 and 2.
So the solution set for the given inequality is:
[-3,2]∪[3,∞), or alternatively,
-3≤x≤2 or x≥3.
http://img208.imageshack.us/img208/5915/1255556868.png
x^3-2x^2-9x-2 ≥ -20
or
f(x)=x^3-2x^2-9x-18 ≥ 0
F(x) can be factored into (x-3)*(x-2)*(x+3) so the roots of f(x) are +3, -3 and 2.
F(x) has a positive leading coefficient, so the graph goes to -∞ on the left, and to +∞ on the right.
We can therefore deduce that for x<-3, f(x) is negative, and for x>3, f(x) is positive.
Also, since the intermediate root is x=2, we can also deduce that f(x) is non-negative between x=-3 and 2.
So the solution set for the given inequality is:
[-3,2]∪[3,∞), or alternatively,
-3≤x≤2 or x≥3.
http://img208.imageshack.us/img208/5915/1255556868.png
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