Question 148921: how do I solve (x-2)(x+1)≥0
related equation: (x-2) (x+1)=0
solutions for the equation: (2,-1)
intervals
(-∞,-1)(-1,2)(2,∞) After this I am confused ?
f(x)=(x-4)(x+2)
Answer by tangitehewagen(7)   (Show Source):
You can put this solution on YOUR website! It looks like you are on the right track although I do not know where f(x) = (x-4)(x+2) came from so lets ignore that and look at the rest of your work.
Lets summarize what you have done so far, okay?
You decided that you needed to solve the equation (x-2)(x+1) = 0 to get your "cut points" which divide up the number line into intervals.
You found the cut points to be x = -1 and x = 2.
Next you used these cut points to divide the interval (-∞,∞) into the subintervals (-∞,-1)(-1,2)(2,∞).
Now you are in the home stretch. This is the point at which you remind yourself of the original problem (x-2)(x+1) ≥ 0.
In other words, you want to know where the product (x-2)(x+1) is either 0 or greater than 0 (Positive), right?
Well, you found that the product will be 0 at x = -1 and x = 2. We will come back to this in a moment.
Lets determine where the product is positive so we go back to the intervals you found earlier. We know that in each of these intervals (x+1)(x-2) < 0 or (x+1)(x-2) > 0 but not both (why?).
To determine the sign of (x+1)(x-2) over each of the intervals, we take a number in each intervals and evaluate (x-2)(x+1) at the number.
So for the interval (-∞,-1), lets use x = -2.
(-2 + 1)(-2 - 2) = (-1)(-4) = 4 which is positive so (x+1)(x-2) > 0 on the interval (-∞,-1).
For the interval (-1,2), lets use x = 0.
(0 + 1)(0 - 2) = 1(-2) = -2 which is negative so (x+1)(x-2) < 0 on the interval (-1,2).
Finally, for the interval (2,∞), lets use x = 3.
(3 + 1)(3 - 2) = (4)(1) = 4 which is positive so (x+1)(x-2) > 0 on the interval (2,∞).
This tells us that (x+1)(x-2) > 0 on the union, (-∞,-1) U (2,∞).
But remember, we also want to know where (x+1)(x-2) = 0 which we found to be at x = -1 and x = 2 so in our union we close the subintervals at -1 and 2 to obtain
(x+1)(x-2) ≥ 0 when x ε (-∞,-1] U [2,∞).
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