mariofelixpasku wrote:how do you get the values of x if you got sthg like this:
4x(x+1) ≥ 3
answers: x>=1/2 and x=< -3/2
Step 1: Rephrase the inequality in terms of 0 and factor the quadratic.
4x(x+1) ≥ 3
4x² + 4x - 3 ≥ 0
(2x + 3)(2x - 1) ≥ 0
Step 2: Test one value to the left and one value to the right of each CRITICAL POINT.
The critical points are the values of x where the left-hand side is equal to 0.
Here, the critical points are x = -3/2 and x = 1/2.
These two points are valid solutions because, in each case, (2x + 3)(2x - 1) = 0.
To determine the range(s) where (2x + 3)(2x - 1) > 0, test one value to each side of x=-3/2 and x=1/2.
Plug x=-2 -- a value to the LEFT of -3/2 -- into (2x + 3)(2x - 1) ≥ 0 :
( 2(-2) + 3 ) ( 2(-2) - 1) ≥ 0
(-1)(-5) ≥ 0
5 ≥ 0.
This works.
x < -3/2 is a viable range.
Plug x=0 -- a value BETWEEN -3/2 and 1/2 -- into (2x + 3)(2x - 1) ≥ 0 :
(2*0 + 3)(2*0 - 1) ≥ 0
(3)(-1) ≥ 0
-3 ≥ 0.
Doesn't work.
-3/2 < x < 1/2 is not a viable range.
Plug x=1 -- a value to the RIGHT of 1/2 -- into (2x + 3)(2x - 1) ≥ 0 :
(2*1 + 3)(2*1 - 1) ≥ 0
(5)(1) ≥ 0
5 ≥ 0.
This works.
x > 1/2 is a viable range.
Thus, the viable ranges are x≤-3/2 and x≥1/2.
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