SOLUTION: I confused trying to solve 2x^2-5x-3>=0. i think you begin by adding 3 to both sides but then having the exponet is throwing me off. i dont know what to do after that. If you could

Question 175328: I confused trying to solve 2x^2-5x-3>=0. i think you begin by adding 3 to both sides but then having the exponet is throwing me off. i dont know what to do after that. If you could please explain it would be much appreciated. thanks!
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
2x^2-5x-3>=0---------------------eq1
(2x+1)(x-3)>=0
Now our drill here is to determine the range of values for x that makes (2x+1)(x-3)>=0. In other words (2x+1)(x-3) has to be either zero or (-)(-) or (+)(+)
We can see by inspection that if x>=3, then
(2x+1)(x-3) is either 0 or positive(+)(+) and eq1 inequality is correct
We can also see by inspection that if x<=-1/2, then
(2x+1)(x-3) is either 0 or positive(-)(-)and eq1 inequality is correct
So, our answer is:
x>=3 or x<=-1/2
This is one of many ways to solve quadratic inequalities
Does this help?----ptaylor
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