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Question 414485: solve the following inequalities.
2x^2+5x-3<0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! solve the following inequalities.
2x^2+5x-3<0
..
The first step is to factor the equation and find its critical points.
2x^2+5x-3<0
(2x-1)(x+3)=0
2x-1=0
x=1/2
x+3=0
x=-3
The critical points are -3 and 1/2
Show these points on a number line and determine the sign (+ or -) of the intervals between these points. One way to do this is to use test points within each interval to determine whether the function is plus or minus. This can be tedius, but there is a shorter way. Here is how to do this shorter method:
For example, for the interval between 1/2 and infinity, any number greater than 1/2 will make this interval positive. Moving to the left between 1/2 and -3, this interval will switch to negative any time the critical point we are going thru comes from a root of odd multiplicity(1,3,5,etc) In this case it came from a root of multiplicity 1,(2x-1),so it switches from + to -. If the point came from an even multiplcity (2,4,6,etc.) the sign would not change. Accordingly, as we go thru point -3 on the number line, the sign switches again from - to + making the interval (-infinity,-3) positive. In summary we now have three intervals. In interval notation they are: (-infinity,-3),(-3,1/2),(1/2,infinity). But only one of these three intervals satisfies the given function. It is the middle interval because the function is negative for x-values within this interval.
Ans: (-3,1/2)
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