# Thread: Inequality Question: Gurus Pls provide your inputs

1. Good post? |

## Inequality Question: Gurus Pls provide your inputs

For what value of x is the following inequality holds true (3x^2-x-2)/x > 0

Ans
SPOILER: is -2/3<x<0 and x>1

IMO there would be two cases

case 1 when both 3x^2-x-2 > 0 and x > 0

case 2 when both 3x^2-x-2 < 0 and x < 0

2. Good post? |
(3x^2-x-2)/x > 0......x=?
==> [(3x+2)(x-1)/x]>0---------(1)
now for above inequality to hold true....either numerator or denominator both are positive or both are negative..\
case 1) x-1>0 ==> x>1//.....(1) = (+ve).(+ve)/(+ve) is +ve..holds true
case 2)0<x<1 ==> (+ve)(-ve)/+ve = -ve..does not hold true...
case 3) 3x+2>0 ==> x>-2/3...holds true only if x<0 because of (case 2)
hence -2/3<x<0..
combining case 3 and case 1,....we have the required answer\

3. Good post? |
I would stick 2 one strategy here...multiply the whole euation by x^2

I get x(3x^2-x-2)>0
2 cases
x>0 and (3x+2)(x-1)>0...draw on a number line...common area x>1

Case 2
x<0 (3x+2)(x-1)<0....Draw on a numberline ....Common area (-2/3, 0)

4. Good post? |

## Interval training

I realize this is an old post, but it's interesting.

If you find the critical points by factoring the polynomial, you will find two values: x=-2/3 and x=1. Including x=0 (which is not possible) these three values make four intervals. Simply test a value in each of these intervals and see if the inequality holds up. I chose x=-1,-1/2,1/2, and 2. If you do this, you will find that x can be -2/3<x<0 OR x>1. Satisfactorily, this corresponds with the solution provided.

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