For what range of values of 'x' will the inequality 15x - 2/x > 1?

I do not understand why 2 cases are considered in the above explanation ?
Only if we had |x| should the 2 cases be considered, am I right ?

Whenever you multiply an inequality by X , you should check for the -ve and +ve values of X, as it inverts the sign.

Thanks for the lightning quick reply, I was not aware of this.

Bunuel you are awesome

Nice alternate solution.

Please note :

\(15x^2-x-2 -> x^2 - x/15 - 2/15 -> x^2 + x/3 - 2x/5 - 2/15 -> (x+1/3)(x-2/5)\)

This works well, however I would suggest to rely on quadratic equation formula: \(-b(+-)\sqrt{D}/2a\), as you don't need to think, how to factorize an equation?

Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect rarely could be the case in GMAT) , for example when D in above formula is irrational, then you might be trying to factorize the fraction forever.)

What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

hi,

can any one explain this with graphical explanation??

Thanks in advance,
Rrsnathan.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

I have the same doubt. Shouldn't the answer be D?

____________
The answer IS D.

After you determine boundary points, you still need to test the potential +ve/-ve ranges per the inequality

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.

I didn't multiply.

\(15x-\frac{2}{x}>1\) --> subtract 1: \(15x-\frac{2}{x}-1>0\) --> put in common denominator: \(\frac{15x^2-x-2}{x}>0\) --> factorize: \(\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0\).

Hope it's clear.

A graphical approach supplements this solution quite well: The roots are the blue dots, the yellow line separates the two cases, and the red lines are where the inequality holds true.

We assume two cases: x>0 and x<0.

For the first case, x>0, we want to see where 15x^2 -x-2 >0. Solving for the roots using the quadratic formula, we find roots when x = 2/5 and x = -1/3. Thus, for x>0, 15x^2 -x-2 >0 when x>2/5.

For the second case, x<0, we want to see where 15x^2 -x-2 <0. Again, using the roots, we know that the 15x^2 -x-2 < 0 when x > -1/3. We also assumed x<0 so we must include that in our solution.

Thus we have x>2/5 and -1/3<x<0. Answer: D

Hi
Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ?
I don't know how to solve this????

Hope this helps.

Can you explain why before and after the "0" you change the sign?
What was the equation that got you to draw this?