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

A. x > 0.4
B. x < 1/3
C. -1/3 < x < 0.4, x > 15/2
D. -1/3 < x < 0, x > 2/5
E. x < -1/3 and x > 2/5

\(15x-\frac{2}{x}>1\) ---> \(\frac{15x^2-x-2}{x}>0\) --> \(\frac{(3x+1)(5x-2)}{x}>0\) -->3 expressions change their signs at: -1/3, 0, 2/5, so 4 ranges: \(x<-\frac{1}{3}\), \(-\frac{1}{3}<x<0\), \(0<x<\frac{2}{5}\) and \(x>\frac{2}{5}\) --> in rightmost range expression is positive: so -+-+ --> \(-\frac{1}{3}<x<0\), and \(x>\frac{2}{5}\).

Now, two options offer the ranges for which given inequality holds true: A (offers the part of the range, but for x>0.4=2/5 given inequality holds true) and D (covers all possible values of x for which the inequality holds true). So if the question asks to solve \(15x-\frac{2}{x}>1\) for the true ranges then the answer is D but if the question asks for which of the following ranges inequality \(15x-\frac{2}{x}>1\) holds true than the answer could be D as well as A.

P.S. You can eliminate options B an C right away as they include zero and in our expression x is in denominator thus it can not be zero.

Check for more about the approach used here: everything-is-less-than-zero-108884.html?hilit=extreme#p868863, here: inequalities-trick-91482.html and here: xy-plane-71492.html?hilit=solving%20quadratic#p841486

Not a good question.
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can someone plz explain me how u go from
15x^2-x-2
into
(3x+1)(5x-2)

can u show me how u do it fast?

thanks.

You factorize by splitting the middle term.
Check out a discussion on this concept:
http://gmatclub.com/forum/hard-factoring-question-109006.html#p870223
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The red parts are not correct/

\(x>0\) --> \((3x+1)(5x-2)>0\) --> roots are -1/3 and 2/5 --> ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: \(x<-\frac{1}{3}\) and \(x>\frac{2}{5}\) --> since we consider \(x>0\) range then the solution for this range is: \(x>\frac{2}{5}\);

\(x<0\) --> \((3x+1)(5x-2)>0\) --> the same roots: -1/3 and 2/5 --> "<" sign indicates that the solution lies between the roots: \(-\frac{1}{3}<x<\frac{2}{5}\) --> since we consider \(x<0\) range then the solution for this range is: \(-\frac{1}{3}<x<0\).

Thus \(15x-\frac{2}{x}>1\) holds true for \(-\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\).

See my 1st and 2nd posts for alternate approach and the links there for theory on this kind of questions.

Hope it helps.
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Basically he moved the 1 over to the LHS, then used x as the common denominator for the LHS

As shown by Bunuel, the range for which the inequality works is -1/3 < x < 0 and x > 2/5. Notice that x cannot be 0 here.

0 is not a root here. Note that the term 'root' is generally used for equations (the value which when substituted for the unknown satisfies the equation). When working with inequalities, you deal with the range of values that work.

The inequality is undefined at 0 but it changes sign around it. So you include it while working on finding the ranges where the expression is positive/negative. To understand how and why you do this, check out the inequalities trick link given by Bunuel above or my post: http://www.veritasprep.com/blog/2012/06 ... e-factors/
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Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot multiply both parts of inequality 15x - (2/x) > 1 by x as you don't know the sign of this unknown: if x>0 you should write 15x^2 - x - 2 > 0 BUT if x<0 you should write 15x^2 - x - 2 < 0 (flip the sign).

Hope it helps.
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Hi Banuel,

When i factorsie 15x^2 - X-2 =0 i got only only 2 roots : -1/3 and 2/5. ( I cross multiplied the denominator x so it become "0" on the RHS.)

So i got only 3 intervals. How did you got 3 roots , especially 0 as a root. please provide me the insight and tell me where did i went wrong.

By the way thank you very much for your links to learn inequalities concepts. It's very useful.

Thanks in advance.

Regards

I understand the way to solve this problem, but I'm still not sure as to why d. is correct and not a. Is is because d is a more "complete" range of correct values?

Thanks!