Question

For what range of values of \(x\), will the inequality \(15x - \frac{2}{x} > 1\) hold?

• A. \(x > 0.4\)
• B. \(x < \frac{1}{3}\)
• C. \(-\frac{1}{3} < x < 0.4, \, x > \frac{15}{2}\)
• D. \(-\frac{1}{3} < x < 0, \, x > \frac{2}{5}\)

Hint:

Move everything to one side, combine over a common denominator, factor the quadratic, then run a sign chart across the three critical points -1/3, 0 and 2/5.

Answer 1

The Correct Option is D

Step 1: Bring every term to one side.
Starting inequality: \(15x - \frac{2}{x} > 1\). Subtract 1 from both sides: \(15x - \frac{2}{x} - 1 > 0\).
Step 2: Combine into a single fraction.
Write everything over the denominator \(x\): \(\frac{15x^2 - x - 2}{x} > 0\).
Step 3: Factor the numerator.
For \(15x^2 - x - 2\), using the quadratic formula with \(a=15, b=-1, c=-2\), the discriminant is \(1 + 120 = 121\), and \(\sqrt{121} = 11\). The roots are \(x = \frac{1 \pm 11}{30}\), giving \(x = \frac{2}{5}\) and \(x = -\frac{1}{3}\). So \(15x^2 - x - 2 = 15\left(x - \frac{2}{5}\right)\left(x + \frac{1}{3}\right)\).
Step 4: Set up the sign chart.
The inequality becomes \(\frac{15\left(x-\frac{2}{5}\right)\left(x+\frac{1}{3}\right)}{x} > 0\), with critical points at \(x = -\frac{1}{3}, 0, \frac{2}{5}\), which split the number line into four regions.
Step 5: Test each region.
For \(x < -\frac{1}{3}\) (say \(x=-1\)): the expression is negative. For \(-\frac{1}{3} < x < 0\) (say \(x=-0.1\)): the expression is positive. For \(0 < x < \frac{2}{5}\) (say \(x=0.1\)): the expression is negative. For \(x > \frac{2}{5}\) (say \(x=1\)): the expression is positive.
Step 6: Pick the regions where the expression is positive.
The inequality holds for \(-\frac{1}{3} < x < 0\) and for \(x > \frac{2}{5}\), which is option (4). The official SNAP 2010 key did not publish a marked answer for this question, so this range was worked out independently using the sign chart method above. Options (1) and (2) miss the sign change caused by dividing by \(x\), and option (3) wrongly stretches the middle region all the way to 0.4 and shifts the second boundary to 7.5, which does not match the factored roots.