Question

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

• A. \(x>0.4\)
• B. \(x<\frac 13\)<
• C. \(-\frac 13<x<0.4, \ x>\frac {15}{2}\)
• D. \(-\frac 13<x<0, \ x>\frac {2}{5}\)

Answer 1

The Correct Option is D

To solve the inequality \(15x-\left(\frac{2}{x}\right)>1\), we need to find the range of values for the variable \(x\). Begin by reorganizing the inequality:

\(15x-\frac{2}{x} > 1\) 

Subtract 1 from both sides to work with a single expression on one side:

\(15x-\frac{2}{x}-1 > 0\)

Combine the terms on the left over a common denominator:

\(\frac{15x^2-1x-2}{x} > 0\)

This can be rearranged as:

\(15x^2-x-2 > 0\)

Factor the numerator:

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

Hence, the inequality becomes:

\(\frac{(3x-2)(5x+1)}{x} > 0\)

To find the intervals where this inequality holds, determine the critical points by analyzing when each factor is zero:

• \(3x-2=0\Rightarrow x=\frac{2}{3}\)
• \(5x+1=0\Rightarrow x=-\frac{1}{5}\)
• Denominator: \(x=0\)

These critical points divide the x-axis into intervals: \(-\infty,-\frac{1}{5}\), \(-\frac{1}{5},0\), \(0,\frac{2}{3}\), \(\frac{2}{3},\infty\)

Examine the sign of \((3x-2)(5x+1)/x\) in each interval:

• \((-\infty,-\frac{1}{5}):\) \((3x-2)\) is negative, \((5x+1)\) is negative, \(x\) is negative. Thus, the overall expression is positive.
• \((-\frac{1}{5},0):\) \((3x-2)\) is negative, \((5x+1)\) is positive, \(x\) is negative. Thus, the overall expression is negative.
• \((0,\frac{2}{3}):\) \((3x-2)\) is negative, \((5x+1)\) is positive, \(x\) is positive. Thus, the overall expression is negative.
• \((\frac{2}{3},\infty):\) \((3x-2)\) and \((5x+1)\) are positive, \(x\) is positive. Thus, the overall expression is positive.

From this analysis, the inequality \(\frac{(3x-2)(5x+1)}{x} > 0\) holds in the intervals: \((-∞,-\frac{1}{5}) \text{ and } (\frac{2}{3},∞)\).

Check if any endpoints solve the initial inequality:

• \(x=0\) makes denominator undefined.
• \(x=-\frac{1}{5}\) and \(x=\frac{2}{3}\) make the expression in the numerator zero, not greater.

Therefore, the solution set is: \(-\frac{1}{5} < x < 0 \text{ and } x > \frac{2}{5}\).