Question

The set of all real values of \(x\) satisfying the inequality \(\frac{7x^2 - 5x - 18}{2x^2 + x - 6}<2\) is

• A. \((-\infty, -\frac{2}{3}] \cup [3, \infty)\)
• B. \((-2, -\frac{2}{3}) \cup (\frac{3}{2}, 3)\)
• C. \((-\infty, -2) \cup (\frac{3}{2}, \infty)\)
• D. \([-\frac{2}{3}, \frac{3}{2})\)

Hint:

When solving rational inequalities, find the roots of the numerator and denominator, and use a sign table.

Answer 1

The Correct Option is B

Solution to the Inequality

We want to solve the inequality:

\( 7x^2 - 5x - 18 + \frac{2x^2 + x - 6}{2} < 2 \)

Subtracting 2 from both sides, we get:

\( 7x^2 - 5x - 18 + \frac{2x^2 + x - 6}{2} - 2 < 0 \)

We combine the terms:

\( 7x^2 - 5x - 18 - 2(2x^2 + x - 6) + 2x^2 + x - 6 < 0 \)

\( 7x^2 - 5x - 18 - 4x^2 - 2x + 12 + 2x^2 + x - 6 < 0 \)

We get:

\( 3x^2 - 7x - 6 + 2x^2 + x - 6 < 0 \)

We factor the numerator and denominator:

\( \frac{(3x + 2)(x - 3)}{(2x - 3)(x + 2)} < 0 \)

The roots of the numerator are \( x = -2, 3 \). The roots of the denominator are \( x = \frac{3}{2}, -2 \).

Now, we build a sign table. We want the intervals where the expression is negative:

• (\( -2, -\frac{3}{2} \)) and ( \( \frac{3}{2}, 3 \) )

The solution to the inequality is the union of these intervals: \( (-2, -\frac{3}{2}) \cup (\frac{3}{2}, 3) \).