Question

The set of all real values of x satisfying the inequation \( \frac{8x^2-14x-9}{3x^2-7x-6}>2 \) is

• A. \( (-\infty, 1) \cup (3, \infty) \)
• B. \( (-\infty, -\frac{2}{3}) \cup (2, \infty) \)
• C. \( (-\frac{2}{3}, 2) \)
• D. \( (-\infty, -\frac{2}{3}) \cup (3, \infty) \) Correct Answer

Hint:

To solve rational inequalities like \( \frac{P(x)}{Q(x)}>k \), first bring all terms to one side to get \( \frac{P(x)}{Q(x)} - k>0 \), then combine into a single fraction \( \frac{N(x)}{D(x)}>0 \). Analyze the signs of \(N(x)\) and \(D(x)\). Find critical points by setting \(N(x)=0\) and \(D(x)=0\). Use these points to test intervals on a number line. Remember that \(D(x) \ne 0\).

Answer 1

The Correct Option is D

Step 1: Rearrange the inequality.
\[ \frac{8x^2-14x-9}{3x^2-7x-6} - 2>0 \] Combine the terms on the left side: \[ \frac{8x^2-14x-9 - 2(3x^2-7x-6)}{3x^2-7x-6}>0 \] \[ \frac{8x^2-14x-9 - 6x^2+14x+12}{3x^2-7x-6}>0 \] \[ \frac{2x^2+3}{3x^2-7x-6}>0 \]
Step 2: Analyze the numerator.
The numerator is \( 2x^2+3 \).
Since \( x^2 \ge 0 \) for all real \( x \), \( 2x^2 \ge 0 \).
Therefore, \( 2x^2+3 \ge 3 \).
This means the numerator is always positive.

Step 3: Analyze the denominator for the inequality to hold.
Since the numerator is always positive, for the fraction \( \frac{2x^2+3}{3x^2-7x-6} \) to be greater than 0, the denominator must also be positive.
So, we require \( 3x^2-7x-6>0 \).

Step 4: Find the roots of the quadratic denominator \( 3x^2-7x-6 = 0 \).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \): \[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-6)}}{2(3)} = \frac{7 \pm \sqrt{49 + 72}}{6} = \frac{7 \pm \sqrt{121}}{6} = \frac{7 \pm 11}{6} \] The roots are \( x_1 = \frac{7-11}{6} = \frac{-4}{6} = -\frac{2}{3} \) and \( x_2 = \frac{7+11}{6} = \frac{18}{6} = 3 \).
So, \( 3x^2-7x-6 = 3(x+\frac{2}{3})(x-3) = (3x+2)(x-3) \).

Step 5: Solve the inequality \( (3x+2)(x-3)>0 \).
This inequality holds when both factors are positive or both factors are negative.
Case 1: \( 3x+2>0 \) and \( x-3>0 \).
This means \( x>-\frac{2}{3} \) and \( x>3 \).
So, \( x>3 \).
Case 2: \( 3x+2<0 \) and \( x-3<0 \).
This means \( x<-\frac{2}{3} \) and \( x<3 \).
So, \( x<-\frac{2}{3} \).
Combining these, the solution is \( x<-\frac{2}{3} \) or \( x>3 \).
In interval notation, this is \( (-\infty, -\frac{2}{3}) \cup (3, \infty) \).
This matches option (4).