Question

$\frac{15-4x}{x^{2}-x-12}<4$

Answer 1

$(-\infty, -\frac{3\sqrt{7}}{2}) \cup (-3, \frac{3\sqrt{7}}{2}) \cup (4, \infty)$

Answer 2

The given inequality is $\frac{15-4x}{x^{2}-x-12}<4$.

Step 1: Move all terms to one side to get a zero on the other side. $\frac{15-4x}{x^{2}-x-12} - 4 < 0$

Step 2: Find a common denominator and combine the terms into a single fraction. The denominator is $x^2 - x - 12$. Factoring the denominator, we get $x^2 - x - 12 = (x-4)(x+3)$. So, the inequality is $\frac{15-4x}{(x-4)(x+3)} - 4 < 0$. The common denominator is $(x-4)(x+3)$. $\frac{15-4x - 4(x-4)(x+3)}{(x-4)(x+3)} < 0$ Expand the term $4(x-4)(x+3)$: $4(x-4)(x+3) = 4(x^2 + 3x - 4x - 12) = 4(x^2 - x - 12) = 4x^2 - 4x - 48$. Substitute this back into the inequality: $\frac{15-4x - (4x^2 - 4x - 48)}{(x-4)(x+3)} < 0$ $\frac{15-4x - 4x^2 + 4x + 48}{(x-4)(x+3)} < 0$ $\frac{-4x^2 + 63}{(x-4)(x+3)} < 0$

Step 3: Multiply by -1 and reverse the inequality sign. $\frac{4x^2 - 63}{(x-4)(x+3)} > 0$

Step 4: Find the critical points by setting the numerator and the denominator equal to zero. Numerator: $4x^2 - 63 = 0 \implies 4x^2 = 63 \implies x^2 = \frac{63}{4} \implies x = \pm\sqrt{\frac{63}{4}} = \pm\frac{\sqrt{9 \times 7}}{2} = \pm\frac{3\sqrt{7}}{2}$. The roots of the numerator are $x = -\frac{3\sqrt{7}}{2}$ and $x = \frac{3\sqrt{7}}{2}$.

Denominator: $(x-4)(x+3) = 0 \implies x-4 = 0$ or $x+3 = 0 \implies x = 4$ or $x = -3$. The roots of the denominator are $x = -3$ and $x = 4$.

The critical points, in increasing order, are $-\frac{3\sqrt{7}}{2}$, $-3$, $\frac{3\sqrt{7}}{2}$, $4$. Approximate values: $\sqrt{7} \approx 2.646$, so $\frac{3\sqrt{7}}{2} \approx \frac{3 \times 2.646}{2} \approx \frac{7.938}{2} \approx 3.969$. The ordered critical points are approximately $-3.969, -3, 3.969, 4$.

Step 5: Analyze the sign of the expression $\frac{4x^2 - 63}{(x-4)(x+3)}$ in the intervals defined by the critical points. The critical points divide the number line into five intervals: $(-\infty, -\frac{3\sqrt{7}}{2})$, $(-\frac{3\sqrt{7}}{2}, -3)$, $(-3, \frac{3\sqrt{7}}{2})$, $(\frac{3\sqrt{7}}{2}, 4)$, $(4, \infty)$.

Let $f(x) = \frac{4x^2 - 63}{(x-4)(x+3)}$. We need to find where $f(x) > 0$. We can use the sign chart method or test points in each interval. The expression is $\frac{4(x - \frac{3\sqrt{7}}{2})(x + \frac{3\sqrt{7}}{2})}{(x-4)(x+3)}$.

• Interval $(-\infty, -\frac{3\sqrt{7}}{2})$: Choose $x=-5$. $f(-5) = \frac{4(-5)^2-63}{(-5-4)(-5+3)} = \frac{100-63}{(-9)(-2)} = \frac{37}{18} > 0$.
• Interval $(-\frac{3\sqrt{7}}{2}, -3)$: Choose $x=-3.5$. $f(-3.5) = \frac{4(-3.5)^2-63}{(-3.5-4)(-3.5+3)} = \frac{4(12.25)-63}{(-7.5)(-0.5)} = \frac{49-63}{3.75} = \frac{-14}{3.75} < 0$.
• Interval $(-3, \frac{3\sqrt{7}}{2})$: Choose $x=0$. $f(0) = \frac{4(0)^2-63}{(0-4)(0+3)} = \frac{-63}{(-4)(3)} = \frac{-63}{-12} = \frac{21}{4} > 0$.
• Interval $(\frac{3\sqrt{7}}{2}, 4)$: Choose $x=3.9$. $f(3.9) = \frac{4(3.9)^2-63}{(3.9-4)(3.9+3)} = \frac{4(15.21)-63}{(-0.1)(6.9)} = \frac{60.84-63}{-0.69} = \frac{-2.16}{-0.69} > 0$.

The intervals where $f(x) > 0$ are $(-\infty, -\frac{3\sqrt{7}}{2})$, $(-3, \frac{3\sqrt{7}}{2})$, and $(4, \infty)$. The points where the denominator is zero ($x=-3$ and $x=4$) must be excluded from the solution. The points where the numerator is zero ($x=\pm\frac{3\sqrt{7}}{2}$) are also excluded because the inequality is strict ($>0$).

The solution set is the union of these intervals: $x \in (-\infty, -\frac{3\sqrt{7}}{2}) \cup (-3, \frac{3\sqrt{7}}{2}) \cup (4, \infty)$.