Question

9(4-1/x) + 5(6-1/x) < 4(9-1/x)

Answer 1

(-1/2, 0)

Answer 2

Let $y = -1/x$. The inequality becomes $9(4^y) + 5(6^y) < 4(9^y)$. Divide by $9^y$ (which is always positive): $9 \left(\frac{4}{9}\right)^y + 5 \left(\frac{6}{9}\right)^y < 4$ $9 \left(\left(\frac{2}{3}\right)^2\right)^y + 5 \left(\frac{2}{3}\right)^y < 4$ $9 \left(\frac{2}{3}\right)^{2y} + 5 \left(\frac{2}{3}\right)^y < 4$

Let $z = \left(\frac{2}{3}\right)^y$. Since $z$ is an exponential term with a positive base, $z > 0$. The inequality in terms of $z$ is: $9z^2 + 5z < 4$ $9z^2 + 5z - 4 < 0$

To find the roots of $9z^2 + 5z - 4 = 0$: Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $z = \frac{-5 \pm \sqrt{5^2 - 4(9)(-4)}}{2(9)} = \frac{-5 \pm \sqrt{25 + 144}}{18} = \frac{-5 \pm \sqrt{169}}{18} = \frac{-5 \pm 13}{18}$ The roots are $z_1 = \frac{-5 + 13}{18} = \frac{8}{18} = \frac{4}{9}$ and $z_2 = \frac{-5 - 13}{18} = \frac{-18}{18} = -1$.

Since the quadratic $9z^2 + 5z - 4$ has a positive leading coefficient (9), the inequality $9z^2 + 5z - 4 < 0$ holds for $z$ between the roots: $-1 < z < \frac{4}{9}$

Considering $z > 0$, we have $0 < z < \frac{4}{9}$. Substitute back $z = \left(\frac{2}{3}\right)^y$: $0 < \left(\frac{2}{3}\right)^y < \frac{4}{9}$

The inequality $\left(\frac{2}{3}\right)^y > 0$ is always true. We solve $\left(\frac{2}{3}\right)^y < \frac{4}{9}$. Since $\frac{4}{9} = \left(\frac{2}{3}\right)^2$, we have: $\left(\frac{2}{3}\right)^y < \left(\frac{2}{3}\right)^2$

Because the base $\frac{2}{3}$ is between 0 and 1, the inequality of the exponents is reversed: $y > 2$

Substitute back $y = -1/x$: $-\frac{1}{x} > 2$

To solve this inequality: $\frac{-1}{x} - 2 > 0$ $\frac{-1 - 2x}{x} > 0$ Multiplying the numerator and denominator by -1 to make the numerator positive: $\frac{1 + 2x}{x} < 0$

This rational inequality is satisfied when the numerator and denominator have opposite signs. The critical points are $x=0$ and $1+2x=0 \implies x=-1/2$. We analyze the sign of $\frac{1 + 2x}{x}$ in the intervals $(-\infty, -1/2)$, $(-1/2, 0)$, and $(0, \infty)$.

1. For $x < -1/2$: Let $x=-1$. $\frac{1+2(-1)}{-1} = \frac{-1}{-1} = 1 > 0$.
2. For $-1/2 < x < 0$: Let $x=-1/4$. $\frac{1+2(-1/4)}{-1/4} = \frac{1-1/2}{-1/4} = \frac{1/2}{-1/4} = -2 < 0$. This interval satisfies the inequality.
3. For $x > 0$: Let $x=1$. $\frac{1+2(1)}{1} = \frac{3}{1} = 3 > 0$.

Therefore, the solution to the inequality is $-1/2 < x < 0$.