Find all real values of $x$ that satisfy $\frac{x(x+1)}{(x-4)^2} \ge 12.$ (Give your answer in interval notation.)
Because $(x-4)^2$ is always nonnegative, we can safety multiply both sides of the inequality by $(x-4)^2$ without changing the direction of the inequality, with the caveat that we cannot have $x = 4$: \[\begin{aligned} x(x+1) &\ge 12(x-4)^2 \\ 0 &\ge 11x^2 - 97x + 192. \end{aligned}\]This quadratic factors as \[0 \ge (x-3)(11x-64),\]which holds if and only if $3 \le x \le \frac{64}{11}.$ However, since $x \neq 4,$ the solutions to the original inequality are given by \[x \in \boxed{[3, 4) \cup \left(4, \frac{64}{11}\right]}\,.\]
[3,4)\cup\le(4,\frac{64}{11}\right]
Level 4
Intermediate Algebra
Problem: Find all real values of $x$ that satisfy $\frac{x(x+1)}{(x-4)^2} \ge 12.$ (Give your answer in interval notation.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality $\frac{x(x+1)}{(x-4)^2} \ge 12$, we first rearrange it to form a single rational expression. This leads to the inequality $\frac{x(x+1)}{(x-4)^2} - 12 \ge 0$. We can express 12 as $\frac{12(x-4)^2}{(x-4)^2}$, allowing us to combine the fractions: $\frac{x(x+1) - 12(x-4)^2}{(x-4)^2} \ge 0$. 

Next, we simplify the numerator: $x(x+1) - 12(x-4)^2$. Expanding $12(x-4)^2$ gives $12(x^2 - 8x + 16) = 12x^2 - 96x + 192$. Thus, the numerator becomes $x^2 + x - (12x^2 - 96x + 192) = -11x^2 + 97x - 192$. 

We need to find the roots of the quadratic $-11x^2 + 97x - 192 = 0$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a = -11$, $b = 97$, and $c = -192$. Calculating the discriminant: $b^2 - 4ac = 97^2 - 4(-11)(-192) = 9409 - 8448 = 961$. The square root of 961 is 31. 

Thus, the roots are $x = \frac{-97 \pm 31}{-22}$. This gives us two roots: $x_1 = \frac{-66}{-22} = 3$ and $x_2 = \frac{-128}{-22} = \frac{64}{11}$. 

Next, we analyze the sign of the rational expression $\frac{-11x^2 + 97x - 192}{(x-4)^2}$. The denominator $(x-4)^2$ is always positive except at $x = 4$, where it is undefined. The critical points are $x = 3$, $x = \frac{64}{11}$, and $x = 4$. 

We test intervals around these points: 
1. For $x < 3$, choose $x = 0$: $\frac{-192}{16} < 0$.
2. For $3 < x < 4$, choose $x = 3.5$: $\frac{-11(3.5)^2 + 97(3.5) - 192}{(3.5-4)^2} > 0$.
3. For $4 < x < \frac{64}{11}$, choose $x = 5$: $\frac{-11(5)^2 + 97(5) - 192}{(5-4)^2} < 0$.
4. For $x > \frac{64}{11}$, choose $x = 6$: $\frac{-11(6)^2 + 97(6) - 192}{(6-4)^2} > 0$.

The expression is non-negative in the intervals $(3, 4)$ and $(\frac{64}{11}, \infty)$. However, since $x = 4$ is excluded, we write the solution in interval notation as $[3, 4) \cup \left(4, \frac{64}{11}\right]$.<|END_OF_LATENT|> The final answer is $[3,4)\cup\le(4,\frac{64}{11}\right]$.