Find all values of $x$ that satisfy \[\frac{x^2}{x+1} \ge \frac{2}{x-1} + \frac{5}{4}.\]
Combining the terms on the right-hand side, we have \[\frac{x^2}{x+1} \ge \frac{5x+3}{4(x-1)}.\]Then, moving all the terms to the left-hand side and combining denominators again, we get \[\begin{aligned} \frac{x^2}{x+1} - \frac{5x+3}{4(x-1)} &\ge 0 \\ \frac{4x^2(x-1)-(x+1)(5x+3)}{(x+1)(x-1)} &\ge 0 \\ \frac{4x^3-9x^2-8x-3}{(x+1)(x-1)} &\ge 0. \end{aligned}\]We try to factor the numerator. Using the rational root theorem to test for rational roots, we see that $x=3$ is a root of $4x^3-9x^2-8x-3.$ Then, doing the polynomial division gives \[4x^3-9x^2-8x-3 = (x-3)(4x^2+3x+1),\]so we have \[\frac{(x-3)(4x^2+3x+1)}{(x+1)(x-1)} \ge 0.\]Since $4x^2+3x+1$ has a positive $x^2$ coefficient, and its discriminant is $3^2 - 4 \cdot 4= -7,$ which is negative, it follows that $4x^2 + 3x + 1 > 0$ for all $x.$ Thus, the above inequality is equivalent to \[f(x) = \frac{x-3}{(x+1)(x-1)} \ge 0.\]We make a sign table for $f(x)$: \begin{tabular}{c|ccc|c} &$x-3$ &$x+1$ &$x-1$ &$f(x)$ \\ \hline$x<-1$ &$-$&$-$&$-$&$-$\\ [.1cm]$-1<x<1$ &$-$&$+$&$-$&$+$\\ [.1cm]$1<x<3$ &$-$&$+$&$+$&$-$\\ [.1cm]$x>3$ &$+$&$+$&$+$&$+$\\ [.1cm]\end{tabular}We see that $f(x) > 0$ when $-1 < x < 1$ or $x > 3.$ Since the inequality is nonstrict, we also include the values of $x$ such that $f(x) = 0,$ that is, only $x = 3.$ Therefore, the solution to the inequality is \[x \in \boxed{(-1, 1) \cup [3, \infty)}.\]
(-1,1)\cup[3,\iny)
Level 4
Intermediate Algebra
Problem: Find all values of $x$ that satisfy \[\frac{x^2}{x+1} \ge \frac{2}{x-1} + \frac{5}{4}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality, first, we need to combine the fractions on the right side. The common denominator for the terms on the right is \(4(x-1)\). Rewriting the right side gives:

\[\frac{2}{x-1} + \frac{5}{4} = \frac{8}{4(x-1)} + \frac{5(x-1)}{4(x-1)} = \frac{8 + 5(x-1)}{4(x-1)} = \frac{5x + 3}{4(x-1)}.\]

Next, we rewrite the original inequality:

\[\frac{x^2}{x+1} \ge \frac{5x + 3}{4(x-1)}.\]

To eliminate the fractions, we multiply both sides by \(4(x+1)(x-1)\), noting that we must consider the sign of the expressions to avoid changing the inequality's direction. This gives:

\[4x^2(x-1) \ge (5x + 3)(x+1).\]

Expanding both sides results in:

\[4x^3 - 4x^2 \ge 5x^2 + 5x + 3 + 3x + 3.\]

This simplifies to:

\[4x^3 - 4x^2 \ge 5x^2 + 8x + 3.\]

Rearranging terms leads to:

\[4x^3 - 9x^2 - 8x - 3 \ge 0.\]

Next, we need to find the roots of the cubic polynomial \(4x^3 - 9x^2 - 8x - 3\). We can use the Rational Root Theorem to test possible rational roots. After testing, we find that \(x = 3\) is a root. We can then factor the polynomial as:

\[(x - 3)(4x^2 + 3x + 1) \ge 0.\]

The quadratic \(4x^2 + 3x + 1\) has a positive discriminant, indicating it has no real roots and is always positive since the leading coefficient is positive. Thus, the sign of the product is determined by \(x - 3\).

The inequality \(x - 3 \ge 0\) gives \(x \ge 3\). 

Next, we must consider the restrictions from the original inequality. The term \(\frac{x^2}{x+1}\) is defined for \(x \neq -1\) and \(\frac{2}{x-1}\) is defined for \(x \neq 1\). Therefore, we exclude \(x = -1\) and \(x = 1\) from our solution set.

Combining the intervals, we find that the solution to the inequality is:

\((-1, 1) \cup [3, \infty).\)<|END_OF_LATENT|> The final answer is $(-1,1)\cup[3,\iny)$.