\begin{align*}
x^{2} - x - 12 & < 0 \\
(x - 4)(x + 3) & < 0
\end{align*}

Critical values | | \(x=-3\) | | \(x=4\) | |

\(x - 4\) | \(-\) | \(-\) | \(-\) | \(\text{0}\) | \(+\) |

\(x + 3\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) |

\(f(x)= (x - 4)(x + 3)\) | \(+\) | \(\text{0}\) | \(-\) | \(\text{0}\) | \(+\) |

From the table we see that \(f(x)\) is less than \(\text{0}\) when \(-3 < x < 4\)

We represent this on a number line:

\begin{align*}
3x^{2} + x - 4 & > 0 \\
(3x + 4)(x - 1) & > 0
\end{align*}

Critical values | | \(x=-\frac{4}{3}\) | | \(x=1\) | |

\(x - 1\) | \(-\) | \(-\) | \(-\) | \(\text{0}\) | \(+\) |

\(3x + 4\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) |

\(f(x)= (3x + 4)(x - 1)\) | \(+\) | \(\text{0}\) | \(-\) | \(\text{0}\) | \(+\) |

From the table we see that \(f(x)\) is greater than \(\text{0}\) when \(x < -\frac{4}{3}\) or when \(x > 1\)

We represent this on a number line:

\[y^{2} + y + 2 < 0\]

There are no real solutions.

The graph lies above the \(x\)-axis and does not cut the \(x\)-axis so the function is never negative. There are no values of \(y\) that will solve this inequality.

\[(3 - t)(1 + t) > 0\]

Critical values | | \(t = -1\) | | \(t = 3\) | |

\(3 - t\) | \(+\) | \(+\) | \(+\) | \(\text{0}\) | \(-\) |

\(1 + t\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) |

\(f(x)= (3 - t)(1 + t)\) | \(-\) | \(\text{0}\) | \(+\) | \(\text{0}\) | \(-\) |

From the table we see that \(f(x)\) is greater than \(\text{0}\) when \(-1 < t < 3\)

We represent this on a number line:

\[s^{2} - 4s + 6 > 0 \\\]

Use the quadratic formula to find critical values:

\begin{align*}
s & = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(6)}}{2(1)} \\
& = \frac{4 \pm \sqrt{16- 24}}{2 } \\
& = \frac{4 \pm \sqrt{-8}}{2}
\end{align*}

Therefore there are no real roots and the graph does not cut the \(x\)-axis. The graph lies above the \(x\)-axis and so this inequality is true for all real values of \(s\).

Use the quadratic formula to find critical values:

\begin{align*}
x & = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(7)(8)}}{2(7)} \\
& = \frac{1 \pm \sqrt{1- 224}}{14} \\
& = \frac{1 \pm \sqrt{-223}}{14}
\end{align*}

Therefore there are no real roots and the graph does not cut the \(x\)-axis. The graph lies above the \(x\)-axis and so this inequality is true for all real values of \(x\).

\begin{align*}
x & \geq -4x^2 \\
4x^2 + x & \geq 0 \\
x(4x + 1) & \geq 0
\end{align*}

Critical values | | \(x = -\frac{1}{4}\) | | \(x = 0\) | |

\(x\) | \(-\) | \(-\) | \(-\) | \(\text{0}\) | \(+\) |

\(4x + 1\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) |

\(f(x)= x(4x + 1)\) | \(+\) | \(\text{0}\) | \(-\) | \(\text{0}\) | \(+\) |

From the table we see that \(f(x)\) is greater than \(\text{0}\) when \(x \leq -\frac{1}{4}\) or \(x \geq 0\).

We can represent this on a number line:

\[2x^{2} + x + 6 \le 0\]

There are no real roots and the graph does not cut the \(x\)-axis. The graph lies above the \(x\)-axis and so this inequality is never true.

\(\dfrac{x}{x-3}<2\), \(x\ne 3\)

We first solve the equation:

\begin{align*}
\dfrac{x}{x-3} & < 2 \\
\dfrac{x}{x-3} - 2 & < 0 \\
\dfrac{x - 2(x-3)}{x-3}& < 0 \\
\dfrac{x - 2x + 6}{x-3}& < 0 \\
\dfrac{-x + 6}{x-3}& < 0 \\
\dfrac{-(x - 6)}{x-3}& < 0 \\
\dfrac{x - 6}{x-3}& > 0 \\
x & = 6
\end{align*}

Critical values | | \(x = 3\) | | \(x = 6\) | |

\(x - 3\) | \(-\) | undef | \(+\) | \(+\) | \(+\) |

\(x - 6\) | \(-\) | \(-\) | \(-\) | \(\text{0}\) | \(+\) |

\(f(x)= x - 6\) | \(+\) | undef | \(-\) | \(\text{0}\) | \(+\) |

From the table we see that \(f(x) > 0\) when \(x < 3 \text{ or } x > 6\) with \(x \ne 3\).

We can represent this on a number line:

\(\dfrac{x^2 + 4}{x - 7} \geq 0\), \(x \ne 7\)

We first simplify the fraction:

\[\dfrac{(x + 2)(x - 2)}{x - 7} \geq 0\]

Critical values | | \(x = -2\) | | \(x = 2\) | | \(x = 7\) | |

\(x + 2\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) | \(+\) | \(+\) |

\(x - 2\) | \(-\) | \(-\) | \(-\) | \(\text{0}\) | \(+\) | \(+\) | \(+\) |

\(x - 7\) | \(-\) | \(-\) | \(-\) | \(-\) | \(-\) | undef | \(+\) |

\(f(x)= \dfrac{(x + 2)(x - 2)}{x - 7}\) | \(-\) | \(\text{0}\) | \(+\) | \(\text{0}\) | \(-\) | undef | \(+\) |

From the table we see that \(f(x)\) is greater than \(\text{0}\) when \(-2 \leq x \leq 2\) and \(x > 7\) with \(x \ne 7\).

We can represent this on a number line:

\(\dfrac{x + 2}{x} - 1 \geq 0\), \(x \ne 0\)

We first simplify the equation:

\begin{align*}
\frac{x + 2}{x} - 1 & \ge 0 \\
\frac{x + 2 - x}{x}& \ge 0 \\
\frac{2}{x} & \ge 0 \\
\text{Therefore } x & > 0
\end{align*}

The solution is \(x > 0\) with \(x \ne 0\).

We can represent this on a number line: