In the following exercises, solve each rational inequality and write the solution in interval notation.

$\frac{1}{2}+\frac{12}{{\mathrm{x}}^2}>\frac{5}{\mathrm{x}}$

Given that the rational inequality is$\frac{1}{2}+\frac{12}{{\mathrm{x}}^2}>\frac{5}{\mathrm{x}}.$

On subtracting $\frac{5}{\mathrm{x}}$on both sides,

localid="1646177383621" $$\begin{gathered} \frac{1}{2}+\frac{12}{{\mathrm{x}}^2}>\frac{5}{\mathrm{x}} \\ \frac{1}{2}+\frac{12}{{\mathrm{x}}^2}-\frac{5}{\mathrm{x}}=0 \\ \frac{{\mathrm{x}}^2+24-10\mathrm{x}}{2{\mathrm{x}}^2}>0 \\ \frac{{\mathrm{x}}^2-6\mathrm{x}-4\mathrm{x}+24}{2{\mathrm{x}}^2}>0 \\ \frac{(\mathrm{x}-4)(\mathrm{x}-6)}{2{\mathrm{x}}^2}>0 \end{gathered}$$

To define this inequality, the denominator should not be zero. One of the critical points is $\mathrm{x}=0.$

To true this inequality, the condition as follows,

$$\begin{gathered} {\mathrm{x}}^2+24>10\mathrm{x} \\ {\mathrm{x}}^2-10\mathrm{x}+24>0 \end{gathered}$$

To factorize the polynomial using AC method. On splitting,

${\mathrm{x}}^2-10\mathrm{x}+24>0$

On solving and factorizing,

$$\begin{gathered} {\mathrm{x}}^2-6\mathrm{x}-4\mathrm{x}+24>0 \\ \mathrm{x}(\mathrm{x}-6)-4(\mathrm{x}-6)>0 \\ (\mathrm{x}-6)(\mathrm{x}-4)>0 \end{gathered}$$

To find the critical points,

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=0 \\ (\mathrm{x}-6)(\mathrm{x}-4)=0 \\ \mathrm{x}=6,4 \end{gathered}$$

Hence, the critical points are$\mathrm{x}=0,4,6.$

Values of $\frac{{\mathrm{x}}^2+24-10}{2{\mathrm{x}}^2}$at different points can be expressed as,

$$\begin{gathered} \mathrm{x}=-1\mathrm{is}\frac{{(-1)}^2+24-10}{2{(-1)}^2}=\frac{15}{2} \\ \mathrm{x}=2\mathrm{is}\frac{{\left(2\right)}^2+24-10}{2{\left(2\right)}^2}=\frac{9}{8} \\ \mathrm{x}=5\mathrm{is}\frac{{\left(5\right)}^2+24-10}{2{\left(5\right)}^2}=\frac{39}{50} \\ \mathrm{x}=7\mathrm{is}\frac{{\left(7\right)}^2+24-10}{2{\left(7\right)}^2}=\frac{9}{98} \end{gathered}$$

The quotient be positive but not even zero for the true inequality. It satisfies the points. Hence,$\mathrm{x}\in (-\infty ,0)\cup (0,4)\cup (6,\infty ).$