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

$\frac{-1}{10x^2+11x-6}\le 0$

In the following exercises, we have to solve rational inequality and write the solution in interval notation.

$\frac{-1}{10x^2+11x-6}\le 0$

The given function is$\frac{-1}{10x^2+11x-6}\le 0$

The inequality is in the correct form.

Factor the denominator.

In the equation the product of first and last term is $60$and second term is $11$. So the possible factors are role="math" localid="1645347549977" $15$and $4$.

$$\begin{gathered} =10x^2+(15-4)x-6 \\ =10x^2+15x-4x-6 \\ =5x(2x+3)-2(2x+3) \\ =(5x-2)(2x+3) \end{gathered}$$

$\frac{-1}{(5x-2)(2x+3)}\le 0$

The quotient is 0 when the numerator is 0.

Since the numerator is always $-1$, the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.

$$\begin{gathered} 5x-2=02x+3=0 \\ x=\frac{2}{5}x=-\frac{3}{2} \end{gathered}$$

The number line is divided into three intervals:

$(-\infty ,-\frac{3}{2})(-\frac{3}{2},\frac{2}{5})(\frac{2}{5},\infty )$

The number $-2$is in the interval $(-\infty ,-\frac{3}{2})$. Test $x=-2$in the expression in the numerator and the denominator.

$\frac{-1}{(5x-2)(2x+3)}=\frac{-1}{\left\{5\right(-2)-2\}\left\{2\right(-2)+3\}}=\frac{-1}{(-10-2)(-4+3)}=\frac{-1}{(-12)(-1)}=-\frac{1}{12}$

The quotient is marked negative in the interval $(-\infty ,-\frac{3}{2})$.

The number $0$is in the interval $(-\frac{3}{2},\frac{2}{5})$. Test $x=0$in the expression in the numerator and the denominator.

$\frac{-1}{(5x-2)(2x+3)}=\frac{-1}{(5\times 0-2)(2\times 0+3)}=\frac{-1}{(-2)\left(3\right)}=\frac{1}{6}$

The quotient is marked positive in the interval $(-\frac{3}{2},\frac{2}{5})$.

The number $1$is in the interval $(\frac{2}{5},\infty )$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{-1}{(5x-2)(2x+3)}=\frac{-1}{(5\times 1-2)(2\times 1+3)}=\frac{-1}{(5-2)(2+3)}=\frac{-1}{\left(3\right)\left(5\right)}=-\frac{1}{15}$The quotient is marked negative in the interval $(\frac{2}{5},\infty )$.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals $(-\infty ,-\frac{3}{2})$and $(\frac{2}{5},\infty )$are solutions.

The solution in interval notation is $(-\infty ,-\frac{3}{2})\cup (\frac{2}{5},\infty )$.