91Ó°ÊÓ

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

$\frac{2}{2x^2+x−15}â‰\yen 0$

The given function is$\frac{2}{2x^2+x−15}â‰\yen 0$

The inequality is in the correct form.

Factor the denominator.

In the equation the product of first and last term is $30$and second term is $1$. So the possible factors are $5$and $6$.

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

$\frac{2}{(2x-5)(x+3)}â‰\yen 0$

The quotient is 0 when the numerator is 0.

Since the numerator is always 2, the quotient cannot be 0.

The quotient will be undefined when the denominator is zero.$$\begin{gathered} 2x-5=0x+3=0 \\ x=\frac{5}{2}x=-3 \end{gathered}$$

The number line is divided into three intervals:

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

The number is in the interval $(-∞,-3)$. Test in the expression in the numerator and the denominator.

$\frac{2}{(2x-5)(x+3)}=\frac{2}{(2×(-4)-5)(-4+3)}=\frac{2}{(-8-5)(-1)}=\frac{2}{(-13)(-1)}=\frac{2}{13}$

The quotient is marked positive in the interval $(-∞,-3)$.

The number $1$is in the interval $(-3,\frac{5}{2})$. Test $x=1$in the expression in the numerator and the denominator.

$\frac{2}{(2x-5)(x+3)}=\frac{2}{(2×1-5)(1+3)}=\frac{2}{(2-5)·4}=\frac{2}{(-3)·4}=\frac{1}{(-3)·2}=\frac{1}{-6}$

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

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

$\frac{2}{(2x-5)(x+3)}=\frac{2}{(2×3-5)(3+3)}=\frac{2}{(6-5)·6}=\frac{2}{1·6}=\frac{1}{3}$

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

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

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