# How do you solve (10y^2-13y-30)/(2y^5-50y^3)<0?

Dec 18, 2016

Solution is $- \infty < y < - 5$ or $- \frac{6}{5} < y < 0$ or $\frac{5}{2} < y < 5$.

#### Explanation:

$\frac{10 {y}^{2} - 13 y - 30}{2 {y}^{5} - 50 {y}^{3}}$

= (10y^2-25y+12y-30)/(2y^3(y^2-25)

= $\frac{5 y \left(2 y - 5\right) + 6 \left(2 y - 5\right)}{2 {y}^{3} \left(y + 5\right) \left(y - 5\right)}$

= $\frac{\left(5 y + 6\right) \left(2 y - 5\right)}{2 {y}^{3} \left(y + 5\right) \left(y - 5\right)}$

Hence, $f \left(y\right) = \frac{10 {y}^{2} - 13 y - 30}{2 {y}^{5} - 50 {y}^{3}} < 0$

$\Leftrightarrow f \left(y\right) = \frac{\left(5 y + 6\right) \left(2 y - 5\right)}{2 {y}^{3} \left(y + 5\right) \left(y - 5\right)} < 0$
$\therefore$ zeros of numerator and denominators are

$\left\{- 5 , - \frac{6}{5} , 0 , \frac{5}{2} , 5\right\}$ and they divide real number line in $6$ intervals.

In $- \infty < y < - 5$, all monomials are negative and hence $f \left(y\right) < 0$

In $- 5 < y < - \frac{6}{5}$, while $\left(y + 5\right)$ is positive, all other monomials are negative and hence $f \left(y\right) > 0$

In $- \frac{6}{5} < y < 0$, while $\left(y + 5\right)$ and $\left(5 y + 6\right)$ are positive, all other monomials are negative and hence $f \left(y\right) < 0$

In $0 < y < \frac{5}{2}$, while $\left(y + 5\right)$, $\left(5 y + 6\right)$ and $y$ are positive, $\left(2 y - 5\right)$ and $\left(y - 5\right)$ are negative and hence $f \left(y\right) > 0$

In $\frac{5}{2} < y < 5$, while $\left(y + 5\right)$, $\left(5 y + 6\right)$, $y$ and $\left(2 y - 5\right)$ are positive, $\left(y - 5\right)$ is negative and hence $f \left(y\right) < 0$

and in $5 < y < \infty$, all monomials are positive and hence $f \left(y\right) > 0$

Hence solution is $- \infty < y < - 5$ or $- \frac{6}{5} < y < 0$ or $\frac{5}{2} < y < 5$.
graph{(10x^2-13x-30)/(2x^5-50x^3) [-10, 10, -2, 2]}