# How do you find the x intercepts for y=16+5x+(3/x)?

Jun 15, 2016

The x-intercepts are the points on the curve where $y = 0$. Solving the equation $\frac{3}{x} + 5 x + 16 = 0$ yields the x values $- 3$ and $- \frac{1}{5}$, so the coordinates of the points are $\left(- 3 , 0\right)$ and $\left(- \frac{1}{5} , 0\right)$.

#### Explanation:

$\frac{3}{x} + 5 x + 16 = 0$

Multiply through by $x$:

$3 + 5 {x}^{2} + 16 x = 0$

And now we have a quadratic equation, and we know how to solve those. There are several methods, including factorisation, but my favourite is the quadratic formula.

Rearranging:

$5 {x}^{2} + 16 x + 3 = 0$

The quadratic formula applies to a quadratic equation in the form $a {x}^{2} + b x + c = 0$, and goes:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case,

$x = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \cdot 5 \cdot 3}}{2 \cdot 5} = \frac{- 16 \pm \sqrt{256 - 60}}{10}$

$= \frac{- 16 \pm 14}{10} = - 3 \mathmr{and} - \frac{1}{5}$