# The rational function f(x)=1/(x-2)-x/(x^2+3) has a unique zero. What is the zero of f(x)?

Jul 4, 2017

$x = - \frac{3}{2}$

#### Explanation:

$f \left(x\right) = \frac{1}{x - 2} - \frac{x}{{x}^{2} + 3}$

The zero of $f$ has $f \left(x\right) = 0$

$\therefore \frac{1}{x - 2} - \frac{x}{{x}^{2} + 3} = 0$

$\frac{1}{x - 2} = \frac{x}{{x}^{2} + 3}$

${x}^{2} + 3 = x \left(x - 2\right)$

${x}^{2} + 3 = {x}^{2} - 2 x$

$3 = - 2 x$

$x = - \frac{3}{2}$

Jul 4, 2017

$- \frac{3}{2.}$

#### Explanation:

The Zeroes of a function $f ,$ is a Set, denoted by ${Z}_{f} ,$ and is defined,

by, Z_f={x in D_f : f(x)=0, where, D_f"=the Domain of "f.

We have, ${D}_{f} = \mathbb{R} - \left\{3\right\} .$

Now, $f \left(x\right) = 0 \Rightarrow \frac{1}{x - 2} - \frac{x}{{x}^{2} + 3} = 0.$

$\Rightarrow \frac{1}{x - 2} = \frac{x}{{x}^{2} + 3} .$

$\Rightarrow {x}^{2} + 3 = x \left(x - 2\right) = {x}^{2} - 2 x .$

$\Rightarrow - 2 x = 3.$

$\Rightarrow x = - \frac{3}{2.}$

$\therefore {Z}_{f} = \left\{- \frac{3}{2}\right\} .$