# How do you solve 3/(z^2-z-2)+18/(z^2-2z-3)=(z+21)/(z^2-z-2) and check for extraneous solutions?

Oct 14, 2016

$z = - 3$
Or
$z = 6$

#### Explanation:

$\frac{3}{{z}^{2} - z - 2} + \frac{18}{{z}^{2} - 2 z - 3} = \frac{z + 21}{{z}^{2} - z - 2}$

$\Rightarrow \frac{3}{{z}^{2} - z - 2} + \frac{18}{{z}^{2} - 2 z - 3} - \frac{z + 21}{{z}^{2} - z - 2} = 0$

To solve this equation we should find the common denominator,
so we have to factorize the denominators of the fractions above.

Let us factorize $\textcolor{b l u e}{{z}^{2} - z - 2}$ and $\textcolor{red}{{z}^{2} - 2 z - 3}$

We can factorize using this method ${X}^{2} + \textcolor{b r o w n}{S} X + \textcolor{b r o w n}{P}$
where $\textcolor{b r o w n}{S}$ is the sum of two real number $a$ and $b$
and
$\textcolor{b r o w n}{P}$ is their product
${X}^{2} + \textcolor{b r o w n}{S} X + \textcolor{b r o w n}{P} = \left(X + a\right) \left(X + b\right)$

$\textcolor{b l u e}{{z}^{2} - z - 2}$
Here,$\textcolor{b r o w n}{S} = - 1 \mathmr{and} \textcolor{b r o w n}{P} = - 2$
so, $a = - 2 \mathmr{and} b = + 1$
Thus,
color(blue)(z^2-z-2=(z-2)(z+1)

Factorize $\textcolor{red}{{z}^{2} - 2 z - 3}$
Here,$\textcolor{b r o w n}{S} = - 2 \mathmr{and} \textcolor{b r o w n}{P} = - 3$
so, $a = - 3 \mathmr{and} b = + 1$
Thus,
color(red)(z^2-2z-3=(z-3)(z+1)

let us start solving the equation:

$\frac{3}{\textcolor{b l u e}{{z}^{2} - z - 2}} + \frac{18}{\textcolor{red}{{z}^{2} - 2 z - 3}} - \frac{z + 21}{\textcolor{b l u e}{{z}^{2} - z - 2}} = 0$

$\Rightarrow \frac{3}{\textcolor{b l u e}{\left(z - 2\right) \left(z + 1\right)}} + \frac{18}{\textcolor{red}{\left(z - 3\right) \left(z + 1\right)}} - \frac{z + 21}{\textcolor{b l u e}{\left(z - 2\right) \left(z + 1\right)}} = 0$

$\Rightarrow \frac{3 \left(\textcolor{red}{z - 3}\right) + 18 \left(\textcolor{b l u e}{z - 2}\right) - \left(z + 21\right) \left(\textcolor{red}{z - 3}\right)}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

$\Rightarrow \frac{3 z - 9 + 18 z - 36 - \left({z}^{2} - 3 z + 21 z - 63\right)}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

$\Rightarrow \frac{3 z - 9 + 18 z - 36 - \left({z}^{2} + 18 z - 63\right)}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

$\Rightarrow \frac{3 z - 9 + 18 z - 36 - {z}^{2} - 18 z + 63}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

$\Rightarrow \frac{3 z - 9 \cancel{+ 18 z} - 36 - {z}^{2} \cancel{- 18 z} + 63}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

$\Rightarrow \frac{- {z}^{2} + 3 z + 18}{\left(z - 2\right) \left(z - 3\right) \left(z + 1\right)} = 0$

As we know a fraction $\textcolor{\mathmr{and} a n \ge}{\frac{m}{n} = 0 \Rightarrow m = 0}$

$- {z}^{2} + 3 z + 18 = 0$

$\textcolor{g r e e n}{\delta} = {\left(3\right)}^{2} - 4 \left(- 1\right) \left(18\right) = 9 + 72 = 81$
Roots are:
${x}_{1} = \frac{- 3 + \sqrt{81}}{2 \left(- 1\right)} = \frac{- 3 + 9}{- 2} = - 3$
${x}_{1} = \frac{- 3 - \sqrt{81}}{2 \left(- 1\right)} = \frac{- 3 - 9}{- 2} = 6$

$- {z}^{2} + 3 z + 18 = 0$
$\left(z + 3\right) \left(z - 6\right) = 0$
$z + 3 = 0 \Rightarrow \textcolor{b r o w n}{z = - 3}$
Or
$z - 6 = 0 \Rightarrow \textcolor{b r o w n}{z = 6}$