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### Section 1-1 : Review : Functions

17. Determine all the roots of $$\displaystyle g\left( w \right) = \frac{{2w}}{{w + 1}} + \frac{{w - 4}}{{2w - 3}}$$.

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Set the function equal to zero and clear the denominator by multiplying by the least common denominator, $$\left( {w + 1} \right)\left( {2w - 3} \right)$$, and then solve the resulting equation.

\begin{align*}\left( {w + 1} \right)\left( {2w - 3} \right)\left( {\frac{{2w}}{{w + 1}} + \frac{{w - 4}}{{2w - 3}}} \right) & = 0\\ 2w\left( {2w - 3} \right) + \left( {w - 4} \right)\left( {w + 1} \right) & = 0\\ 5{w^2} - 9w - 4 & = 0\end{align*}

This quadratic doesn’t factor so we’ll need to use the quadratic formula to get the solution.

$w = \frac{{9 \pm \sqrt {{{\left( { - 9} \right)}^2} - 4\left( 5 \right)\left( { - 4} \right)} }}{{2\left( 5 \right)}} = \frac{{9 \pm \sqrt {161} }}{{10}}$

So, it looks like this function has the following two roots,

$\frac{{9 + \sqrt {161} }}{{10}} = 2.168858\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\frac{{9 - \sqrt {161} }}{{10}} = - 0.368858$

Recall that because we started off with a function that contained rational expressions we need to go back to the original function and make sure that neither of these will create a division by zero problem in the original function. Neither of these do and so they are the two roots of this function.