Question

How do you evaluate \[\dfrac{3q+1}{3{{q}^{2}}+11q-4}-\dfrac{5q+4}{2{{q}^{2}}+7q-4}\]

Answer 1

Hint: In this question we are provided with polynomials and so we have to factorise the polynomials \[3{{q}^{2}}+11q-4\] and \[2{{q}^{2}}+7q-4\] to get the factors and then we have to write it in the place of the denominator and check whether they can get cancelled so that we can simplify directly. Now we can simplify the terms we obtain to get the required solution.

Complete step by step solution:
We know the method of factorisation so the quadratic equations in the denominator can be factored as below,
\[\begin{align}
  & 3{{q}^{2}}+11q-4 \\
 & \Rightarrow 3{{q}^{2}}+12q-q-4 \\
 & \Rightarrow 3q(q+4)-1(q+4) \\
 & \Rightarrow (3q-1)(q+4) \\
\end{align}\]
Similarly,
 \[\begin{align}
  & 2{{q}^{2}}+7q-4 \\
 & \Rightarrow 2{{q}^{2}}+8q-q-4 \\
 & \Rightarrow 2q(q+4)-1(q+4) \\
 & \Rightarrow (2q-1)(q+4) \\
\end{align}\]
So the question can be written as follows,
\[\dfrac{3q+1}{3{{q}^{2}}+11q-4}-\dfrac{5q+4}{2{{q}^{2}}+7q-4}\]
\[\Rightarrow \dfrac{3q+1}{(3q-1)(q+4)}-\dfrac{5q+4}{(2q-1)(q+4)}\]
\[\Rightarrow \dfrac{1}{(q+4)}-\dfrac{5q+4}{(2q-1)(q+4)}\]
Now let’s take the LCM of the denominator to get the following,
We get the LCM of \[(q+4)\] and \[(2q-1)(q+4)\]as \[(2q-1)(q+4)\]. So we get can write the above equation as follows,
\[\Rightarrow \dfrac{(2q-1)-(5q+4)}{(2q-1)(q+4)}\]
\[\begin{align}
  & \Rightarrow \dfrac{2q-1-5q-4}{(2q-1)(q+4)} \\
 & \\
\end{align}\]
\[\Rightarrow \dfrac{-3q-5}{(2q-1)(q+4)}\]
\[\Rightarrow \dfrac{-(3q+5)}{(2q-1)(q+4)}\]
The given problem is simplified to get the following answer.
\[\dfrac{3q+1}{3{{q}^{2}}+11q-4}-\dfrac{5q+4}{2{{q}^{2}}+7q-4}\] = \[\dfrac{-(3q+5)}{(2q-1)(q+4)}\]

Note: We should always be careful while solving these problems because it's very important here to factorise the quadratic equation properly to get the correct factors otherwise it would lead to wrong answers. Please be attentive with the signs while answering this type of question as a small change in question can get you different values in solution. The method followed here is one of the simplest forms, we also can directly tae LCM of the denominator but then our solution becomes more complicated so it is always better to factorise the given quadratic expressions to simplify the problem.