Question

How do you simplify \[\dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}\]?

Answer 1

Hint: This question is from the topic of algebra. In this question, we will solve the term given in the question. In solving this question, we will first factorize the term \[3{{x}^{2}}+19x-14\]. After that, we will factorize the term \[9{{x}^{2}}-12x+4\]. After that, we will take a common factor from the denominators of the given term. After solving the further term, we will get our answer.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to solve the term which is given in the question. The term which we have to solve is
\[\dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}\]
So, let us first factorize the term \[3{{x}^{2}}+19x-14\].
Before factoring, let us understand how to factorize a quadratic equation.
If a quadratic equation is given as \[a{{x}^{2}}+bx+c=0\], then their roots will be \[x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\] and the factorization of the quadratic equation will be \[\left( x-\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right) \right)\left( x-\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right) \right)\].
So, let us find out the factorization of equation \[3{{x}^{2}}+19x-14=0\].
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-19\pm \sqrt{{{\left( 19 \right)}^{2}}-4\times 3\times \left( -14 \right)}}{2\times 3}=\dfrac{-19\pm \sqrt{361+168}}{2\times 3}=\dfrac{-19\pm 23}{6}\]
From the above, we can write the values of x as
\[x=\dfrac{-19+23}{6}=\dfrac{4}{6}=\dfrac{2}{3}\] and
\[x=\dfrac{-19-23}{6}=\dfrac{-42}{6}=-7\]
So, we get the values of x as \[\dfrac{2}{3}\] and \[-7\]. So, we can write the factorization of the equation \[3{{x}^{2}}+19x-14=0\] can be written as
\[\left( x-\left( \dfrac{2}{3} \right) \right)\left( x-\left( -7 \right) \right)=0\]
The above can also be written as \[\left( 3x-2 \right)\left( x+7 \right)=0\].
Similarly, we can write the factorization of equation \[9{{x}^{2}}-12x+4=0\].
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-\left( -12 \right)\pm \sqrt{{{\left( -12 \right)}^{2}}-4\times 9\times 4}}{2\times 9}=\dfrac{12\pm \sqrt{144-144}}{18}=\dfrac{12}{18}=\dfrac{2}{3}\]
As this equation has only one root, so it will be square. So, we can write the equation \[9{{x}^{2}}-12x+4=0\] as
\[{{\left( x-\dfrac{2}{3} \right)}^{2}}=0\]
\[\Rightarrow {{\left( 3x-2 \right)}^{2}}=0\]
The above can also be written as
\[\left( 3x-2 \right)\left( 3x-2 \right)=0\]
Hence, we can say that the factorization of the equation \[3{{x}^{2}}+19x-14\] is \[\left( 3x-2 \right)\left( x+7 \right)\] and the factorization of the equation \[9{{x}^{2}}-12x+4\] is \[\left( 3x-2 \right)\left( 3x-2 \right)\].
Now, we can write the equation that we have to solve as
\[\dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{5x}{\left( 3x-2 \right)\left( x+7 \right)}-\dfrac{1}{\left( 3x-2 \right)\left( 3x-2 \right)}\]
Now, taking the out common factor from the denominator, we can write
\[\Rightarrow \dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{5x}{\left( x+7 \right)}-\dfrac{1}{\left( 3x-2 \right)} \right]\]
The above can also be written as
\[\Rightarrow \dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{5x}{\left( x+7 \right)}\times \dfrac{\left( 3x-2 \right)}{\left( 3x-2 \right)}-\dfrac{1}{\left( 3x-2 \right)}\times \dfrac{\left( x+7 \right)}{\left( x+7 \right)} \right]\]
Now, the denominator is the same, then we can add the numerator easily.
\[\Rightarrow \dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{5x\left( 3x-2 \right)-\left( x+7 \right)}{\left( x+7 \right)\left( 3x-2 \right)} \right]\]
\[\Rightarrow \dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{15{{x}^{2}}-10x-x-7}{\left( x+7 \right)\left( 3x-2 \right)} \right]\]
The above can also be written as
\[\Rightarrow \dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}=\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{15{{x}^{2}}-11x-7}{\left( x+7 \right)\left( 3x-2 \right)} \right]\]
So, we have simplified the term \[\dfrac{5x}{3{{x}^{2}}+19x-14}-\dfrac{1}{9{{x}^{2}}-12x+4}\] and get the answer as \[\dfrac{1}{\left( 3x-2 \right)}\left[ \dfrac{15{{x}^{2}}-11x-7}{\left( x+7 \right)\left( 3x-2 \right)} \right]\].

Note: We should have a better knowledge in the topic of algebra to solve this type of question easily. We should remember how to find the roots and how to do the factorization of any quadratic equation. Suppose, we have a given an equation as \[a{{x}^{2}}+bx+c=0\], then the roots of the equation will be \[x=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\] and \[x=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}\]. And, the factorization of the equation \[a{{x}^{2}}+bx+c=0\] will be \[\left( x-\left( \dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} \right) \right)\left( x-\left( \dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a} \right) \right)=0\].