Question

Solve the following for the value of x: $\dfrac{2x+3}{2x-3}+\dfrac{2x-3}{2x+3}=\dfrac{17}{4}$

Answer 1

Hint: To solve this question we need to have the concept of fraction and LCM. In this question we are supposed to find the value of $x$ when the expression in terms of $x$ is equated with the fraction given in Right Hand Side in this problem. The first step is to find the LCM of the fraction in LHS forming a quadratic equation and thereafter solving for the value of $x$.

Complete step by step answer:
The question ask us to solve the expression given in this question which is $\dfrac{2x+3}{2x-3}+\dfrac{2x-3}{2x+3}=\dfrac{17}{4}$. To solve we will firstly find the LCM of the expression which is in the left hand side of the equation given to us. So to find the LCM of the expression we will multiply the denominator of both the fractions in the LHS. Hence on multiplying each of the denominator we get:
$\Rightarrow \dfrac{2x+3}{2x-3}+\dfrac{2x-3}{2x+3}=\dfrac{17}{4}$
\[\Rightarrow \dfrac{\left( 2x+3 \right)\left( 2x+3 \right)+\left( 2x-3 \right)\left( 2x-3 \right)}{\left( 2x-3 \right)\left( 2x+3 \right)}=\dfrac{17}{4}\]
\[\Rightarrow \dfrac{{{\left( 2x+3 \right)}^{2}}+{{\left( 2x-3 \right)}^{2}}}{\left( 2x-3 \right)\left( 2x+3 \right)}=\dfrac{17}{4}\]
For calculating the above expression the formula we will use for the expansion are ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$. Considering $a=2x$ and $b=3$, applying the above formula we get:
\[\Rightarrow \dfrac{\left( {{\left( 2x \right)}^{2}}+{{3}^{2}}+2\times 2x\times 3 \right)+\left( {{\left( 2x \right)}^{2}}+{{3}^{2}}-2\times 2x\times 3 \right)}{\left( {{\left( 2x \right)}^{2}}-{{3}^{2}} \right)}=\dfrac{17}{4}\]
On further calculating we get:
\[\Rightarrow \dfrac{\left( 4{{x}^{2}}+9+12x \right)+\left( 4{{x}^{2}}+9-12x \right)}{\left( 4{{x}^{2}}-9 \right)}=\dfrac{17}{4}\]
The term in the above expression get cancelled resulting in the equation:
\[\Rightarrow \dfrac{4{{x}^{2}}+9+4{{x}^{2}}+9}{4{{x}^{2}}-9}=\dfrac{17}{4}\]
\[\Rightarrow \dfrac{8{{x}^{2}}+18}{4{{x}^{2}}-9}=\dfrac{17}{4}\]
On cross multiplying the terms we get:
\[\Rightarrow 4\left( 8{{x}^{2}}+18 \right)=17\left( 4{{x}^{2}}-9 \right)\]
\[\Rightarrow 32{{x}^{2}}+72=68{{x}^{2}}-153\]
\[\Rightarrow 32{{x}^{2}}-68{{x}^{2}}=-153-72\]
\[\Rightarrow -36{{x}^{2}}=-225\]
On multiplying the terms with \[-1\]in both the
\[\Rightarrow 36{{x}^{2}}-225=0\]
The above expression are the square of \[6\] and \[15\]respectively;
\[\Rightarrow {{\left( 6x \right)}^{2}}-{{15}^{2}}=0\]
The above expression on expansion and on equating with zero we get:
\[\Rightarrow \left( 6x-15 \right)\left( 6x+15 \right)=0\]
On equating with zero we get:
\[x=\pm \dfrac{15}{6}\]
On further converting the fraction into the lowest term we get:
\[x=\pm \dfrac{5}{2}\]
$\therefore $ The equation $\dfrac{2x+3}{2x-3}+\dfrac{2x-3}{2x+3}=\dfrac{17}{4}$ is correct for \[x=\pm \dfrac{5}{2}\].

Note: To solve the question we need to remember the formula of expansion which says ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$. To solve the problem these formulas are used in many steps.