Question

Solve: $\dfrac{{x - 3}}{{x + 3}} - \dfrac{{x + 3}}{{x - 3}} = \dfrac{{48}}{7}$.

Answer 1

Hint: To solve the given question, we will first take the LCM on the left hand side of the equation given in the question and simplify it using identities. After this, we will cross multiply the equation and again simplify it. After simplifying the equation, we will get a quadratic equation in $x$. To find the roots of this quadratic equation, we will use the factorization method. In this method, we express $a{x^2} + bx + c = 0$ as the product of two linear factors, say $\left( {px + q} \right)$ and $\left( {rx + s} \right)$, where $p$, $q$, $r$ and $s$ are real numbers such that $p \ne 0$ and $r \ne 0$. We will get linear equations in the form of $px + q = 0$ or, $rx + s = 0$. Solving these linear equations, we get the possible roots of the given quadratic equation as $x = - \dfrac{p}{q}$ and $x = - \dfrac{s}{r}$.

Complete step by step solution:
Given equation is,
$\dfrac{{x - 3}}{{x + 3}} - \dfrac{{x + 3}}{{x - 3}} = \dfrac{{48}}{7}$
Take LCM on left-hand side
$ \Rightarrow \dfrac{{\left( {x - 3} \right)\left( {x - 3} \right) - \left[ {\left( {x + 3} \right)\left( {x + 3} \right)} \right]}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{48}}{7}$
It can also be written as,
$ \Rightarrow \dfrac{{{{\left( {x - 3} \right)}^2} - \left[ {{{\left( {x + 3} \right)}^2}} \right]}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \dfrac{{48}}{7}$
Now, we will use identities to simply the above written equation: ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ and ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$.
\[ \Rightarrow \dfrac{{{x^2} - 6x + 9 - \left( {{x^2} + 6x + 9} \right)}}{{{x^2} - 3x + 3x - 9}} = \dfrac{{48}}{7}\]
\[ \Rightarrow \dfrac{{{x^2} - 6x + 9 - {x^2} - 6x - 9}}{{{x^2} - 9}} = \dfrac{{48}}{7}\]
On addition and subtraction of like terms, we get
\[ \Rightarrow \dfrac{{ - 12x}}{{{x^2} - 9}} = \dfrac{{48}}{7}\]
On cross-multiplication, we get
\[ \Rightarrow 7\left( { - 12x} \right) = 48\left( {{x^2} - 9} \right)\]
On multiplication of terms, we get
\[ \Rightarrow - 84x = 48{x^2} - 432\]
Shift all the terms on one side
\[ \Rightarrow 48{x^2} + 84x - 432 = 0\]
On dividing by $12$, we get
\[ \Rightarrow 4{x^2} + 7x - 36 = 0\]
Let us express $4{x^2} + 7x - 36 = 0$ as the product of two linear factors.
$ \Rightarrow 4{x^2} + 16x - 9x - 36 = 0$
$ \Rightarrow 4x\left( {x + 4} \right) - 9\left( {x + 4} \right) = 0$
Now, we have two factors $\left( {4x - 9} \right)$ and $\left( {x + 4} \right)$
$ \Rightarrow \left( {4x - 9} \right)\left( {x + 4} \right) = 0$
Solving these linear equations, we get roots of $4{x^2} + 7x - 36 = 0$
$ \Rightarrow x = - 4$ or $x = \dfrac{9}{4}$
Therefore, $x = - 4$ or $x = \dfrac{9}{4}$ are the two roots of the given equation.
So, the correct answer is “ $x = - 4$ or $x = \dfrac{9}{4}$”.

Note: A quadratic equation of the form $a{x^2} + bx + c = 0$ where $a \ne 0$ can be solved by many methods. Such as, completing the square, factorization method, by using the quadratic formula, etc. Doesn’t matter with which method we solve any quadratic equation because the roots of the quadratic equation will remain the same.