Question

Solve for x: $ \dfrac{{x + 2}}{{x - 2}} + \dfrac{{x - 4}}{{x + 4}} = 6;\,\left( {x \ne 2, - 4} \right) $

Answer 1

Hint: We have a sum of two rational functions in x on the left side of the equation. So, we simplify the sum by taking the LCM of the denominators and multiplying the numerators and denominators of both the fractions by the same expression so that the value remains unchanged. Now, we add the numerators after making the denominators of the fraction common and continue with the simplification. Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease.

Complete step by step solution:
We would use the method of transposition and addition of rational expressions to find the value of x in $ \dfrac{{x + 2}}{{x - 2}} + \dfrac{{x - 4}}{{x + 4}} = 6 $ . Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding the value of the required parameter.
Now, there is a sum of rational expressions on the left side of the equation. So, we take the LCM of the denominators and make the denominator of both the rational expressions equal.
So, \[\dfrac{{x + 2}}{{x - 2}} \times \left( {\dfrac{{x + 4}}{{x + 4}}} \right) + \dfrac{{x - 4}}{{x + 4}} \times \left( {\dfrac{{x - 2}}{{x - 2}}} \right) = 6\]
Now, opening the brackets in numerator and simplifying, we get,
 $ \Rightarrow $ \[\dfrac{{\left( {x + 2} \right)\left( {x + 4} \right) + \left( {x - 2} \right)\left( {x - 4} \right)}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = 6\]
 $ \Rightarrow $ \[\dfrac{{{x^2} + 6x + 8 + {x^2} - 6x + 8}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = 6\]
Cancelling the like terms with opposite signs,
 $ \Rightarrow $ \[\dfrac{{2{x^2} + 16}}{{\left( {x - 2} \right)\left( {x + 4} \right)}} = 6\]
Now, cross multiplying the terms, we get,
 $ \Rightarrow $ \[2{x^2} + 16 = 6\left( {x - 2} \right)\left( {x + 4} \right)\]
Opening the brackets and evaluating the product of terms, we get,
 $ \Rightarrow $ \[2{x^2} + 16 = 6\left( {{x^2} + 2x - 8} \right)\]
 $ \Rightarrow $ \[2{x^2} + 16 = 6{x^2} + 12x - 48\]
Adding up like terms, we get,
 $ \Rightarrow $ \[6{x^2} + 12x - 48 - 2{x^2} - 16 = 0\]
 $ \Rightarrow $ \[4{x^2} + 12x - 64 = 0\]
Dividing both sides of the equation by $ 4 $ ,
 $ \Rightarrow $ \[{x^2} + 3x - 16 = 0\]
Now, we can find the value of x by solving the quadratic equation \[{x^2} + 3x - 16 = 0\] using the quadratic formula.
The quadratic formula can be employed for solving an equation only if we compare the given equation with the standard form of quadratic equation.
Comparing with standard quadratic equation $ a{x^2} + bx + c = 0 $
Here, $ a = 1 $ , $ b = 3 $ and $ c = - 16 $ .
Now, using the quadratic formula, we get the roots of the equation as:
 $ x = \dfrac{{( - b) \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Substituting the values of a, b, and c in the quadratic formula, we get,
 $ x = \dfrac{{ - \left( 3 \right) \pm \sqrt {{{\left( 3 \right)}^2} - 4 \times 1 \times \left( { - 16} \right)} }}{{2 \times 1}} $
 $ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {9 + 64} }}{2} $
 $ \Rightarrow x = \dfrac{{ - 3 \pm \sqrt {73} }}{2} $
So, $ x = \dfrac{{ - 3 + \sqrt {73} }}{2} $ and $ x = \dfrac{{ - 3 - \sqrt {73} }}{2} $ are the roots of the equation \[{x^2} + 3x - 16 = 0\].
So, the roots of the equation \[{x^2} + 3x - 16 = 0\] are: $ x = \dfrac{{ - 3 + \sqrt {73} }}{2} $ and $ x = \dfrac{{ - 3 - \sqrt {73} }}{2} $ .
So, the correct answer is “ $ x = \dfrac{{ - 3 + \sqrt {73} }}{2} $ and $ x = \dfrac{{ - 3 - \sqrt {73} }}{2} $ ”.

Note: Linear equations in one variable can be solved by a transposition method with ease. There are various methods that can be employed to solve a quadratic equation like completing the square method, using quadratic formula and by splitting the middle term. Using the quadratic formula gives us the roots of the equation directly with ease. Quadratic equations may also be solved by a hit and trial method if the roots of the equation are easy to find. We must take care of calculations while doing such questions.