Question

How do solve $\dfrac{x}{{x - 2}} + \dfrac{{2x}}{{4 - {x^2}}} = \dfrac{5}{{x + 2}}$?

Answer 1

In this question we have to solve the given equation, first we have to simplify the equation to get a quadratic equation and then solve the quadratic equation using quadratic formula which is stated as, In the polynomial l$a{x^2} + bx + c$, where "$a$", "$b$", and “$c$" are real numbers and the Quadratic Formula is derived from the process of completing the square, and is formally stated as:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, by substituting the values in the formula and further simplification, we will get the required result.

Complete step by step solution:
Given equation is $\dfrac{x}{{x - 2}} + \dfrac{{2x}}{{4 - {x^2}}} = \dfrac{5}{{x + 2}}$,
Now rewrite the given equation as,
$\Rightarrow \dfrac{x}{{x - 2}} - \dfrac{{2x}}{{{x^2} - 4}} = \dfrac{5}{{x + 2}}$,
Now the denominator in the second term i.e., ${x^2} - 4$is in the form of algebraic identity, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$, so here ${x^2} - 4 = \left( {x - 2} \right)\left( {x + 2} \right)$, again rewriting the equation, we get,
$\Rightarrow \dfrac{x}{{x - 2}} - \dfrac{{2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{5}{{x + 2}}$,
Now taking the L.C.M i.e., multiplying both sides with $\left( {x - 2} \right)\left( {x + 2} \right)$, we get,
$\Rightarrow \dfrac{{x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{{2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{5\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}$,
Now simplifying we get,
$\Rightarrow \dfrac{{{x^2} + 2x - 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{5\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}$,
Now eliminating the like terms , we get,
$\Rightarrow {x^2} = 5x - 10$,
Now simplifying by taking all terms to the left hand side, we get,
$\Rightarrow {x^2} - 5x + 10 = 0$,
Now this is a quadratic equation, we will solve this by using the quadratic formula which is stated as,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$,
So, here $a = 1$,$b = - 5$and $c = 10$, now substituting the values in the formula we get,
$\Rightarrow x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( {10} \right)} }}{{2\left( 1 \right)}}$,
Now by simplifying, we get,
$\Rightarrow x = \dfrac{{5 \pm \sqrt {25 - 40} }}{2}$,
Again by further simplification we get,
$\Rightarrow x = \dfrac{{5 \pm \sqrt { - 15} }}{2}$,
We know that, ${i^2} = - 1$, so, now the value of $x$will be,
$\Rightarrow x = \dfrac{{5 \pm \sqrt {15 \times {i^2}} }}{2}$,
Now by simplifying we get,
$\Rightarrow x = \dfrac{{5 \pm \sqrt {15} i}}{2}$,
So, the values of $x$are $\dfrac{{5 \pm \sqrt {15} i}}{2}$.

Final Answer:
$\therefore$The value of $x$ when the given equation $\dfrac{x}{{x - 2}} + \dfrac{{2x}}{{4 - {x^2}}} = \dfrac{5}{{x + 2}}$is solved will be equal to $\dfrac{{5 \pm \sqrt {15} i}}{2}$.

Note:
Quadratic equation formula is a method of solving quadratic equations, but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of$x$by using the above formula. Also we should always convert the coefficient of${x^2} = 1$, to easily solve the equation by this method, and there are other methods to solve such kind of solutions, other method used to solve the quadratic equation is by factoring method, in this method we should obtain the solution factorising quadratic equation terms. In these types of questions, we can solve by using quadratic formula i.e., $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.