Question

Solve the following equation for $x$:
$\dfrac{1}{{2a + b + 2x}} = \dfrac{1}{{2a}} + \dfrac{1}{b} + \dfrac{1}{{2x}}$

Answer 1

Hint: First we will bring the terms containing x to LHS, then when we take LCM of the whole equation, we will get a quadratic in x. Then, by solving that quadratic for x, we will have the required answer.

Complete step-by-step answer:
We have with us the equation:-
$\dfrac{1}{{2a + b + 2x}} = \dfrac{1}{{2a}} + \dfrac{1}{b} + \dfrac{1}{{2x}}$
Bringing all the terms containing x to LHS only, we will get:-
$\dfrac{1}{{2a + b + 2x}} - \dfrac{1}{{2x}} = \dfrac{1}{{2a}} + \dfrac{1}{b}$
Now, taking LCM on the LHS, we will get:-
$\dfrac{{2x - (2a + b + 2x)}}{{2x(2a + b + 2x)}} = \dfrac{1}{{2a}} + \dfrac{1}{b}$
Now, simplifying by opening the bracket in the numerator of LHS, we will get:-
$\dfrac{{2x - 2a - b - 2x}}{{2x(2a + b + 2x)}} = \dfrac{1}{{2a}} + \dfrac{1}{b}$
Now, simplifying by clubbing the like terms in the numerator of LHS, we will get:-
$\dfrac{{ - 2a - b}}{{2x(2a + b + 2x)}} = \dfrac{1}{{2a}} + \dfrac{1}{b}$
Now, taking the negative sign common from the numerator of LHS and taking LCM on the RHS, we will get:-
$\dfrac{{ - (2a + b)}}{{2x(2a + b + 2x)}} = \dfrac{{b + 2a}}{{2ab}}$
Since the numerator of both the sides is same, we can cancel it out and write the expression as:-
$\dfrac{{ - 1}}{{2x(2a + b + 2x)}} = \dfrac{1}{{2ab}}$
Now, we will cross multiply the terms to get:-
$ - 2ab = 2x(2a + b + 2x)$
Simplifying it further to get:-
$ - 2ab = 4ax + 2bx + 4{x^2}$
We can rewrite it as:-
$4ax + 2bx + 4{x^2} + 2ab = 0$
Rewriting in a simplified form will lead us to:-
$4{x^2} + (4a + 2b)x + 2ab = 0$
We can also take 2 common and get:-
$2{x^2} + (2a + b)x + ab = 0$
The equation above is clearly a quadratic equation in x. We know that if we have an equation: $a{x^2} + bx + c = 0$, then $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
Using this formula in $2{x^2} + (2a + b)x + ab = 0$, we will get:-
\[x = \dfrac{{ - 2a - b \pm \sqrt {{{(2a + b)}^2} - 4 \times 2 \times ab} }}{{2 \times 2}}\]
Simplifying the values as much as we can:-
\[ \Rightarrow x = \dfrac{{ - 2a - b \pm \sqrt {{{(2a + b)}^2} - 8ab} }}{8}\]
Now using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we will get:-
\[ \Rightarrow x = \dfrac{{ - 2a - b \pm \sqrt {4{a^2} + {b^2} + 4ab - 8ab} }}{4}\]
Combining the like terms in this, we will get:-
\[ \Rightarrow x = \dfrac{{ - 2a - b \pm \sqrt {4{a^2} + {b^2} - 4ab} }}{4}\]
Now using the formula ${(a - b)^2} = {a^2} + {b^2} - 2ab$, we will get:-
\[ \Rightarrow x = \dfrac{{ - 2a - b \pm \sqrt {{{(2a - b)}^2}} }}{4}\]
\[ \Rightarrow x = \dfrac{{ - 2a - b \pm (2a - b)}}{4}\]
So, the values are \[x = \dfrac{{ - 2a - b + 2a - b}}{4} = - \dfrac{{2b}}{4} = - \dfrac{b}{2}\]
and \[x = \dfrac{{ - 2a - b - 2a + b}}{4} = - \dfrac{{4a}}{4} = - a\].

Note: Remember that if we cancel out any factor which is common, it must be always equal to non zero, because we cannot cancel out zero.
Fun fact:- The name Quadratic comes from "quad" meaning square, because the variable gets squared (like ${x^2}$).
It is also called an "Equation of Degree 2" (because of the "2" on the x).