Question

Solve the following for the value of x,
$\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.

Answer 1

Hint: To solve this question we will first take the term $\dfrac{1}{2x}$ from the RHS to the LHS. After that we will take the LCM on both the sides of the obtained equation. After that we will form a quadratic in the x and solve it using the quadratic formula i.e. for $a{{x}^{2}}+bx+c=0$ the solution of this equation are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Hence in this way we will get the two values of x as our answer.

Complete step-by-step answer:
We are given the expression,
$\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$
And we have to solve for the value of x,
So first of all we will take the term $\dfrac{1}{2x}$ from RHS of the equation to the LHS, so we get
$\dfrac{1}{2a+b+2x}-\dfrac{1}{2x}=\dfrac{1}{2a}+\dfrac{1}{b}$
Now taking the LCM on both the sides of the above expression, we get
$\begin{align}
  & \dfrac{2x-2x-b-2x}{\left( 2a+b+2x \right)\left( 2x \right)}=\dfrac{2a+b}{\left( 2a \right)\left( b \right)} \\
 & \dfrac{-\left( 2a+b \right)}{\left( 2a+b+2x \right)\left( 2x \right)}=\dfrac{2a+b}{\left( 2a \right)\left( b \right)} \\
\end{align}$
Cancelling out 2a + b from both sides we get,
$\dfrac{-1}{\left( 2a+b+2x \right)\left( 2x \right)}=\dfrac{1}{\left( 2a \right)\left( b \right)}$
Now cross multiplying we get,
$-2ab=4ax+2bx+4{{x}^{2}}$
Arranging the equation such that a quadratic in x is formed, we get
$4{{x}^{2}}+\left( 4a+2b \right)x+2ab=0$
Dividing the whole equation by 2, we get
$2{{x}^{2}}+\left( 2a+b \right)x+ab=0$
Now using the quadratic formula i.e. for the equation $a{{x}^{2}}+bx+c=0$, its solutions are given by,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So applying this in equation
$2{{x}^{2}}+\left( 2a+b \right)x+ab=0$ we get
\[x=\dfrac{-\left( 2a+b \right)\pm \sqrt{{{\left( 2a+b \right)}^{2}}-8ab}}{4}\]
Expanding the term ${{\left( 2a+b \right)}^{2}}$, we have
\[\begin{align}
  & x=\dfrac{-\left( 2a+b \right)\pm \sqrt{4{{a}^{2}}+4ab+{{b}^{2}}-8ab}}{4} \\
 & x=\dfrac{-\left( 2a+b \right)\pm \sqrt{4{{a}^{2}}-4ab+{{b}^{2}}}}{4} \\
 & x=\dfrac{-\left( 2a+b \right)\pm \sqrt{{{\left( 2a-b \right)}^{2}}}}{4} \\
 & x=\dfrac{-\left( 2a+b \right)\pm \left( 2a-b \right)}{4} \\
\end{align}\]
So we get
$\begin{align}
  & x=\dfrac{-2a-b+2a-b}{4}\,\,or\,x=\,\dfrac{-2a-b-2a+b}{4} \\
 & x=\dfrac{-2b}{4}\,\,or\,x=\,\dfrac{-4a}{4} \\
 & x=\dfrac{-b}{2}\,\,or\,x=\,-a \\
\end{align}$
Hence we get two values of x as $-\dfrac{b}{2},-a$ for the condition $\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.

Note: You need to note that we cannot solve this question by some other shortcut method if you try to think it in some other way it will just waste your time in exam so try to solve these kind of problems by basic approach and by performing each and every step otherwise you may end up making a mistake and wrong answer.