Question
Solve the following quadratic equation:
$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$

Answer

$\frac{1}{\text{2a}+\text{b}+\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}+\frac{1}{\text{2x}}$
$\Rightarrow\frac{1}{\text{2a}+\text{b}+\text{2x}}-\frac{1}{\text{2x}}=\frac{1}{\text{2a}}+\frac{1}{\text{b}}$
$\Rightarrow\frac{\text{2x}-(\text{2a}+\text{b}+\text{2x})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{\text{2x}-\text{2a}-\text{b}-\text{2x}}{\text{4ax}+\text{2bx}+\text{4x}^2}=\frac{\text{b}+\text{2a}}{\text{2ab}{}}$
$\Rightarrow\frac{-(\text{2a}+\text{b})}{\text{2x}(\text{2a}+\text{b}+\text{2x})}=\frac{\text{b}+\text{2a}}{\text{2ab}}$
$\Rightarrow\frac{-1}{\text{x}(\text{2a}+\text{b}+\text{2x})}=\frac{1}{\text{ab}}$
$\Rightarrow -ab = 2ax + bx + 2x^2$
$\Rightarrow 2x^2 + bx + 2ax + ab = 0$
$\Rightarrow 2x^2 + 2ax + bx + ab = 0$
$\Rightarrow 2x(x + a) + b(x + a) = 0$
$\Rightarrow (x + a)(2x + b) = 0$
$\Rightarrow x + a = 0 or 2x + b = 0$
⇒ x = -a or $\text{x}=\frac{-\text{b}}{2}$

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