Question

Solve the following quadratic equations by factorization: \[\frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x}\]

Answer 1

\[\frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x}\]

\[ \Rightarrow \frac{1}{2a + b + 2x} - \frac{1}{2a} = \frac{1}{b} + \frac{1}{2x}\]

\[ \Rightarrow \frac{2a - \left( 2a + b + 2x \right)}{\left( 2a + b + 2x \right)\left( 2a \right)} = \frac{2x + b}{2bx}\]

\[ \Rightarrow \frac{- b - 2x}{4 a^2 + 2ab + 4ax} = \frac{2x + b}{2bx}\]

\[ \Rightarrow \frac{- 1\left( 2x + b \right)}{4 a^2 + 2ab + 4ax} = \frac{2x + b}{2bx}\]

\[ \Rightarrow - 2bx\left( 2x + b \right) = \left( 4 a^2 + 2ab + 4ax \right)\left( 2x + b \right)\]

\[ \Rightarrow \left( 4 a^2 + 2ab + 4ax \right)\left( 2x + b \right) + 2bx\left( 2x + b \right) = 0\]

\[ \Rightarrow \left( 2x + b \right)\left( 4 a^2 + 2ab + 4ax + 2bx \right) = 0\]

\[ \Rightarrow 2x + b = 0 \text { or } 4 a^2 + 2ab + \left( 4a + 2b \right)x = 0\]

\[ \Rightarrow x = - \frac{b}{2} \text { or } x = - \frac{4 a^2 + 2ab}{4a + 2b}\]

\[ \Rightarrow x = - \frac{b}{2} \text { or } x = - \frac{a\left( 4a + 2b \right)}{4a + 2b}\]

\[ \Rightarrow x = - \frac{b}{2} \text { or } x = - a\]

Hence, the factors are \[- a\] and \[- \frac{b}{2}\].