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Question
Solve the following quadratic equation by factorization: \[\frac{a}{x - b} + \frac{b}{x - a} = 2\]
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Solution
\[\frac{a}{x - b} + \frac{b}{x - a} = 2\]
\[\Rightarrow \frac{ax - a^2 + bx - b^2}{\left( x - a \right)\left( x - b \right)} = 2\]
\[ \Rightarrow ax - a^2 + bx - b^2 = 2 x^2 - 2bx - 2ax + 2ab\]
\[ \Rightarrow 2 x^2 - 2bx - 2ax + 2ab - ax + a^2 - bx + b^2 = 0\]
\[ \Rightarrow 2 x^2 + x\left[ - 2b - 2a - a - b \right] + a^2 + b^2 + 2ab = 0\]
\[ \Rightarrow 2 x^2 - 3x\left[ a + b \right] + \left( a + b \right)^2 = 0\]
\[\Rightarrow 2 x^2 - 2\left( a + b \right)x - \left( a + b \right)x + \left( a + b \right)^2 = 0\]
\[ \Rightarrow 2x\left[ x - \left( a + b \right) \right] - \left( a + b \right)\left[ x - \left( a + b \right) \right] = 0\]
\[ \Rightarrow \left[ 2x - \left( a + b \right) \right]\left[ x - \left( a + b \right) \right] = 0\]
So, the value of x will be \[x = \frac{a + b}{2}, a + b\]
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