Question
Solve the following quadratic equations by factorization:
$\frac{\text{a}}{\text{x}-\text{b}}+\frac{\text{b}}{\text{x}-\text{a}}=2$
$\frac{\text{a}}{\text{x}-\text{b}}+\frac{\text{b}}{\text{x}-\text{a}}=2$
✓
Answer
$\frac{\text{a}}{\text{x}-\text{b}}+\frac{\text{b}}{\text{x}-\text{a}}=2$
$\Rightarrow\frac{\text{a}(\text{x}-\text{a})+\text{b}(\text{x}-\text{b})}{(\text{x}-\text{a})(\text{x}-\text{b})}=2$
$\Rightarrow ax - a^2 + bx - b^2 = 2x^2 - 2ax - 2bx + 2ab$
$\Rightarrow 2x^2 - 2ax - ax - 2bx - bx + a^2 + b^2 + 2ab = 0$
$\Rightarrow 2x^2 - 3x(a + b) + (a + b)^2 = 0$
$\Rightarrow 2x^2 - 2x(a + b) - x(a + b) + (a + b)^2= 0$
$\Rightarrow 2x[x - (a + b)] - (a + b)[x - (a + b)] = 0$
$\Rightarrow [2x - (a + b)] [(x - a + b)] = 0$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{2}$ or $x = a + b$
$\Rightarrow\frac{\text{a}(\text{x}-\text{a})+\text{b}(\text{x}-\text{b})}{(\text{x}-\text{a})(\text{x}-\text{b})}=2$
$\Rightarrow ax - a^2 + bx - b^2 = 2x^2 - 2ax - 2bx + 2ab$
$\Rightarrow 2x^2 - 2ax - ax - 2bx - bx + a^2 + b^2 + 2ab = 0$
$\Rightarrow 2x^2 - 3x(a + b) + (a + b)^2 = 0$
$\Rightarrow 2x^2 - 2x(a + b) - x(a + b) + (a + b)^2= 0$
$\Rightarrow 2x[x - (a + b)] - (a + b)[x - (a + b)] = 0$
$\Rightarrow [2x - (a + b)] [(x - a + b)] = 0$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{2}$ or $x = a + b$
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