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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-\text{a}}{\text{x}-\text{b}}+\frac{\text{x}-\text{b}}{\text{x}-\text{a}}=\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}$
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Answer

$\frac{\text{x}-\text{a}}{\text{x}-\text{b}}+\frac{\text{x}-\text{b}}{\text{x}-\text{a}}=\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}$
$\Rightarrow\frac{\text{x}-\text{a}}{\text{x}-\text{b}}-\frac{\text{a}}{\text{b}}=\frac{\text{b}}{\text{a}}-\frac{\text{x}-\text{b}}{\text{x}-\text{a}}$
$\Rightarrow\frac{\text{bx}-\text{ab}-\text{ax}+\text{ab}}{\text{b}(\text{x}-\text{b})}=\frac{\text{bx}-\text{ab}-\text{ax}+\text{ab}}{\text{a}(\text{x}-\text{a})}$
$\Rightarrow\frac{\text{bx}-\text{ax}}{\text{b}(\text{x}-\text{b})}=\frac{\text{bx}-\text{ax}}{\text{a}(\text{x}-\text{a})}$
$\Rightarrow\frac{\text{bx}-\text{ax}}{\text{b}(\text{x}-\text{b})}-\frac{\text{bx}-\text{ax}}{\text{a}(\text{x}-\text{a})}=0$
$\Rightarrow(\text{bx}-\text{ax})\bigg(\frac{1}{\text{b}(\text{x}-\text{b})}-\frac{1}{\text{a}(\text{x}-\text{a})}\bigg)=0$
$\Rightarrow(\text{b}\text{x}-\text{ax})\bigg(\frac{\text{a}(\text{x}-\text{a})-\text{b}(\text{x}-\text{b})}{\text{a}\text{b}(\text{x}-\text{a})(\text{x}-\text{b})}\bigg)=0$
$\Rightarrow\frac{\text{x}(\text{b}-\text{a})(\text{ax}-\text{bx}-\text{a}^2+\text{b}^2)}{\text{ab}(\text{x}-\text{a})(\text{x}-\text{b})}=0$
$\because\text{x}\neq\text{a},\text{x}\neq\text{b}$
(Division by Zero is not possible)
$\therefore\text{x}(\text{ax}-\text{bx}-\text{a}^2+\text{b}^2)=0$
Either $\text{x}=0$
or $\text{ax}-\text{bx}=\text{a}^2-\text{b}^2$
$\Rightarrow(\text{a}-\text{b})\text{x}=(\text{a}+\text{b})(\text{a}-\text{b})$
$\Rightarrow\text{x}=\frac{(\text{a}+\text{b})(\text{a}-\text{b})}{\text{a}-\text{b}}=\text{a}+\text{b}$
Hence, $\text{x}=0,\text{a}+\text{b}$

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