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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b}),$ $\text{x}\neq\frac{1}{\text{a}},\ \frac{1}{\text{b}}$
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Answer

$\frac{\text{a}}{(\text{ax}-\text{1})}+\frac{\text{b}}{(\text{bx}-\text{1})}=(\text{a}+\text{b})$
$\Rightarrow\Big[\frac{\text{a}}{\text{ax}-1}-\text{b}\Big]+\Big[\frac{\text{b}}{\text{bx}-1}-\text{a}\Big]=0$
$\Rightarrow\Big[\frac{\text{a}-\text{abx}+\text{b}}{\text{ax}-1}\Big]+\Big[\frac{\text{b}-\text{abx}+\text{a}}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})\Big[\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}\Big]=0$
$\Rightarrow(\text{a}-\text{abx}+\text{b})$ or $\frac{1}{\text{ax}-1}+\frac{1}{\text{bx}-1}=0$
$\Rightarrow\text{abx}=\text{a}+\text{b}$ or $\frac{1}{\text{ax}-1}=-\frac{1}{\text{bx}-1}$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} - 1 = -\text{ax} + 1$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{bx} + \text{ax} = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}(\text{b} + \text{a}) = 2$
$\Rightarrow\text{x}=\frac{\text{a}+\text{b}}{\text{ab}}$ or $\text{x}=\frac{2}{\text{a}+\text{b}}$

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