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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equation:
$\frac{\text{a}}{(\text{x}-\text{b})}+\frac{\text{b}}{(\text{x}-\text{a})}=2,$ $\text{x}\neq-\text{b},\ \text{a}$
✓

Answer

The given equation:
$\Big(\frac{\text{a}}{\text{x}-\text{b}}-1\Big)+\Big(\frac{\text{b}}{\text{x}-\text{a}}-1\Big)=0$
$\Rightarrow\frac{(\text{a}-\text{x}+\text{b})}{(\text{x}-\text{b})}+\frac{(\text{b}-\text{x}+\text{a})}{(\text{x}-\text{a})}=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{1}{(\text{x}-\text{b})}+\frac{1}{(\text{x}-\text{a})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})\Big[\frac{\text{2x}-(\text{a}+\text{b})}{(\text{x}-\text{a})(\text{x}-\text{b})}\Big]=0$
$\Rightarrow(\text{a}-\text{x}+\text{b})[\text{2x}-(\text{a}+\text{b})]=0$
$\Rightarrow\text{x}=(\text{a}+\text{b})$ or $\text{x}=\frac{(\text{a}+\text{b})}{2}$
Hence, (a + b) and $\frac{(\text{a}+\text{b})}{2}$ is the root of the given equation.

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