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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-1}{2\text{x}+1}+\frac{2\text{x}+1}{\text{x}-1}=\frac{5}{2},$ $\text{x}\neq-\frac{1}{2},1$
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Answer

$\frac{\text{x}-1}{2\text{x}+1}+\frac{2\text{x}+1}{\text{x}-1}=\frac{5}{2},$ $\text{x}\neq-\frac{1}{2},1$
$\Rightarrow\frac{(\text{x}-1)(\text{x}-1)+(2\text{x}+1)(2\text{x}+1)}{(2\text{x}+1)(\text{x}-1)}=\frac{5}{2}$
$\Rightarrow\frac{(\text{x}-1)^2+(2\text{x}+1)^2}{2\text{x}^2-2\text{x}+\text{x}-1}=\frac{5}{2}$
$\Rightarrow\frac{\text{x}^2-2\text{x}+1+4\text{x}^2+4\text{x}+1}{2\text{x}^2-\text{x}-1}=\frac{5}{2}$
$\Rightarrow\frac{5\text{x}^2+2\text{x}+2}{2\text{x}^2-\text{x}-1}=\frac{5}{2}$
[$\because$ $(a + b)^2 = a^2 + b^2 + 2ab, (a - b)^2 = a^2 + b^2 - 2ab]$
$\Rightarrow 2(5x^2 + 2x + 2) = 5(2x^2 - x - 1)$
$\Rightarrow 10x^2 + 4x + 4 = 10x^2 - 5x - 5$
$\Rightarrow 4x + 5x + 4 +5 = 0$
$\Rightarrow 9x + 9 = 0$
$\Rightarrow 9x = -9$
$\Rightarrow x = -1$
$\therefore$ x = -1 is the only root for the given equation

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