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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equation:
$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$ $\text{x}\neq\frac{1}{3},\ \frac{-3}{2}$
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Answer

$3\Big(\frac{\text{3x}-1}{\text{2x}+3}\Big)-2\Big(\frac{\text{2x}+3}{\text{3x}-1}\Big)=5$
Taking $\frac{\text{3x}-1}{\text{2x}+3}=\text{y},$ we have
$\text{3y}-\frac{2}{\text{y}}=5$
$\Rightarrow\frac{\text{3y}^2-2}{\text{y}}=5$
$\Rightarrow 3y^2 - 2 = 5y$
$\Rightarrow 3y^2 - 5y - 2 = 0$
$\Rightarrow 3y^2 - 6y + y - 2 = 0$
$\Rightarrow 3y(y - 2) + 1(y - 2) = 0$
$\Rightarrow (y - 2)(3y + 1) = 0$
$\Rightarrow y - 2 = 0$ or $3y + 1 = 0$
$\Rightarrow y = 2$ or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{3x}-1}{\text{2x}+3}=2$ or $\frac{\text{3x}-1}{\text{2x}+3}=\frac{-1}{3}$
$\Rightarrow 3x - 1 = 4x + 6$ or $9x - 3 = -2x - 3$
$\Rightarrow x = -7$ or $11x = 0$
$\Rightarrow x = -7$ or $x = 0$

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