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Quadratic Equations — Maths STD 10 — Question

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

$3\Big(\frac{3\text{x}-1}{2\text{x}+3}\Big)-2\Big(\frac{2\text{x}+3}{3\text{x}-1}\Big)=5$
Let $\Big(\frac{3\text{x}-1}{2\text{x}+3}\Big)=\text{y},$ then $\Big(\frac{2\text{x}+3}{3\text{x}-1}\Big)=\frac{1}{\text{y}}$
$\therefore3\text{y}-\frac{2}{\text{y}}=5$
$\Rightarrow3\text{y}^2-2=5\text{y}$
$\Rightarrow3\text{y}^2-5\text{y}-2=0$
$\Rightarrow3\text{y}^2-6\text{y}+\text{y}-2=0$
$\Rightarrow(\text{y}-2)(3\text{y}+1)=0$
Either $\text{y}-2=0,$ then $\text{y}=2$
or $3\text{y}+1=0,$ then $3\text{y}=-1$
$\Rightarrow\text{y}=\frac{-1}{3}$
When $\text{y}=2,$ then
$\frac{3\text{x}-1}{2\text{x}+3}=2$
$\Rightarrow3\text{x}-1=2(2\text{x}+3)$
$\Rightarrow3\text{x}-1=4\text{x}+6$
$\Rightarrow3\text{x}-4\text{x}=6+1$
$\Rightarrow-\text{x}=7$
$\Rightarrow\text{x}=-7$
If $\text{y}=\frac{-1}{3},$ then
$\frac{3\text{x}-1}{2\text{x}+3}=\frac{-1}{3}$
$\Rightarrow9\text{x}-3=-2\text{x}-3$
$\Rightarrow9\text{x}+2\text{x}=-3+3$
$\Rightarrow11\text{x}=0$
$\Rightarrow\text{x}=0$
$\Rightarrow\text{x}=0,7$

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