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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equation:
$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3,$ $\text{x}\neq\frac{-1}{2},\ \frac{3}{4}$
✓

Answer

$\Big(\frac{\text{4x}-3}{\text{2x}+1}\Big)-10\Big(\frac{\text{2x}+1}{\text{4x}-3}\Big)=3$
Taking $\frac{\text{4x}-3}{\text{2x}+1}=\text{y},$
we have $\text{y}-\frac{10}{\text{y}}=3$
$\Rightarrow\frac{\text{y}^2-10}{\text{y}}=3$
$\Rightarrow y^2 - 10 = 3y $
$\Rightarrow y^2 - 3y - 10 = 0 $
$\Rightarrow y^2 - 5y + 2y - 10 = 0 $
$\Rightarrow y(y - 5) + 2(y - 5) = 0 $
$\Rightarrow (y - 5)(y + 2) = 0 $
$\Rightarrow y - 5 = 0$ or $y + 2 = 0$
$\Rightarrow y = 5$ or $y = -2$
$\Rightarrow\frac{\text{4x}-3}{\text{2x}+1}=5$ or $\frac{\text{4x}-3}{\text{2x}+1}=-2$
$\Rightarrow 4x - 3 = 10x + 5$ or $4x - 3 = -4x - 2$
$\Rightarrow 6x = -8$ or $8x = 1$
$\Rightarrow\text{x}=\frac{-4}{3}$ or $\text{x}=\frac{1}{8}$

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