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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equation:
$\Big(\frac{\text{x}}{\text{x}}+1\Big)^2-5\Big(\frac{\text{x}}{\text{x}+1}\Big)+6=0,$ $\text{x}\neq-1$
✓

Answer

Putting $\Big(\frac{\text{x}}{\text{x}+1}\Big)=\text{y},$ the given equation become.
$\Rightarrow y^2 - 5y + 6 = 0 $
$\Rightarrow y^2 - 3y - 2y + 6 = 0 $
$\Rightarrow y(y - 3) - 2(y - 3) = 0 $
$\Rightarrow (y - 3)(y - 2) = 0 $
$\Rightarrow y - 3 = 0$ or $y - 2 = 0$
$\Rightarrow y = 3$ or $y = 2$
Case I:
$y = 3$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=3 $
$\Rightarrow 3x + 3 = x $
$\Rightarrow 3x - x = 3 $
$\Rightarrow 2x = 3$
$\Rightarrow\text{x}=\frac{-3}{2}$
Case II:
$y = 2$
$\Rightarrow\frac{\text{x}}{\text{x}+1}=2 $
$\Rightarrow 2x + 2 = x $
$\Rightarrow 2x - x = -2$
$\Rightarrow x = -2$
Hence, $\frac{-3}{2},\ -2$ are the roots of the given equation.

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