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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-3}{\text{x}+3}-\frac{\text{x}+3}{\text{x}-3}=\frac{48}{7},$ $\text{x}\neq3,\text{x}\neq-3$
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Answer

$\frac{\text{x}-3}{\text{x}+3}-\frac{\text{x}+3}{\text{x}-3}=\frac{48}{7}$
$\Rightarrow\frac{\text{x}-3}{\text{x}+3}-\frac{\text{x}+3}{\text{x}-3}=\frac{48}{7}$
$\Rightarrow\frac{(\text{x}-3)^2-(\text{x}+3)^2}{(\text{x}+3)(\text{x}-3)}=\frac{48}{7}$
$\Rightarrow\frac{\text{x}^2-6\text{x}+9-\text{x}^2-6\text{x}-9}{\text{x}^2-9}=\frac{48}{7}$
$\Rightarrow\frac{-12\text{x}}{\text{x}^2-9}=\frac{48}{7}$
$\Rightarrow 48x^2 + 84x - 432 = 0$
$\Rightarrow 4x^2 + 7x - 36 = 0 (Dividing by 12)$
$\Rightarrow 4x^2 + 16x - 9x - 36 = 0$
$\begin{cases}\because36\times4=-144\\\therefore-144=16\times-9\\7=16-9\end{cases}$
$\Rightarrow 4x(x + 4) - 9(x + 4) = 0$
$\Rightarrow (x + 4)(4x - 9) = 0$
Either x + 4 = 0, then x = -4
or 4x - 9 = 0, then 4x = 9
$\Rightarrow\text{x}=\frac{9}{4}$
$\therefore$ Roots are -4, $\frac{9}{4}$

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