Solve the following quadratic equations by factorization: x+3 x-2… - Vidyadip

Question

Solve the following quadratic equations by factorization:$\frac{\text{x+3}}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=\frac{17}{4}$

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Answer

We have,
$\frac{\text{x+3}}{\text{x}-2}-\frac{1-\text{x}}{\text{x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}(\text{x}+3)-(\text{x}-2)(1-\text{x})}{\text{x}(\text{x}-2)}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+3\text{x}-(\text{x}-\text{x}^2-2+2\text{x})}{\text{x}^2-2\text{x}}=\frac{17}{4}$
$\Rightarrow\frac{\text{x}^2+3\text{x}-\text{x}+\text{x}^2+2-2\text{x}}{\text{x}^2-2\text{x}}=\frac{17}{4}$
$\Rightarrow\frac{2\text{x}^2+2}{\text{x}^2-2\text{x}}=\frac{17}{4}$
$\Rightarrow 4(2x^2 + 2) = 17(x^2 - 2x)$
$\Rightarrow 8x^2 + 8 = 17x^2 - 34x$
$\Rightarrow 8x^2 + 8 = 17x^2 - 34x$
$\Rightarrow (17 - 8)x^2 - 34x - 8 = 0$
$\Rightarrow 9x^2 - 34x - 8 = 0$
[$9 \times -8 = -72 \Rightarrow -72 = 36 \times 2$ and $-34 = -36 + 2$]
$\Rightarrow 9x^2 - 36x + 2x - 8 = 0$
$\Rightarrow 9x(x - 4) + 2(x - 4) = 0$
$\Rightarrow (x - 4)(9x + 2) = 0$
$\Rightarrow (x - 4) = 0$ or $9x + 2 = 0$
$\Rightarrow x = 4$ or $\text{x}=\frac{-8}{2}$
$\therefore$ $x = 4$ and $\text{x}=\frac{-8}{2}$ are the two roots of the given equation.

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