Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}-3}+\frac{2}{\text{x}-2}=\frac{8}{\text{x}},$ $\text{x}\neq0, 2, 3$
$\frac{1}{\text{x}-3}+\frac{2}{\text{x}-2}=\frac{8}{\text{x}},$ $\text{x}\neq0, 2, 3$
✓
Answer
$\frac{1}{\text{x}-3}+\frac{2}{\text{x}-2}=\frac{8}{\text{x}}$
$\Rightarrow\frac{(\text{x}-2)+2(\text{x}-3)}{(\text{x}-3)(\text{x}-2)}=\frac{8}{\text{x}}$
$\Rightarrow\frac{\text{x}-2+2\text{x}-6}{\text{x}^2-2\text{x}-3\text{x}+6}=\frac{8}{\text{x}}$
$\Rightarrow\frac{3\text{x}-8}{\text{x}^2-5\text{x}+6}=\frac{8}{\text{x}}$
$\Rightarrow x(3x - 8) = 8(x^2 - 5x + 6)$
$\Rightarrow 3x^2 - 8x = 8x^2 - 40x + 48$
$\Rightarrow 5x^2 - 32x + 48 = 0$
$\Rightarrow 5x^2 - 20x - 12x + 48 = 0$
$\Rightarrow 5x(x - 4) - 12(x - 4) = 0$
$\Rightarrow (5x - 12)(x - 4) = 0$
$\Rightarrow 5x - 12 = 0$ or $x - 4 = 0$
$\Rightarrow\text{x}=\frac{12}{5}$ or $x = 4$
Hence, the factors 4 and $\frac{12}{5}$
$\Rightarrow\frac{(\text{x}-2)+2(\text{x}-3)}{(\text{x}-3)(\text{x}-2)}=\frac{8}{\text{x}}$
$\Rightarrow\frac{\text{x}-2+2\text{x}-6}{\text{x}^2-2\text{x}-3\text{x}+6}=\frac{8}{\text{x}}$
$\Rightarrow\frac{3\text{x}-8}{\text{x}^2-5\text{x}+6}=\frac{8}{\text{x}}$
$\Rightarrow x(3x - 8) = 8(x^2 - 5x + 6)$
$\Rightarrow 3x^2 - 8x = 8x^2 - 40x + 48$
$\Rightarrow 5x^2 - 32x + 48 = 0$
$\Rightarrow 5x^2 - 20x - 12x + 48 = 0$
$\Rightarrow 5x(x - 4) - 12(x - 4) = 0$
$\Rightarrow (5x - 12)(x - 4) = 0$
$\Rightarrow 5x - 12 = 0$ or $x - 4 = 0$
$\Rightarrow\text{x}=\frac{12}{5}$ or $x = 4$
Hence, the factors 4 and $\frac{12}{5}$
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