Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}},$ $\text{x}\neq0$
$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}},$ $\text{x}\neq0$
✓
Answer
We have, $\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}},$ $\text{x}\neq0$
$\Rightarrow\frac{(\text{x}-1)+2(\text{x}-2)}{(\text{x}-2)(\text{x}-1)}=\frac{6}{\text{x}}$
$\Rightarrow\frac{\text{x}-1+2\text{x}-4}{\text{x}^2-2\text{x}-\text{x}+2}=\frac{6}{\text{x}}$
$\Rightarrow\frac{3\text{x}-5}{\text{x}^2-3\text{x}+2}=\frac{6}{\text{x}}$
$\Rightarrow x(3x - 5) = 6(x^2 - 3x + 2)$
$\Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12$
$\Rightarrow 3x^2 - 18x + 5x + 12 = 0$
$\Rightarrow 3x^2 - 13x + 12 = 0$
[$\because$ 3 × 18 = 36
⇒ -9x - 4 and $-13 = -9 - 4]$
$\Rightarrow 3x^2 - 9x - 4x + 12 = 0$
$\Rightarrow 3x(x - 3) - 4(x - 3) = 0$
$\Rightarrow (x - 3)(3x - 4) = 0$
$\Rightarrow x = 3$ or $\text{x}=\frac{4}{3}$
$\therefore$ x = 3 and $\text{x}=\frac{4}{3}$ are the two roots of the given equation.
$\Rightarrow\frac{(\text{x}-1)+2(\text{x}-2)}{(\text{x}-2)(\text{x}-1)}=\frac{6}{\text{x}}$
$\Rightarrow\frac{\text{x}-1+2\text{x}-4}{\text{x}^2-2\text{x}-\text{x}+2}=\frac{6}{\text{x}}$
$\Rightarrow\frac{3\text{x}-5}{\text{x}^2-3\text{x}+2}=\frac{6}{\text{x}}$
$\Rightarrow x(3x - 5) = 6(x^2 - 3x + 2)$
$\Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12$
$\Rightarrow 3x^2 - 18x + 5x + 12 = 0$
$\Rightarrow 3x^2 - 13x + 12 = 0$
[$\because$ 3 × 18 = 36
⇒ -9x - 4 and $-13 = -9 - 4]$
$\Rightarrow 3x^2 - 9x - 4x + 12 = 0$
$\Rightarrow 3x(x - 3) - 4(x - 3) = 0$
$\Rightarrow (x - 3)(3x - 4) = 0$
$\Rightarrow x = 3$ or $\text{x}=\frac{4}{3}$
$\therefore$ x = 3 and $\text{x}=\frac{4}{3}$ are the two roots of the given equation.
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