Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}, \qquad x\ne 0,1,2 \]Solution
Given:
\[ \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x} \]Multiplying both sides by \(x(x-1)(x-2)\):
\[ x(x-1)+2x(x-2)=6(x-1)(x-2) \] \[ x^2-x+2x^2-4x=6(x^2-3x+2) \] \[ 3x^2-5x=6x^2-18x+12 \] \[ 3x^2-13x+12=0 \]Factorizing:
\[ 3x^2-9x-4x+12=0 \] \[ 3x(x-3)-4(x-3)=0 \] \[ (x-3)(3x-4)=0 \]Therefore,
\[ x-3=0 \quad \text{or} \quad 3x-4=0 \] \[ x=3 \quad \text{or} \quad x=\frac{4}{3} \]Both values satisfy the condition \(x\ne 0,1,2\).