Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac16 \]Solution
Using
\[ \frac{1}{(x-a)(x-b)} = \frac{1}{a-b} \left( \frac{1}{x-a} – \frac{1}{x-b} \right) \]we get
\[ \frac{1}{(x-1)(x-2)} = \frac{1}{x-2} – \frac{1}{x-1} \] \[ \frac{1}{(x-2)(x-3)} = \frac{1}{x-3} – \frac{1}{x-2} \] \[ \frac{1}{(x-3)(x-4)} = \frac{1}{x-4} – \frac{1}{x-3} \]Adding, the middle terms cancel:
\[ \frac{1}{x-4} – \frac{1}{x-1} = \frac16 \] \[ \frac{(x-1)-(x-4)} {(x-4)(x-1)} = \frac16 \] \[ \frac{3} {(x-4)(x-1)} = \frac16 \]Cross-multiplying:
\[ 18=(x-4)(x-1) \] \[ x^2-5x+4=18 \] \[ x^2-5x-14=0 \]Factorizing:
\[ x^2-7x+2x-14=0 \] \[ x(x-7)+2(x-7)=0 \] \[ (x-7)(x+2)=0 \]Therefore,
\[ x=7 \quad \text{or} \quad x=-2 \]