Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}, \qquad x\ne 1,-1 \]Solution
Given:
\[ \frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6} \]Taking LCM on the left side:
\[ \frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)} =\frac{5}{6} \]Using the identity \(a^2-b^2=(a-b)(a+b)\):
\[ \frac{\big[(x+1)-(x-1)\big]\big[(x+1)+(x-1)\big]} {x^2-1} =\frac{5}{6} \] \[ \frac{(2)(2x)}{x^2-1} =\frac{5}{6} \] \[ \frac{4x}{x^2-1} =\frac{5}{6} \]Cross-multiplying:
\[ 24x=5(x^2-1) \] \[ 24x=5x^2-5 \] \[ 5x^2-24x-5=0 \]Factorizing:
\[ 5x^2-25x+x-5=0 \] \[ 5x(x-5)+1(x-5)=0 \] \[ (5x+1)(x-5)=0 \]Therefore,
\[ 5x+1=0 \quad \text{or} \quad x-5=0 \] \[ x=-\frac{1}{5} \quad \text{or} \quad x=5 \]Both values satisfy the condition \(x\ne 1,-1\).