Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x+1}{x-1}+\frac{x-2}{x+2} = 4-\frac{2x+3}{x-2}, \qquad x\ne 1,-2,2 \]Solution
Given:
\[ \frac{x+1}{x-1}+\frac{x-2}{x+2} = 4-\frac{2x+3}{x-2} \]Multiplying both sides by \((x-1)(x+2)(x-2)\):
\[ (x+1)(x+2)(x-2) +(x-2)^2(x-1) = 4(x-1)(x+2)(x-2) -(2x+3)(x-1)(x+2) \]Simplifying and expanding:
\[ 2x^3-3x^2-6x = 2x^3-5x^2-14x+20 \] \[ 2x^2+8x-20=0 \] \[ x^2+4x-10=0 \]Using factorization through surds:
\[ x^2+4x-10 = (x+2-\sqrt{14})(x+2+\sqrt{14}) \]Therefore,
\[ x+2-\sqrt{14}=0 \] or \[ x+2+\sqrt{14}=0 \] \[ x=-2+\sqrt{14} \] or \[ x=-2-\sqrt{14} \]Both values satisfy the conditions \(x\ne1,-2,2\).