Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}, \qquad x\ne 3,-3 \]Solution
Given:
\[ \frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7} \]Taking LCM on the left side:
\[ \frac{(x-3)^2-(x+3)^2}{(x+3)(x-3)}=\frac{48}{7} \]Using \(a^2-b^2=(a-b)(a+b)\):
\[ \frac{\big[(x-3)-(x+3)\big]\big[(x-3)+(x+3)\big]}{x^2-9} =\frac{48}{7} \] \[ \frac{(-6)(2x)}{x^2-9} =\frac{48}{7} \] \[ \frac{-12x}{x^2-9} =\frac{48}{7} \]Cross-multiplying:
\[ -84x=48(x^2-9) \] \[ -84x=48x^2-432 \] \[ 48x^2+84x-432=0 \]Dividing by 12:
\[ 4x^2+7x-36=0 \]Factorizing:
\[ 4x^2+16x-9x-36=0 \] \[ 4x(x+4)-9(x+4)=0 \] \[ (x+4)(4x-9)=0 \]Therefore,
\[ x+4=0 \quad \text{or} \quad 4x-9=0 \] \[ x=-4 \quad \text{or} \quad x=\frac{9}{4} \]Both values satisfy the condition \(x\ne 3,-3\).