Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4}, \qquad x\ne 5,-5 \]Solution
Given:
\[ \frac{5+x}{5-x}-\frac{5-x}{5+x}=\frac{15}{4} \]Taking LCM on the left side:
\[ \frac{(5+x)^2-(5-x)^2}{(5-x)(5+x)} =\frac{15}{4} \]Using the identity \(a^2-b^2=(a-b)(a+b)\):
\[ \frac{\big[(5+x)-(5-x)\big]\big[(5+x)+(5-x)\big]} {25-x^2} =\frac{15}{4} \] \[ \frac{(2x)(10)}{25-x^2} =\frac{15}{4} \] \[ \frac{20x}{25-x^2} =\frac{15}{4} \]Cross-multiplying:
\[ 80x=15(25-x^2) \] \[ 80x=375-15x^2 \] \[ 15x^2+80x-375=0 \] \[ 3x^2+16x-75=0 \]Factorizing:
\[ 3x^2+25x-9x-75=0 \] \[ x(3x+25)-3(3x+25)=0 \] \[ (3x+25)(x-3)=0 \]Therefore,
\[ 3x+25=0 \quad \text{or} \quad x-3=0 \] \[ x=-\frac{25}{3} \quad \text{or} \quad x=3 \]Both values satisfy the condition \(x\ne5,-5\).