Solve the Following Quadratic Equation by Factorization
Question:
Solve the quadratic equation:
\[ \frac{16}{x}-1=\frac{15}{x+1}, \qquad x\ne 0,-1 \]Solution
Given:
\[ \frac{16}{x}-1=\frac{15}{x+1} \]Multiplying both sides by \(x(x+1)\), we get:
\[ 16(x+1)-x(x+1)=15x \] \[ 16x+16-x^2-x=15x \] \[ -x^2+15x+16=15x \] \[ -x^2+16=0 \] \[ x^2-16=0 \]Using the identity \(a^2-b^2=(a-b)(a+b)\):
\[ (x-4)(x+4)=0 \]Therefore,
\[ x-4=0 \quad \text{or} \quad x+4=0 \] \[ x=4 \quad \text{or} \quad x=-4 \]Both values satisfy the condition \(x\ne0,-1\).