Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{a}{x-b}+\frac{b}{x-a}=2, \qquad x\ne a,b \]Solution
Given:
\[ \frac{a}{x-b}+\frac{b}{x-a}=2 \]Taking LCM on the left side:
\[ \frac{a(x-a)+b(x-b)} {(x-a)(x-b)} =2 \] \[ \frac{(a+b)x-(a^2+b^2)} {(x-a)(x-b)} =2 \]Cross-multiplying:
\[ (a+b)x-(a^2+b^2) = 2(x-a)(x-b) \] \[ (a+b)x-(a^2+b^2) = 2x^2-2(a+b)x+2ab \] \[ 2x^2-3(a+b)x+(a^2+2ab+b^2)=0 \] \[ 2x^2-3(a+b)x+(a+b)^2=0 \]Factorizing:
\[ 2x^2-2(a+b)x-(a+b)x+(a+b)^2=0 \] \[ 2x\bigl(x-(a+b)\bigr) -(a+b)\bigl(x-(a+b)\bigr)=0 \] \[ \bigl(x-(a+b)\bigr)\bigl(2x-(a+b)\bigr)=0 \]Therefore,
\[ x-(a+b)=0 \] or \[ 2x-(a+b)=0 \] \[ x=a+b \] or \[ x=\frac{a+b}{2} \]