Solve the Following Quadratic Equation by Factorization
Question:
\[ (a+b)^2x^2-4abx-(a-b)^2=0 \]Solution
Given,
\[ (a+b)^2x^2-4abx-(a-b)^2=0 \]Since
\[ (a-b)^2=(a+b)^2-4ab \]Substituting, we get
\[ (a+b)^2x^2-4abx-(a+b)^2+4ab=0 \] \[ (a+b)^2(x^2-1)-4ab(x-1)=0 \] \[ (x-1)\Big[(a+b)^2(x+1)-4ab\Big]=0 \]Now,
\[ (a+b)^2(x+1)-4ab \] \[ =(a+b)^2x+(a+b)^2-4ab \] \[ =(a+b)^2x+(a-b)^2 \]Therefore,
\[ (x-1)\Big[(a+b)^2x+(a-b)^2\Big]=0 \]Hence,
\[ x-1=0 \] or \[ (a+b)^2x+(a-b)^2=0 \] \[ x=1 \] or \[ x=-\frac{(a-b)^2}{(a+b)^2} \]