Solve the Following Quadratic Equation by Factorization
Question:
\[ \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c} \]Solution
Given,
\[ \frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c} \]Multiplying both sides by \((x-a)(x-b)(x-c)\), we get
\[ a(x-b)(x-c)+b(x-a)(x-c) = 2c(x-a)(x-b) \]Expanding:
\[ a(x^2-bx-cx+bc) + b(x^2-ax-cx+ac) = 2c(x^2-ax-bx+ab) \] \[ (a+b)x^2 – 2abx – c(a+b)x + 2abc = 2cx^2 – 2c(a+b)x + 2abc \] \[ (a+b-2c)x^2 + (a+b-2c)cx – 2abx =0 \] \[ x\Big[(a+b-2c)x+c(a+b-2c)-2ab\Big]=0 \] \[ x\Big[(a+b-2c)\{x+c\}-2ab\Big]=0 \]Now,
\[ (a+b-2c)x+(ac+bc-2c^2)-2ab \] \[ =(a+b-2c)(x-a-b+c) \]Therefore,
\[ x(x-a-b+c)=0 \]Hence,
\[ x=0 \quad \text{or} \quad x=a+b-c \]