Solve the following quadratic equation by factorization:

$\frac{a}{x-b}+\frac{b}{x-a}=2, x ≠ a, b$

Given:

Given quadratic equation is $\frac{a}{x-b}+\frac{b}{x-a}=2, x ≠ a, b$.

To do:

We have to solve the given quadratic equation.

Solution:

$\frac{a}{x-b}+\frac{b}{x-a}=2$

$\frac{a(x-a)+b(x-b)}{(x-a)(x-b)}=2$

$\frac{ax-a^2+bx-b^2}{x^2-ax-bx+ab}=2$

$(ax-a^2+bx-b^2)=2(x^2-ax-bx+ab)$

$ax-a^2+bx-b^2=2x^2-2ax-2bx+2ab$

$2x^2+(-2a-2b-a-b)x+2ab+a^2+b^2=0$

$2x^2+(-3a-3b)x+(a+b)^2=0$

$2x^2-2(a+b)x-(a+b)x+(a+b)^2=0$

$2x(x-(a+b))-(a+b)(x-(a+b))=0$

$(2x-(a+b))(x-(a+b))=0$

$2x-(a+b)=0$ or $x-(a+b)=0$

$2x=(a+b)$ or $x=a+b$

$x=\frac{a+b}{2}$ or $x=a+b$

The values of $x$ are $\frac{a+b}{2}$ and $a+b$.