Question

Solve the following quadratic equations by factorization method :
$\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

Answer 1

Given equation:

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

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

$⟹[x^2+a^2-2xa+x^2+b^2+(-2xb\left)\right]ab=(a^2+b^2)(x^2+ab-ax-bx)$

$⟹\left[x^2\right(2ab)-x(2ab+2ba)+(a^{3}b+b^{3}a\left)\right]=\left(x^2\right(a^2+b^2)-x(a^{2}b+b^{2}a)+(a^{3}b+b^{3}a\left)\right)$

$\therefore a^{\prime }=(a-b{)}^2$

$b^{\prime }=a^{2}b+b^{2}a$

$c^{\prime }=0$

$\therefore x^2(a-b{)}^2+(a^{2}b+b^{2}a)x=0$

$\therefore x\left[x\right(a-b{)}^2+(a^{2}b+b^{2}a)]=0$

$\therefore x=0$,

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

$⟹x=\frac{ab(a+b)}{(a-b{)}^2}$

$\therefore$ Roots of equation are: $0,\frac{ab(a+b)}{(a-b{)}^2}$