Question

Solve the following quadratic equation by factorization: \frac { a } { x - b } + \frac { b } { x - a } = 2 , x \neq a , b x−b a ​ + x−a b ​ =2,x  ​ =a,b.

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\:\dfrac{a}{x - b} + \dfrac{b}{x - a} = 2$

can be rewritten as

$\rm :\longmapsto\:\dfrac{a}{x - b} + \dfrac{b}{x - a} - 2 = 0$

can be rewritten as

$\rm :\longmapsto\:\dfrac{a}{x - b} + \dfrac{b}{x - a} - 1 - 1 = 0$

can be re-arranged as

$\rm :\longmapsto\:\bigg[\dfrac{a}{x - b} - 1\bigg] + \bigg[\dfrac{b}{x - a} - 1\bigg]= 0$

$\rm :\longmapsto\:\bigg[\dfrac{a - (x - b)}{x - b}\bigg] + \bigg[\dfrac{b - (x - a)}{x - a}\bigg]= 0$

$\rm :\longmapsto\:\bigg[\dfrac{a - x + b}{x - b}\bigg] + \bigg[\dfrac{b - x + a}{x - a}\bigg]= 0$

can further rewritten as after taking common,

$\rm :\longmapsto\:(a + b - x)\bigg[\dfrac{1}{x - a} + \dfrac{1}{x - b} \bigg] = 0$

$\rm :\longmapsto\:(a + b - x)\bigg[\dfrac{x - b + x - a}{(x - a)(x - b)} \bigg] = 0$

$\rm :\longmapsto\:(a + b - x)\bigg[\dfrac{2x - b - a}{(x - a)(x - b)} \bigg] = 0$

$\rm :\longmapsto\:a + b - x = 0 \: \: \: or \: \: \: 2x - a - b = 0$

$\bf\implies \:x = a + b \: \: \: or \: \: \: x = \dfrac{a + b}{2}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac