Question

solve a/x-a+b/x-b=2c/x-C​

Answer 1

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

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

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

$\rm :\longmapsto\:\dfrac{ax - ab + bx - ab}{(x - a)(x - b)} = \dfrac{2c}{x - c}$

$\rm :\longmapsto\:\dfrac{ax + bx - 2ab}{ {x}^{2} - ax - bx + ab } = \dfrac{2c}{x - c}$

$\rm :\longmapsto\:(x - c)(ax + bx - 2ab) = 2c( {x}^{2} - ax - bx + ab)$

$\rm \: {ax}^{2}+{bx}^{2} -2abx-cax-cbx+ 2abc={2cx}^{2}-2acx-2bcx+2abc$

$\rm \: {ax}^{2}+{bx}^{2} -2abx-cax-cbx={2cx}^{2}-2acx-2bcx$

$\rm \: {ax}^{2}+{bx}^{2} -2abx-cax-cbx - {2cx}^{2} + 2acx + 2bcx = 0$

$\rm \: {ax}^{2}+{bx}^{2} -2abx + cax + cbx - {2cx}^{2}= 0$

$\rm \: {(a + b - 2c)x}^{2} + x(-2ab + ca + cb)= 0$

$\rm \:x \bigg((a + b - 2c)x -2ab + ca + cb\bigg)= 0$

$\rm :\implies\:x = 0 \: \: \: or \: x = \dfrac{2ab - ac - bc}{a + b - 2c}$

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