Question

x /x-1 + x -1/x = 2½
simplfy

please tell the solution​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2}, \: \: \: \: \: x \ne \: 0,1$

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

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 2\dfrac{1}{2}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {(x - 1)}^{2} }{x(x - 1)} = \dfrac{5}{2}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + {x }^{2} + {1}^{2} - 2x}{x(x - 1)} = \dfrac{5}{2}$

$\: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

$\rm :\longmapsto\:\dfrac{ {2x}^{2} - 2x + 1 }{ {x}^{2} - x} = \dfrac{5}{2}$

$\rm :\longmapsto\: {4x}^{2} - 4x + 2 = {5x}^{2} - 5x$

$\rm :\longmapsto\: {x}^{2} - x - 2 = 0$

$\rm :\longmapsto\: {x}^{2} - 2x + x - 2 = 0$

$\rm :\longmapsto\:x(x - 2) + 1(x - 2) = 0$

$\rm :\longmapsto\:(x - 2)(x + 1) = 0$

$\bf\implies \:x = 2 \: \: \: or \: \: - \: 1$

Verification :-

When x = 2

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x - 1}{x}$

$\rm :\longmapsto\: = \: \dfrac{ 2}{ 2 - 1} + \dfrac{ 2 - 1}{ 2}$

$\rm :\longmapsto\: = \: 2 + \dfrac{ 1}{ 2}$

$\rm :\longmapsto\: = \: \dfrac{5}{2}$

Hence, x = 2 is solution of given equation.

When x = - 1

$\rm :\longmapsto\:\dfrac{x}{x - 1} + \dfrac{x - 1}{x}$

$\rm :\longmapsto\: = \: \dfrac{ - 1}{ - 1 - 1} + \dfrac{ - 1 - 1}{ - 1}$

$\rm :\longmapsto\: = \: \dfrac{1}{2} + 2$

$\rm :\longmapsto\: = \: \dfrac{5}{2}$

Hence, x = - 1 is a solution of given equation.

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