Question

x/x+1+x+1/x= 2(1/6) Answer this question properly then I will mark u as a brainlist.

Answer 1

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

Given equation is

$\rm :\longmapsto\:\dfrac{x + 1}{x} + \dfrac{x}{x + 1} = 2\dfrac{1}{6}$

$\rm :\longmapsto\:\dfrac{ {(x + 1)}^{2} + {x}^{2} }{x(x + 1)} = \dfrac{12 + 1}{6}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1 + 2x + {x}^{2} }{ {x}^{2} + x} = \dfrac{12 + 1}{6}$

$\rm :\longmapsto\:\dfrac{ 2{x}^{2} + 1 + 2x}{ {x}^{2} + x} = \dfrac{13}{6}$

$\rm :\longmapsto\:6( {2x}^{2} + 2x + 1) = 13( {x}^{2} + x)$

$\rm :\longmapsto\:{12x}^{2} + 12x + 6 = 13{x}^{2} + 13x$

$\rm :\longmapsto\: 13{x}^{2} + 13x - 12 {x}^{2} - 12x - 6 = 0$

$\rm :\longmapsto\: {x}^{2} + x - 6 = 0$

$\rm :\longmapsto\: {x}^{2} + 3x - 2x - 6 = 0$

$\rm :\longmapsto\:x(x + 3) - 2(x + 3) = 0$

$\rm :\longmapsto\:(x + 3)(x - 2) = 0$

$\rm\implies \:x = - 3 \: \: or \: \: x = 2$

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MORE TO KNOW:-

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

Answer 2

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

Given equation is

$\rm :\longmapsto\:\dfrac{x + 1}{x} + \dfrac{x}{x + 1} = 2\dfrac{1}{6}$

$\rm :\longmapsto\:\dfrac{ {(x + 1)}^{2} + {x}^{2} }{x(x + 1)} = \dfrac{12 + 1}{6}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1 + 2x + {x}^{2} }{ {x}^{2} + x} = \dfrac{12 + 1}{6}$

$\rm :\longmapsto\:\dfrac{ 2{x}^{2} + 1 + 2x}{ {x}^{2} + x} = \dfrac{13}{6}$

$\rm :\longmapsto\:6( {2x}^{2} + 2x + 1) = 13( {x}^{2} + x)$

$\rm :\longmapsto\:{12x}^{2} + 12x + 6 = 13{x}^{2} + 13x$

$\rm :\longmapsto\: 13{x}^{2} + 13x - 12 {x}^{2} - 12x - 6 = 0$

$\rm :\longmapsto\: {x}^{2} + x - 6 = 0$

$\rm :\longmapsto\: {x}^{2} + 3x - 2x - 6 = 0$

$\rm :\longmapsto\:x(x + 3) - 2(x + 3) = 0$

$\rm :\longmapsto\:(x + 3)(x - 2) = 0$

$\rm\implies \:x = - 3 \: \: or \: \: x = 2$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW:-

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