Question

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Answer 1

Answer:

answer is -12/13 is correct answer

Answer 2

Question :- Solve for x :-

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

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

Given equation is

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

can be rewritten as on taking LCM, we get

$\rm \: \dfrac{ {x}^{2} + {(x + 1)}^{2} }{(x + 1)x} = \dfrac{25}{12}$

$\rm \: \dfrac{ {x}^{2}+ {x}^{2} + 1 + 2x}{ {x}^{2} + x} = \dfrac{25}{12}$

$\rm \: \dfrac{ 2{x}^{2} + 1 + 2x}{ {x}^{2} + x} = \dfrac{25}{12}$

$\rm \: 25( {x}^{2} + x) = 12( {2x}^{2} + 1 + 2x)$

$\rm \: 25{x}^{2} + 25x = {24x}^{2} + 12 + 24x$

$\rm \: 25{x}^{2} + 25x - {24x}^{2} - 12 - 24x = 0$

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

On splitting the middle terms, we get

$\rm \: {x}^{2} + 4x - 3x - 12 = 0$

$\rm \: x(x + 4) - 3(x + 4) = 0$

$\rm \: (x + 4)(x - 3) = 0$

$\rm \: x + 4 = 0 \: \: \: or \: \: \: x - 3 = 0$

$\rm\implies \:x = - 4 \: \: \: or \: \: \: x = 3$

$\rule{190pt}{2pt}$

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