Question

$\frac{x}{x + 1} + \frac{x + 1}{x} = 2 \times \frac{1}{2}$
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Answer 1

GIVEN :-

$\\ \sf \: \dfrac{x}{x + 1} + \dfrac{x + 1}{x} = 2 \times \dfrac{1}{2}$

TO FIND :-

• Value of x.

SOLUTION :-

$\\ \sf \: \dfrac{x}{x + 1} + \dfrac{x + 1}{x} = \cancel2 \times \dfrac{1}{ \cancel2}$

Cross multiplying,

$\\ \sf \: \dfrac{ {x}^{2} + ( {x + 1)}^{2} }{x(x + 1)} = 1 \\ \\ \\ \sf \: {x}^{2} + ( {x + 1)}^{2} = x(x + 1) \\ \\ \\ \sf \: {x}^{2} + {x}^{2} + 2x + 1 = {x}^{2} + x \\ \\ \\ \sf \: 2 {x}^{2} + 2x + 1 = {x}^{2} + x \\ \\ \\ \sf \: 2 {x}^{2} - {x}^{2} + 2x - x + 1 = 0 \\ \\ \\ \underline{ \sf \: {x}^{2} + x + 1 = 0}$

Solving by Quadratic Formula ,

$\\ \boxed{ \sf \: roots = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}}$

Here ,

• a → Coefficient of x².
• b → Coefficient of x.
• c → Constant.

We have ,

• a = 1
• b = 1
• c = 1

$\\ \sf \: roots = \dfrac{ - 1 \pm \sqrt{ {1}^{2} - 4(1)(1)} }{2(1)} \\ \\ \\ \sf \: roots = \dfrac{ - 1 \pm \sqrt{1 - 4} }{2} \\ \\ \\ \sf \: roots = \dfrac{1 \pm \sqrt{3} }{2} \\ \\ \\ \boxed{\sf \: roots = \dfrac{1 + \sqrt{3} }{2} \: , \: \dfrac{1 - \sqrt{3} }{2} }$

Hence ,

$\\ \underline{\boxed{\sf \: x = \dfrac{1 + \sqrt{3} }{2} \: , \: \dfrac{1 - \sqrt{3} }{2} }}$