Math, asked by teresanag123, 1 month ago


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

Answers

Answered by Anonymous
5

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} }}

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