Math, asked by mohinidas715, 3 months ago

solve this equation
x/x-1 + x-1/x =5/2​

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Answers

Answered by djs974897
1

Answer:

1. x =(5-√41)/4=-0.351

2. x =(5+√41)/4= 2.851

Step-by-step explanation:

Answered by Flaunt
27

\sf\huge\bold{\underline{\underline{{Solution}}}}

\sf \longmapsto \dfrac{x}{x - 1}  +  \dfrac{x - 1}{x}  =  \dfrac{5}{2}

Taking LCM

\sf \longmapsto \dfrac{ {x}^{2}  +  {(x - 1)}^{2} }{x(x - 1)}  =  \dfrac{5}{2}

Identity used here :

(a-b)²=a²+b²-2ab

\sf \longmapsto \dfrac{ {x}^{2}  +  {x}^{2} + 1 - 2x }{ {x}^{2} - x }  =  \dfrac{5}{2}

\sf \longmapsto \dfrac{2 {x}^{2}  - 2x + 1}{ {x}^{2}  - x}  =  \dfrac{5}{2}

Now,cross multiply to both sides:

\sf \longmapsto2(2 {x}^{2}  - 2x + 1) = 5( {x}^{2}  - x)

\sf \longmapsto4 {x}^{2}  - 4x + 2 = 5 {x}^{2}  - 5x

\sf \longmapsto5 {x}^{2}  - 4 {x}^{2}  - 5x + 4x = 2

\sf \longmapsto {x}^{2}  - x = 2

\sf \longmapsto {x}^{2}  - x - 2 = 0

Now,it is in the form of a quadratic equation so, simply factorise it.

\sf \longmapsto {x}^{2}  - 2x + x - 2 = 0

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

\sf \longmapsto(x  + 1)(x - 2) = 0

 \sf \bold{x =  - 1,x = 2 }

Check :

At x= -1

\sf \longmapsto \dfrac{ - 1}{ - 2} ( + ) \dfrac{ - 1 - 1}{ - 1}

\sf \longmapsto \dfrac{1}{2}  + 2

\sf \longmapsto \dfrac{1 + 4}{2} =  \dfrac{5}{2}

LHS=RHS (Verified)

At x=2

\sf \longmapsto \dfrac{2}{2 - 1}  +  \dfrac{2 - 1}{2}

\sf \longmapsto2 +  \dfrac{1}{2}

\sf \longmapsto \dfrac{4 + 1}{2}  =  \dfrac{5}{2}

LHS=RHS(Verified)

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