Math, asked by krrish0324gmailcom, 1 year ago


 \frac{x - 1}{x}  +  \frac{x}{x - 1}  = 4
solve for x ; x does't =0/1

Answers

Answered by laxman10201969
0
Hello ,

Here is your answer , Mark it as the Brainliest
if it helps

 \frac{x - 1}{x}  +  \frac{x}{x - 1}  = 4 \\
The L.C.M. of x and x-1 is x^2 -x . So,
 \frac{x - 1(x - 1) + x(x - 1)}{ {x}^{2} - x }  = 4 \\
Multiply both sides by x^2 - x
(x - 1)(x - 1) +  {x}^{2}  - x = 4( {x}^{2}  - x)
Use the identity (a-b)^2
 {x}^{2}  - 2x + 1 +  {x}^{2}  - x = 4 {x}^{2}  - 4x
2 {x}^{2}   -  4 {x}^{2}   - 3x + 4x + 1 = 0 \\
 - 2 {x}^{2}  + x + 1 = 0
The formula to solve a quadratic equation
a {x}^{2}  + bx + c = 0
is
x =  \frac{ - b + or - \sqrt{ ( {b}^{2} - 4ac)}  }{2a}
So here ,
x =   \frac{ - 1 + or -  \sqrt{1  -  \: (4  \times  - 2 \times 1)} }{ - 4} \\
x =  \frac{  - 1 +  or - \sqrt{9} }{ - 4}
Now taking + sign answer is
x =  \frac{ - 1 + 3}{ - 4} \\ x =   - 0.5
Now taking the - sign answer is
x =  \frac{ - 1 -3}{4}  \\ x =  1



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