solve the equation

Question

(1/x-2)²+(2/x-2)-15=0

solve the above equation

Answer 1

Always get the unknowns out of the denominator.  We can do this by using logs or expanding and simplifying. This solution presumes (1/x    -     2)^2  +  (2/x   -   2) - 15 = 0.  If I'm wrong, please learn a valuable lesson regarding parenthesis.  (1/x-2)^2 is ENTIRELY different from 1/(x-2)^2.   Additionally, (2/x    -   2) is entirely different from 2/(x - 2).  I presume you meant what you typed, but I'd bet a ton of money that you made some parenthetical errors.  Let my exercise NOT BE IN VAIN!!!  My solution presumes you meant what you typed  ;-)

 

Parenthesis are EVERYTHING!!!!

 

Lets expand and simplify:

 

X^-2 -4X^-1 + 4 + 2X^-1 -2 -15 = 0

 

Collect like terms:

 

x^-2 -2X^-1 - 13 = 0

 

Multiply by x^2:

 

1 -2X - 13X^2 = 0      (now rearrange and change sign) 

 

13X^2 + 2X - 1 = 0      (now place in standard form)

 

x^2 +2/13X - 1/13 = 0   (Now solve via quadratic eqtn: [-b± sqrt(b^2-4ac) ]/(2a)

 

x= [-2/13 ± sqrt (4/169 + 4/13) ]/2

 

x= -1/13 ± sqrt(4/169 + 52/169)/2

x= -1/13 ± sqrt(56/169)/2

x= -1/13 ± 0.29792  (approx) = (.2109, -0.36474)

 

 

Answer 2

      ^{1  )} -----  +     2          - 15 =0

(  X -2 )^2        ( X-2)

 

   1        = Y

  X -2

 

  then :  Y^2 + 2Y - 15 =0

 

            ( Y + 5 ) ( Y -3) =0

 

          Y = -5  ,   Y =3

 

        1     = -5     -5X +10 =1  X = 9/5

        X -2 

 

         1        = 3     3X -6 = 1    X = 7/3

         X - 2             

Answer 3

(1/(x-2))^2+(2/(x-2))-15=0

 

Let u = 1/(x-2), then:

 

u^2 + 2u - 15 =0

 

(u + 5)(u - 3) = 0

 

Using zero product property:

 

u + 5 = 0 or u - 3 = 0

 

1/(x-2) = -5 or 1/(x-2) = 3

 

1 = -5x + 10 or 1 = 3x - 6

 

5x = 9 or 3x = 7

 

x = 9/5 or x = 7/3