College Algebra

Question

1 a) Solve: the square root of x + 2 - x = 2 (only the x + 2 is under the square root)

 

b) Solve:       10       =       1       

                 4x - 4             1 - x

 

c)  Solve: 10x^-2 + 33^-1 -7 = 0

 

 

Answer 1

b). 10/(4x-4) = 1/(1-x)

 

     10(1-x) = (1)(4x-4)

 

      10 - 10x = 4x - 4

 

          -14x = -14

 

              x = 1

 

     However, 1 does not check in the original equation.

 

     So, there is NO SOLUTION.  

 

c).  10(x^-1)² + 33(x^-1) - 7 = 0

 

      Let u = x^-1.  Then, 10u²+33u-7=0

 

                                 (5u-1)(2u+7) = 0

 

                                    u = 1/5 or u = -7/2

 

      But, u = x^-1 = 1/x

 

      So, 1/x = 1/5 or -7/2

 

      Therefore, x = 5 or -2/7

 

 

Answer 2

1)  √ (X + 2) -X = 2

   

         √ (X + 2) = X + 2

 

    Square both sides

 

       X + 2  = (X + 2)²

 

        X + 2 = X² + 4X + 4

 

              0 = X² + 3X + 2

 

              0 = (X + 2)(X + 1)

 

    X + 2 = 0   ;   X + 1 = 0

 

       X = -2    ;    X = -1