SOLUTION: Find the real solutions of the equations. A. x^2 - 3x - sqrt{x^2 - 3x} = 2 B..3x^(4/3) + 5x^(2/3) = 0

    (A)  x^2 - 3x - sqrt{x^2 - 3x} = 2.



The domain of equation (A) is the set of real numbers x such that

    x^2 - 3x >= 0,  x*(x-3) >= 0,   x <= 0  OR  x >= 3.    (1)


Introduce new variable t = .


Then equation (A)  takes the form

    t^2 - t = 2.    (2)


We consider  as a non-negative value; 
so, we look for the solution of equation (2) with  t >= 0.


Write equation (2) in the standard form quadratic equation

    t^2 - t - 2 = 0.


Factor left side

    (t-2)*(t+1) = 0.


Two roots are  t= 2  and  t= -1.

We are looking for non-negative t, so we ignore  t= -1.


Now consider t= 2


It means   = 2,  x^2 - 3x = 2^2 ,  x^2 - 3x = 4,  

                           x^2 - 3x - 4 = 0,  (x-4)*(x+1) = 0,  x= 4  or  x= -1.

          These potential solutions are in the domain of equation (A), 
          so they are valid solutions to equation (A).


ANSWER.  Equation (A)  has two solutions:  x= -1  and  x= 4.


You may check that the answer is correct, by substituting these values into equation (A).