SOLUTION: Find the number of real -number solutions of each equation 1. x^2+8=0 2. 3x^2-9x=-5 3. x^2+4x=-4

(1)  x^2 + 8 = 0.

     This equation has no real solutions, 

     because its left side is always positive (strictly greater than zero,
     therefore, it can not be equal to zero.

     The number of real solutions is zero.



(2)  3x^3 - 9x = -5.

     Write this quadratic equation in the standard form

         3x^2 - 9x + 6 = 0.

     
     Use the discriminant of this equation

         d = b^2 - 4ac = (-9)^2 - 4*3*6 = 81 - 72 = 9.

     The discriminant is positive number - hence, this equation has two real solutions.



(3)  x^2 + 4x = -4.

     Write this quadratic equation in the standard form

         x^2 + 4x + 4 = 0.


     Left side is a perfect square  (x+2)^2.

     So, your equation has the form

         (x+2)^2 = 0.


     It has one root x = -2.

     The number of solutions is 1.