SOLUTION: I have to evaluate the discriminate of each equation. Tell how many solutions each equation has and whether the solutions are real or imaginary. 2x^2 +x +28 = 0, and 6x^2 - 2x

Algebra ->  Rational-functions -> SOLUTION: I have to evaluate the discriminate of each equation. Tell how many solutions each equation has and whether the solutions are real or imaginary. 2x^2 +x +28 = 0, and 6x^2 - 2x       Log On


   

Question 458533: I have to evaluate the discriminate of each equation. Tell how many solutions each equation has and whether the solutions are real or imaginary.
2x^2 +x +28 = 0, and 6x^2 - 2x + 5 = 0
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Rule:

Ax² + Bx + C = 0

1. Substitute A, B, and C in the discriminant formula

       Discriminant = B² - 4AC
2. If the discriminant is a positive number then the equation
   has two real solutions.

3. If the discriminant is a negative number then the equation
   has two conjugate imaginary solutions.
  
4. If the discriminant is zero then the equation
   has one real solution.

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2x² + x + 28 = 0

A = 2, B = 1, C = 28

Substitute A=2, B=1, and C=28 in the discriminant formula

       Discriminant = B² - 4AC
       Discriminant = 1² - 4(2)(28)
       Discriminant = 1 - 224
       Discriminant = -223

-223 is a negative number, so the equation has two 
conjugate imaginary solutions.

-----------------------------

6x² - 2x + 5 = 0

A = 6, B = -2, C = 5

Substitute A=6, B=-2, and C=5 in the discriminant formula

       Discriminant = B² - 4AC
       Discriminant = (-2)² - 4(6)(5)
       Discriminant = 4 - 120
       Discriminant = -116

-116 is a negative number, so the equation has two 
conjugate imaginary solutions.

Edwin