SOLUTION: I wasn't sure where to put this. Consider only the discriminant, b^2-4ac, to determine whether one real-number solutions, or two different imaginary-number solutions exist. x

Algebra ->  Algebra  -> Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> SOLUTION: I wasn't sure where to put this. Consider only the discriminant, b^2-4ac, to determine whether one real-number solutions, or two different imaginary-number solutions exist. x      Log On

Ad: You enter your algebra equation or inequality - Algebrator solves it step-by-step while providing clear explanations. Free on-line demo .
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!
Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

   

Question 458756: I wasn't sure where to put this.
Consider only the discriminant, b^2-4ac, to determine whether one real-number solutions, or two different imaginary-number solutions exist.
x^2+3x+7=0
thanks for the help.
Answer by math-vortex(472) About Me  (Show Source):
You can put this solution on YOUR website!
Begin by determining a, b, and c in your equation. The standard form for a quadratic equation is ax%5E2%2Bbx%2Bc=0.
In your equation, a=1, b=3, and c=7. Evaluate b%5E2-4ac using these values for a, b, and c. So, b%5E2-4ac=%283%29%5E2-4%281%29%287%29=9-28=-18
Important Math Fact:
If b%5E2-4ac%3C0 (negative), then there are two imaginary-number solutions.
If b%5E2-4ac%3E0 (positive), then there are two real-number solutions.
If b%5E2-4ac=0, then there is one real-number solution.
Your discriminant is negative, so you have two non-real (imaginary number) solutions.
*******
NOTE:
If you forget this math fact, think about the quadratic formula x=%28b%2B-+sqrt%28b%5E2-4ac%29%29%2F2a.
If the value under the square root is negative, then the solution cannot be a real number. If the value under the square root is positive, "+" or "-" after "-b" gives us two real number solutions. If the number under the square root is zero, then we just get one real-number solution, because the square root of zero is zero.