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
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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(648) (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 .
In your equation, a=1, b=3, and c=7. Evaluate using these values for a, b, and c. So,
Important Math Fact:
If (negative), then there are two imaginary-number solutions.
If (positive), then there are two real-number solutions.
If , 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 .
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.
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