SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; jus

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; jus      Log On


   

Question 36135: use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
z^2 + z + 1 = 0
Found 2 solutions by rapaljer, stanbon:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
If b%5E2+-+4ac+%3E+0 then there will be TWO distinct real roots.
If b%5E2+-+4ac+%3E=+0 then there will be a double (real) root.
If b%5E2+-+4ac+%3C+0 then there will be NO real roots, but there will be two complex roots.

In your example, z%5E2+%2B+z%2B1=0 a=1, b= 1, and c= 1, so b%5E2+-+4ac+=+1-+4%2A1%2A1=-3. Therefore, there will be NO real roots, the roots will be complex.

R^2 at SCC

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
z^2 + z + 1 = 0
a=1, b=1, c=1
discriminant = b^2-4ac
= 1-4=-3
Since the discrinant is negative
the equation has no Real Number zeroes.
It has two complex roots.
Cheers,
Stan H.