SOLUTION: Determine the nature of the solutions of the equation. x^2 + 3 = 0 Does it have: 2 imaginary solutions 2 real solutions 1 real solutions I do not know how to determin

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Determine the nature of the solutions of the equation. x^2 + 3 = 0 Does it have: 2 imaginary solutions 2 real solutions 1 real solutions I do not know how to determin      Log On


   

Question 131774: Determine the nature of the solutions of the equation.
x^2 + 3 = 0
Does it have:
2 imaginary solutions
2 real solutions
1 real solutions
I do not know how to determine this.
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
One way is to use the quadratic formula and calculate the discriminant.

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

The discriminant is the b%5E2-4%2Aa%2Ac part under the radical.

If b%5E2-4%2Aa%2Ac%3E0, there are two real and unequal roots.

If b%5E2-4%2Aa%2Ac=0, there are two real and equal roots. (Also known as one real root with a multiplicity of two. There is never just one real solution to a quadratic, there are always two, it's just that sometimes the two are equal. Google "Fundamental Theorem of Algebra" for more information.)

If b%5E2-4%2Aa%2Ac%3C0, there is a conjugate pair of complex roots.

So how does that apply to your equation?

x%5E2+%2B+3+=+0+. There is no 'x' term in this equation. That means the b coefficent in the quadratic formula is 0. You could actually write your equation as:

x%5E2+%2B+0x+%2B+3=0

Let's look at the discriminant:

b%5E2-4%2Aa%2Ac
0%5E2-4%2A1%2A3=-12%3C0

Since the discriminant is < 0, there is a conjugate pair of complex roots of the form a%2Bbi and a-bi where i%5E2=-1. In the case of your problem, the a in a%2Bbi will turn out to be 0, and b=sqrt%283%29.