Algebra Help: I think I am correct here

Working ahead and want to impress my teacher before school :D
Decide how many solutions this problem has

x^2+3=0
Just to understand better put 0 for b: X^2+0+3
Now put into quadratic form: -0+/- sqrt -3-4(0)(3)
Simplify: 0+ sqrt -3
0- sqrt -3
Considering there can be no - in radicand, the answer to this question is 0 solutions.

x^2 = -3
x = ±i√3 => where i^2 = -1

Edit: How about the Discriminant?
The expression in the square root, D = b^2 - 4ac , is called the discriminant, because its value can be used to discriminate between the different types of solutions that are possible:
1) D > 0 results in two real solutions:
a) If D is a perfect square (like 4, 9 64, 100, etc.) then you get two rational solutions.
b) If D is a not perfect square then you get two irrational solutions.

2) If D = 0 then you get a single real (and rational since 0 is a perfect square) solution.

3) If D < 0 then you get two complex solutions. (Note: you only get imaginary solutions if the discriminant is negative and b = 0!)

so in this case:
D = 0 - 4*1*3 = -12 < 0 => No real solutions.

You are welcome, I'm glad it worked out for you and best of luck in your Math classes.

Regards.

Report Abuse

I'm not sure if you have gone over sqrt(-1) = i (aka imaginary numbers) yet, but that would give you an answer in the form of a complex number. If you haven't gone over it yet, done worry about it.

Even shorter:

x^2 = -3

x = sqrt(-3)

x = +/- i*sqrt(3)

There are two imaginary roots but no real roots.

x^2+3 = 0
x^2 = -3
x = +/- sqrt -3

No real number can fill in, so this equation has no solution.