Determine the nature of the roots of the following quadratic equation. x² + 4x +4 = 0 anyone know how to answer this?

Question

anonymous · Answer

Ok, you know the quadratic formula?

cucacula · Answer

what is the Quadratic formula? @freewilly922

anonymous · Answer

there are two basic natures of roots they are probably looking for. Real and imaginary roots. Imaginary roots result from a square root of a negative number somewhere. 
$$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

anonymous · Answer

a good word to google is "discriminant"

anonymous · Answer

So the quadratic formula will give you the roots for any quadratic equation... the only way to get a complex root is for the square root part to be taking the square root of a negative number.

cucacula · Answer

The answer is 2 ?? @freewilly922

anonymous · Answer

That would be the number of roots, but that's not what was asked. The question asked about the nature of the roots. Whether they are REAL or COMPLEX I assume.

anonymous · Answer

So Real roots would have $$\sqrt{\text{positive#}}$$ and Complex Roots would have 
$$\sqrt{\text{negative#}}$$

anonymous · Answer

For example if the quadratic formula gave you 
$$\frac{-(9)\pm\sqrt{(9)^2-4(10)(10)}}{(2)(10)}$$
You would end up with the square root of -319 in the fraction and that would give you complex roots. 

$$\frac{-(9)\pm\sqrt{(9)^2-4(10)(1)}}{(2)(10)}$$
on the other hand would give you the square root of 41 and that would leave you with real roots.

As was mentioned earlier, $$$b^2-4ac$$ is generally called the disciminant and when it is negative, you have complex roots and when positive you have Real roots.

cucacula · Answer

so (9)² - 4(10)(1) = 41 which b² - 4ac > 0 right? @freewilly922

anonymous · Answer

Yes for my example. That is correct:)  so that one has two -real- roots.

anonymous · Answer

Can you now do this with the problem they gave you?

anonymous · Answer

$$b^2 -4ac > 0: \text{2 real roots}$$
$$b^2 -4ac < 0: \text{2 complex roots}$$
$$b^2 -4ac = 0: \text{1 real root*}$$
* I think there is a special name for this root. Like it is degree 2 or something.

cucacula · Answer

so the asnwer is real and equal :) Thanks @freewilly922