Multiplying Complex Numbers: Understanding the Two Methods

[Square1]

Ok there is no way I am writing out all the work of this question using a keyboard, and my scanner chose today not to work ( yes, it chose to be an idiot and not work *VERY* grumpy face) so I can't upload a picture of my work. If I were to type out the following it think it would be very difficult to read SOOO I will try to make a description as friendly as possible.

I have the equation x^2 + 3x + 7
It has complex roots x = -2/3 + i√19 and x = -2/3 - i√19

I try evaluating these roots by plugging them back in. I try it two ways, and I am only trying out right now the positive root. I am having a problem already at the x^2 term.

THE PROBLEM:
To multiply the positive root by itself, I first try the "from the ground up" method of just distributing, and eventually making a substitution of -1 for i^2. After making that substitution, the point is that I end up subtracting the term that had the i^2, from 9/4. After simplifying, I have real component -10/4 and complex component (-6i√19)/4

The second way I try to evaluate x^2 term is by using the multiplication definition of complex numbers (a + bi)(c + di) = (ac - bd) + (bc + ad)i . This results in me ADDING the same kind of complex part to the same real part...and

 

[willem2]

Both methods should work, everything you've computed so far is correct. If you add (3x+7) to what you've computed for x^2 you get 0.

 

[Square1]

Sorry i misclicked and posted before I was finished. The new thread is right above.

 

[haruspex]

I have the equation x^2 + 3x + 7
It has complex roots x = -2/3 + i√19 and x = -2/3 - i√19

Try -3/2 ± i√(19)/2

 

[pierce15]

Why is this in number theory? By the way, like haruspex noticed, you are working with the wrong roots, by using the quadratic formula you should have got
$$\frac{-3}{2}\pm\frac{i\sqrt{19}}{2}$$

P.S. just practicing typing in latex