How to find quadratic equation when the complex roots are given?

Question

For example, $x= -4+0.4i$, $x= -4 - 0.4i$

How do I find remove the complex $i$ and find the quadratic formula ?

Someone told me that the formula is $x^2$ - (sum of roots)$x$ + (product of roots) = $0$

But this formula does not resonate with me well as it is out of the blue and I don't understand it

Answer 1

Remember how you solve a quadratic equation by factoring?

$$(x-r_1)(x-r_2) = 0 $$ $$x -r_1 = 0 \;\;\;\;\;\textrm{ or }\;\;\;\;\;x-r_2 = 0$$ $$x = r_1 \;\;\;\;\;\textrm{ or }\;\;\;\;\;x = r_2$$

So just reverse this process. Simplify the product on the first line if you want to, which then becomes

$$x^2 - r_1x -r_2x + r_1r_2 = 0$$ $$x^2 - (r_1 + r_2)x + r_1r_2 = 0$$

Answer 2

The formula your friend gave you is a special case of Vieta's formulas. So, using this formula, we find that the quadratic equation is $x^2-(-4+0.4i+-4-0.4i)x+(-4+0.4i)(-4-0.4i)=x^2+8x+16.16$. Multiply this equation by a number $a$ to get all the possible parabolas with these two roots.