Answer:
- Over the real numbers there are no numbers to add to -2 and multiply to 3
- Over the complex set of numbers, the numbers that add to -2 and multiply to 3 are
[tex]x=-1-i\sqrt{2} , y=-1+i\sqrt{2}[/tex]
[tex]x=-1+i\sqrt{2} , y=-1-i\sqrt{2}[/tex]
Step-by-step explanation:
Let [tex]x[/tex] and [tex]y[/tex] be the numbers.
We know that they add to -2, so [tex]x+y=-2[/tex] equation (1).
We know that they multiply to 3, so [tex]xy=3[/tex] equation (2).
To find our numbers, let's solve our system of equations step-by-step:
Step 1. Solve for [tex]y[/tex] in equation (1):
[tex]y=-2-x[/tex] equation (3)
Step 2. Replace equation (3) in equation (2) and solve for [tex]x[/tex]:
[tex]x(-2-x)=3[/tex]
[tex]-x^{2} -2x-3=0[/tex]
Using the quadratic formula [tex]x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex] with a=-1, b=-2, and c=-3 we get that
[tex]x=-1-i\sqrt{2}[/tex] equation (4)
[tex]x=-1+i\sqrt{2}[/tex] equation (5)
As you can see, the system of equation doesn't have a solution over the reals, but if you need a complex answer, let's keep on.
Step 4. Replace equations (4) and (5) in equation (3) and solve for [tex]y[/tex]:
- For [tex]x=-1-i\sqrt{2}[/tex]
[tex]y=-2-(-1-i\sqrt{2} )[/tex]
[tex]y=-2+1+i\sqrt{2}[/tex]
[tex]y=-1+i\sqrt{2}[/tex]
- For [tex]x=-1+i\sqrt{2}[/tex]
[tex]y=-2-(-1+i\sqrt{2} )[/tex]
[tex]y=-1-i\sqrt{2}[/tex]
The solutions of our system of equations are:
[tex]x=-1-i\sqrt{2} , y=-1+i\sqrt{2}[/tex]
[tex]x=-1+i\sqrt{2} , y=-1-i\sqrt{2}[/tex]
