What 2 numbers multiplied equal 12 and added together equal 0?

Let One of the numbers be X

Then another will be 12/X

Now their Sum= 0

So X+X/12=0

X^2+X=0

X(X+1)=0

X=0 0r X=-1

So other may be 0 or --12

Let xy be the two numbers.

xy = 12

x + y = 0

Solve for y in the second equation.

y = -x

Substitute that value of y in the first equation.

x(-x) = 12

-x² = 12

Multiply both sides by -1

x² = -12

Square root both sides

x = ±√(-12)

x = ±√(12i²)

x = ±√(2² · 3 · i²)

x = ±2i√3

x = 2i√3 or x = -2i√3

Substitute that value of x in the second equation.

x + y = 0

2i√3 + y = 0

y = -2i√3

or

x + y = 0

-2i√3 + y = 0

y = 2i√3

The two numbers are 2i√3 and -2i√3.

Are you sure you didnt mean when multiplied equals 0 and adds to 12? In that case it would be 12 and 0 :)

The numbers are complex:  (-√-12)  and  (√-12)

(-√-12) • (√-12)  =  (-i√12) • (i√12)  =  -i² • (√12) • (√12)  =  (√12) • (√12)  =  12

(-√-12) + (√-12)  =  0

Since they equal zero it would have to be the same two numbers, one negative and one positive to cancel each other out... but no numbers multiply to 12

3 * -3 = -9

4 * -4 = -16

you probably factored wrong.

The only factors of 12 are:

12x1 = 12

6x2 = 12

4x3 = 12

There are no way any of these can add to 0.

let the two numbers be: x and y

then...

xy = 12

x + y = 0

solve for x: x = - y

substitute:

(- y) * (y) = 12

- y² = 12

y² = - 12

y = ± 2i√3

the two numbers are: (2i√3) and (- 2i√3)

check...do something !!

2 complex

x^2 + 12 = 0----> x = .....(2 complex nmbers)