What has a product of 16 and sum of -2?

Answer 1

Using two numbers, there is no answer. And for those that tried with three numbers, I'm sorry but 2 * -2 * -2 = 8, not 16...

Answer 2

xy = 16
x + y = -2

Solve the first equation for either x or y:

y = 16/x

Plug in the second equation:

x + 16/x = -2

Multiply through by x:

x^2 + 16 = -2x
x^2 - 2x + 16 = 0

Unfortunately, it looks like there's no answer that doesn't require complex numbers, because the discriminant:

(-2)^2 - 4(1)(16) = 4 - 64 = -60

is negative.

Are you sure you posted the problem correctly?

Hmm. I guess Oddy's answer works. I just assumed that you were only allowed to use 2 numbers instead of 3.

Answer 3

n² + 2n + 16 = 0

Unfortunately to say, just because there are no solutions in the real number field, there are no solutions at all is a cop out. There was no condition placed on this question seeking 2 real numbers. In fact it is sad that they are called real numbers for complex numbers are very real as well. It is just that they have far different properties to the numbers we call real numbers.

There is a solution!

The solution is:
n1 = -1 + i√15 and n2 = -1 - i√15 where i =√(-1)

n1 + n2 = -2

n1 * n2 = (-1 + i√15)(-1 - i√15)
= (-1)² - (i√15)²
= 1 - - 15
= 16

So these two numbers most certainly fulfill the required conditions

To deny the existence of complex numbes is like the ancient mathematicians who denied the existence of irrational numbers and the mediaeval mathematicians who denied the existence of negative numbers and the 19th and early 20th century mathematicians who denied the existence of countable and uncountable infinities!

Answer 4

As long as you allow more than 2 numbers, there is an infinite number of solutions to this problem (intuitively because you have more degrees of freedom than constraints).
A solution with 3 numbers: 4, -3+sqrt(5), -3-sqrt(5). In this example, I've fixed 4 and found the two other numbers such that their sum is -6 and thir product is 4: this is done by finding the roots of polynomial of degree 2.
Another one: -2, -sqrt(8), +sqrt(8)

Answer 5

Let x be the "what"...
So x*x=16 or
x^2=16 (eq1)
and x+x=-2 or
2x=-2 (eq2)
Solve eq2.
2x=-2
x=-1.
Solve eq1.
x^2=16
x= 4, or -4, none of these match, therefore I would say there isn't a real number answer.
Using imaginary numbers? Only if they let you use two different complex numbers that are conjugates... and are technically not the exact same "what". Hope that helps. Good luck!

Answer 6

The numbers are: 2, 2, -2, -2, -1, -1

Product = 2 * 2 * -2 * -2 * -1 * -1 = 16
Sum = 2+2-2-2-1-1 = -2

Answer 7

2, -2, -2

2+ (-2) + (-2) = -2

2(-2)(-2) = 16

Answer 8

xy=16
x+y=-2
x=-y-2
(-y-2)y=16
-y^2-2y-16=0
y^2+2y+16=0 determinent is 4-64=-60<0
There is no solution.

Answer 9

2,2,-2,-2,-1,-1