To solve:
x+y = 4
x*y = 13
x=4-y, so
(4-y)*y = 13 iff
y^2 - 4y + 13 = 0 iff
(y-2)^2 = -9 iff
y-2 = +- 3i
y = 2 +- 3i
e.g., x = 2+3i, y = 2-3i
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.
6 and 13
They are 9 and 4
6 and 7
1, 3 The above solution is wrong! 1+3=4 ok. But what about product? 1x3 = 3 and not 13. So the answer is wrong. The right answer is 2+3i and 2-3i. Of course the numbers are not real numbers but they are complex.
If the sum of 2 numbers is 25 and their product is 156, the 2 numbers would be 12 and 13.
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.
Take 13 and 17 as an example. Sum is 13 + 17 ie 30, product is 13 x 17 = 221
9 and 13
The numbers are: 16 and -3
The numbers are four and nine.
The two numbers are 13 and 18
13 and 7
six, seven 7 + 6 = 13 (sum) 7 - 6 = 1 (difference) 7 x 6 = 42 (product)
6 and 7
6 and 13
12 and 13.