Are there two integers with a product of -12 and a sum of -3?
To find two integers with a product of -12 and a sum of -3, we can start by listing the factors of 12: 1, 2, 3, 4, 6, and 12.
Since the product is negative, one of the integers must be positive, and the other must be negative. Seeing that the sum is negative, the larger number should be the negative one.
Let's list the possible pairings:
-1 * 12 = -12 (sum = 11)
-2 * 6 = -12 (sum = 4)
-3 * 4 = -12 (sum = 1)
None of these pairings have a sum of -3, so there are no two integers with a product of -12 and a sum of -3.
To find out if there are two integers that satisfy the given conditions, we can use algebraic equations. Let's assume the two integers to be x and y.
According to the given conditions, the product of the integers is -12, so we have the equation: x * y = -12.
The sum of the integers is -3, so we have the equation: x + y = -3.
To solve this system of equations, we need to find the values of x and y that satisfy both equations simultaneously.
To solve for one variable, we can isolate it in either equation. Let's isolate y in the second equation. Subtracting x from both sides, we get:
y = -3 - x.
Now we substitute this expression for y into the first equation:
x * (-3 - x) = -12.
Expanding and rearranging the equation, we have:
-3x - x^2 = -12.
Rearranging again, we get a quadratic equation:
x^2 + 3x - 12 = 0.
Factoring or using the quadratic formula, we find the solutions for x:
(x + 4)(x - 3) = 0. Therefore, x = -4 or x = 3.
Now we substitute these values of x back into either of the original equations to solve for y.
For x = -4, using the first equation, we have:
(-4) * y = -12.
Simplifying, we find y = 3.
For x = 3, using the first equation, we get:
3 * y = -12.
Simplifying, we find y = -4.
So, there are two sets of integers that satisfy the given conditions: (-4, 3) and (3, -4).
You could go through all the possibilities, after all there are only a limited number of cases, but
let one integer be x and the other one y
x+y = -3
y = -3-x
product = xy = -12
x(-3-x) = -12
-3x - x^2 + 12 = 0
x^2 + 3x - 12 = 0
x = (-3 ± √(9 - 4(1)(-12)) )/2
= (-3 ± √57)/2
which is irrational, so no two integers exist with your stated property