For two consecutive positive even integers, the product of the smaller and twice the larger is 160. Find the integers.
the integers are 2k and 2k+2
(2k)*2(2k+2) = 160
k(k+1) = 20
20=4*5, so k=4
The numbers are 8 and 10
To solve this problem, let's represent the smaller even integer as x, and the larger even integer as (x + 2), where we add 2 to get the next consecutive even integer.
According to the problem, the product of the smaller integer and twice the larger integer is 160. Mathematically, we can express this as:
x * 2(x + 2) = 160
Now let's solve this equation step by step:
1. Distribute 2 to (x + 2):
2x(x + 2) = 160
2. Expand the expression:
2x^2 + 4x = 160
3. Move 160 to the other side of the equation:
2x^2 + 4x - 160 = 0
4. Simplify the equation by dividing every term by 2:
x^2 + 2x - 80 = 0
5. To factorize the quadratic equation, we need two numbers whose product is -80 and sum is 2. After some trial and error, we find that the numbers are 10 and -8.
So, the factored equation becomes:
(x + 10)(x - 8) = 0
6. Apply the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero:
x + 10 = 0 or x - 8 = 0
7. Solve each equation separately:
For x + 10 = 0:
x = -10
For x - 8 = 0:
x = 8
Since we are looking for positive even integers, we exclude the negative value x = -10.
Therefore, the smaller even integer x = 8, and the larger even integer is x + 2 = 8 + 2 = 10.
Hence, the two consecutive positive even integers are 8 and 10.