The sum of the squares of two consecutive positive even integers is one hundred sixty-four. Find the two integers.
n^2 + (n+2)^2 = 164
2n^2 + 4n - 160 = 0
n^2 + 2n - 80 = 0
(n+10)(n-8) = 0
...
x 1st + even integer
x+2 2nd + even integer
x^2+(x+2)^2= 164
x^2+x^2+4x+4-164=0
1/2(2x^2+4x-160=0)
x^2+2x-80=0
(x+10)(x-8)=0
x=-10 & x=8, disregard -10 since we are looking for positive integers...
so, x=8 (1st even)
x+2= 10 (2nd even)
Let's assume the first positive even integer is x.
According to the problem, the next consecutive positive even integer would be (x + 2).
The sum of their squares is given as 164.
So, we set up the equation:
x^2 + (x + 2)^2 = 164.
Expanding the equation:
x^2 + (x^2 + 4x + 4) = 164.
Combining like terms:
2x^2 + 4x + 4 = 164.
Subtracting 164 from both sides:
2x^2 + 4x - 160 = 0.
Dividing the entire equation by 2:
x^2 + 2x - 80 = 0.
Let's factorize the quadratic equation:
(x + 10)(x - 8) = 0.
Setting each factor to zero:
x + 10 = 0, or x - 8 = 0.
Solving for x:
x = -10 or x = 8.
Since we are looking for positive even integers, x = 8.
Therefore, the two consecutive positive even integers are 8 and 10.
To find the two consecutive positive even integers, we can set up an equation based on the given information.
Let's assume that the first even integer is represented by the variable x. Since the two integers are consecutive, the second even integer can be represented by x + 2.
According to the problem, the sum of the squares of these two consecutive even integers is 164. Therefore, we have the equation:
x^2 + (x + 2)^2 = 164
To solve this equation, we can follow these steps:
1. Expand the equation:
x^2 + (x + 2)^2 = 164
x^2 + (x^2 + 4x + 4) = 164
2x^2 + 4x + 4 = 164
2. Subtract 164 from both sides of the equation to set it equal to zero:
2x^2 + 4x + 4 - 164 = 0
2x^2 + 4x - 160 = 0
3. Divide the equation by 2 to simplify it:
x^2 + 2x - 80 = 0
4. Factor the quadratic equation:
(x + 10)(x - 8) = 0
5. Solve for x by setting each factor equal to zero:
x + 10 = 0 --> x = -10 (not a positive even integer)
x - 8 = 0 --> x = 8
Since we are looking for positive even integers, we can discard the solution x = -10.
Therefore, the first even integer is x = 8. The second even integer, x + 2, is 8 + 2 = 10.
So, the two consecutive positive even integers are 8 and 10.