the square of the sum of two consecutive positive even integer is greater than the sum of their square by 48. Find two integers.
x and x+2
x^2 + (x+2)^2 = (2x+2)^2 - 48
x^2 + x^2 + 4 x + 4 = 4 x^2 + 8 x + 4 - 48
2 x^2 + 4 x - 48 = 0
x^2 + 2 x - 24 = 0
(x -4)(x+6) = 0
x = 4 and x+2 = 6
Let's assume the two consecutive positive even integers as x and x + 2.
The square of the sum of these two integers can be represented as (x + (x + 2))^2.
The sum of their squares can be represented as x^2 + (x + 2)^2.
According to the given information, we can write the equation as:
(x + (x + 2))^2 > x^2 + (x + 2)^2 + 48.
Expanding and simplifying the equation, we get:
4x^2 + 12x + 4 > x^2 + 4x^2 + 4 + 4x + 48.
Combining like terms and simplifying the equation further, we have:
4x^2 + 12x + 4 > 5x^2 + 4x^2 + 4 + 4x + 48.
4x^2 + 12x + 4 > 9x^2 + 4x + 52.
Moving all the terms to the left side, we get:
5x^2 - 8x - 48 > 0.
Solving this quadratic inequality, we find that x < -4 or x > 4/5.
Since the given numbers are consecutive positive even integers, x must be positive. Therefore, we can consider x > 4/5.
Let's choose x = 2.
Substituting the value of x in the equation, we get:
(2 + (2 + 2))^2 > 2^2 + (2 + 2)^2 + 48.
(4)^2 > 4 + 4^2 + 48.
16 > 4 + 16 + 48.
This inequality is true, so we have found our solution.
The two integers are 2 and 4.
To solve this problem, let's represent the two consecutive positive even integers. Let the first even integer be "x" and the second even integer be "x + 2" (since they are consecutive).
The square of the sum of these two numbers is (x + x + 2)^2 = (2x + 2)^2.
The sum of their squares is x^2 + (x + 2)^2.
According to the problem, the square of the sum of the two consecutive even integers is greater than the sum of their squares by 48. So we can write the equation as:
(2x + 2)^2 = x^2 + (x + 2)^2 + 48.
Expanding the equation, we have:
4x^2 + 8x + 4 = x^2 + x^2 + 4x + 4 + 48.
Simplifying the equation further:
4x^2 + 8x + 4 = 2x^2 + 4x + 52.
Rearranging the equation by subtracting 2x^2, 4x, and 56 from both sides:
2x^2 + 4 = 0.
Dividing both sides by 2:
x^2 + 2 = 0.
Subtracting 2 from both sides:
x^2 = -2.
This equation has no real solutions since the square of any real number cannot be negative.
Therefore, there are no two consecutive positive even integers that satisfy the given condition.