One positive integer is 2 less than twice another. The sum of their squares is 745. Find the integers
Let's say one of the positive integers is x. The other positive integer would be 2x - 2.
According to the given information, we have the equation:
x^2 + (2x - 2)^2 = 745
Expanding the expression and simplifying, we get:
x^2 + 4x^2 - 8x + 4 = 745
5x^2 - 8x + 4 = 745
5x^2 - 8x - 741 = 0
We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, factoring is not possible, and completing the square is quite tedious. So, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Using the formula with a = 5, b = -8, and c = -741, we have:
x = (-(-8) ± √((-8)^2 - 4(5)(-741))) / (2(5))
x = (8 ± √(64 + 14820)) / 10
x = (8 ± √14884) / 10
x = (8 ± 122) / 10
So, we have two possible values for x:
1) x = (8 + 122) / 10 = 13
2) x = (8 - 122) / 10 = -11.4
Since we are looking for positive integers, we discard the second solution.
Therefore, the two positive integers are 13 and 2(13) - 2 = 24. The integers are 13 and 24.