Find two consecutive positive integers such that the sum of their squares is 25.
Let the first integer be x. The next consecutive positive integer is x + 1.
The sum of their squares equals 25. We write this as::
x^2 + (x + 1)^2
Expanding, we get:
x^2 + x^2 + 2x + 1 = 25
Group like terms:
2x^2 + 2x + 1 = 25
Subtract 25 from each side:
2x^2 + 2x - 24 = 0
Simplify by dividing each side by 2:
x^2 + x - 12 = 0
Using our quadratic calculator, we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3.
This means, our next positive integer is 3 + 1 = 4. So we have (3, 4) as our answers.
Let's check our work:
3^2 + 4^2 = 9 + 16 = 25
Let the first integer be x. The next consecutive positive integer is x + 1.
The sum of their squares equals 25. We write this as::
x^2 + (x + 1)^2
Expanding, we get:
x^2 + x^2 + 2x + 1 = 25
Group like terms:
2x^2 + 2x + 1 = 25
Subtract 25 from each side:
2x^2 + 2x - 24 = 0
Simplify by dividing each side by 2:
x^2 + x - 12 = 0
Using our quadratic calculator, we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3.
This means, our next positive integer is 3 + 1 = 4. So we have (3, 4) as our answers.
Let's check our work:
3^2 + 4^2 = 9 + 16 = 25