one positive integer is 3 less than twice another.The sum of their squares is 482.Find the integers?
Let one integer be x
then the other is 2x-3
x^2 + (2x-3)^2 = 482
expand and solve as a quadratic.
Hint: it actually factors, one is (5x+43) which of course would yield a non-integer solution.
So the solution must come from the other factor.
To solve this problem, let's use algebraic representation.
Let's assume that one positive integer is represented by "x" and another positive integer is represented by "y."
From the given information, we can write the following equations:
1) One positive integer is 3 less than twice another:
x = 2y - 3
2) The sum of their squares is 482:
x^2 + y^2 = 482
Now, let's substitute equation 1 into equation 2:
(2y - 3)^2 + y^2 = 482
Simplifying this equation:
4y^2 - 12y + 9 + y^2 = 482
5y^2 - 12y - 473 = 0
Now, we have a quadratic equation. We can solve this equation using factoring, completing the square, or the quadratic formula.
Factoring this quadratic equation, we have:
(5y + 19)(y - 25) = 0
Setting each factor to zero and solving for y, we get:
5y + 19 = 0 --> 5y = -19 --> y = -19/5
y - 25 = 0 --> y = 25
Since we're looking for positive integers, we can discard the negative solution. Therefore, y = 25.
Now, substitute this value back into equation 1 to find x:
x = 2(25) - 3
x = 50 - 3
x = 47
So, the two positive integers are 47 and 25.