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One positive integer is 4 less than twice another. The sum of their squares is 772 . Find the integers

User Isidat
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The positive integers are 21 and 38, where one integer is 4 less than twice the other. The sum of their squares is 772, satisfying the given conditions.

Let's denote the two positive integers as x and y, with x > y. According to the given information:

1. x is 4 less than twice y: x = 2y - 4.

2. The sum of their squares is 772:
\(x^2 + y^2 = 772\).

Now, substitute the expression for x from the first equation into the second:


\[(2y - 4)^2 + y^2 = 772\]

Expand and simplify:


\[4y^2 - 16y + 16 + y^2 = 772\]

Combine like terms:


\[5y^2 - 16y + 16 = 772\]

Subtract 772 from both sides:


\[5y^2 - 16y - 756 = 0\]

Factor the quadratic equation:

(5y + 36)(y - 21) = 0

This gives two possible solutions for y:
\(y = -(36)/(5)\) (extraneous) or y = 21.

Substitute y = 21 back into the first equation to find x:

x = 2(21) - 4 = 38

Therefore, the two positive integers are 21 and 38.

User Jay Hu
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