Asked by kwante
one positive integer is 3 less than twice another.The sum of their squares is 482.Find the integers?
Answers
Answered by
Reiny
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.
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.
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