SOLUTION: Solve and verify your answer.
The sum of the reciprocals of two consecutive even integers is 9 / 40. Find the integers (smaller and larger integers).
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The sum of the reciprocals of two consecutive even integers is 9 / 40. Find the integers (smaller and larger integers).
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Question 1130446: Solve and verify your answer.
The sum of the reciprocals of two consecutive even integers is 9 / 40. Find the integers (smaller and larger integers). Found 2 solutions by ankor@dixie-net.com, greenestamps:Answer by ankor@dixie-net.com(22740) (Show Source):
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The sum of the reciprocals of two consecutive even integers is 9 / 40. + =
Find the integers (smaller and larger integers).
multiply equation by 40n(n+2), cancel the denominators
40(n+2) + 40n = 9n(n+2)
40n + 80 + 40n = 9n^2 + 18n
80n + 80 = n^2 + 18n
Form a quadratic equation on the right
0 = 9n^2 + 18n - 80n - 80
n^2 - 62 - 80 = 0
Use the quadratic formula a=9; b=-62; c=-80
the positive integer solution
n = 8, and therefore 10 are the integers
:
:
see if that works + =
common denominator is 40 + +
A formal algebraic solution as shown by the other tutor is of course valid. But if an algebraic solution is not required, solve the problem by logical reasoning and a bit of mental arithmetic, knowing that the numbers are even integers.
Half the sum of the reciprocals is 9/80, which is about 1/9. So try 8 and 10 as the two consecutive even integers. They work; so they are the answers.
And here is another way to solve the problem using logical reasoning. The reciprocals of the two even integers, added together, yield a fraction with denominator 40. So a more advanced student might also solve the problem by finding two consecutive even integers whose least common multiple is 40; the answer is again 8 and 10.
