SOLUTION: The sum of the recipricals of 2 consectutive even integars is 11/60. Find the integars.

Question 256684: The sum of the recipricals of 2 consectutive even integars is 11/60. Find the integars.
Found 2 solutions by stanbon, vleith:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The sum of the reciprocals of 2 consecutive even integers is 11/60. Find the integers.
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Let the integers be:
1st 2x
2nd 2x+2
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Equation:
1/(2x) + 1/(2x+2) = 11/60
Multiply thru by 2 to get:
---
1/x + 1/(x+1) = 11/30
(x+1 + x)/(x(x+1)) = 11/30
(2x+1)/(x^2+x) = 11/30
Cross-multiply
60x + 30 = 11x^2 + 11x
11x^2 - 49x - 30 = 0
(x-5)(11x+6) = 0
Positive solution:
x = 5
--------------
1st number: 2x = 10
2nd number: 2x+2 = 12
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Cheers,
Stan H.

Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
let the two integers be represented by x and (x+2)
Then their reciprocals are and
You are told the sum of the reciprocals is 11/60

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=14884 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 10, -1.09090909090909. Here's your graph:

So one integer is 10 and the next one is 12
Check your answer. Does 1/10 + 1/12 = 11/60???
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