SOLUTION: find two consecutive positive odd integers given that the sum of their reciprocal is 24/143

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Question 953711: find two consecutive positive odd integers given that the sum of their reciprocal is 24/143
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
n=first integer; n+2=second integer
1%2Fn%2B1%2F%28n%2B2%29=24%2F143 Find common denominator
%28%28n%2B2%29%2F%28n%2B2%29%29%281%2Fn%29%2B%28n%2Fn%29%281%2Fn%2B2%29=24%2F143 Multiply each side by 143(n)(n+2)
143%28%28n%2B2%29%2Bn%29=24%28n%29%28n%2B2%29
286n%2B286=24n%5E2%2B48n Subtract (286n+286) from each side.
0=24n%5E2-238n-286
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 24n%5E2%2B-238n%2B-286+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-238%29%5E2-4%2A24%2A-286=84100.

Discriminant d=84100 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--238%2B-sqrt%28+84100+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%28-238%29%2Bsqrt%28+84100+%29%29%2F2%5C24+=+11
n%5B2%5D+=+%28-%28-238%29-sqrt%28+84100+%29%29%2F2%5C24+=+-1.08333333333333

Quadratic expression 24n%5E2%2B-238n%2B-286 can be factored:
24n%5E2%2B-238n%2B-286+=+24%28n-11%29%2A%28n--1.08333333333333%29
Again, the answer is: 11, -1.08333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+24%2Ax%5E2%2B-238%2Ax%2B-286+%29

ANSWER 1: The first integer is 11
n+2=13 ANSWER 2: The second integer is 13
CHECK:
1%2F11%2B1%2F13=24%2F143
13%2F143%2B11%2F143=24%2F143
24%2F143=24%2F143