Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on positive solution.
Positive solution – AIME I, 1990
Find the positive solution to
\(\frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\)
- is 107
- is 13
- is 840
- cannot be determined from the given information
Key Concepts
Integers
Divisibility
Algebra
Check the Answer
Answer: is 13.
AIME I, 1990, Question 4
Elementary Algebra by Hall and Knight
Try with Hints
here we put \(x^{2}-10x-29=p\)
\(\frac{1}{p}+\frac{1}{p-16}-\frac{2}{p-40}=0\)
or, (p-16)(p-40)+p(p-40)-2p(p-16)=0
or, -64p+(40)(16)=0
or, p=10
or, 10=\(x^{2}-10x-29\)
or, (x-13)(x+3)=0
or, x=13 positive solution.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA