Categories

# Positive solution | AIME I, 1990 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Positive solution.

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

But try the problem first…

Source

AIME I, 1990, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here we put $$x^{2}-10x-29=p$$

$$\frac{1}{p}+\frac{1}{p-16}-\frac{2}{p-40}=0$$

Second Hint

or, (p-16)(p-40)+p(p-40)-2p(p-16)=0

or, -64p+(40)(16)=0

or, p=10

Final Step

or, 10=$$x^{2}-10x-29$$

or, (x-13)(x+3)=0

or, x=13 positive solution.

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.