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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

Integers

Divisibility

Algebra

Check the Answer

But try the problem first…

Answer: is 13.

Source
Suggested Reading

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.

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