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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

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.

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