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


  • is 107
  • is 13
  • is 840
  • cannot be determined from the given information

Key Concepts




Check the Answer

But try the problem first…

Answer: is 13.

Suggested Reading

AIME I, 1990, Question 4

Elementary Algebra by Hall and Knight

Try with Hints

First hint

here we put \(x^{2}-10x-29=p\)


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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