Competition problem — Algebra — Maths Olympiad Prep

Library / /423 of 860

Algebra Difficulty 5.2 AIME, harder Find the answer

For any positive real numbers $a$ and $b$ , define $a \circ b=a+b+2 \sqrt{a b}$ . Find all positive real numbers $x$ such that $x^{2} \circ 9x=121$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since $a \circ b=(\sqrt{a}+\sqrt{b})^{2}$ , we have $x^{2} \circ 9x=(x+3\sqrt{x})^{2}$ . Moreover, since $x$ is positive, we have $x+3\sqrt{x}=11$ , and the only possible solution is that $\sqrt{x}=\frac{-3+\sqrt{53}}{2}$ , so $x=\frac{31-3\sqrt{53}}{2}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.