Competition problem — Algebra — Maths Olympiad Prep

Library / /136 of 860

Algebra Difficulty 4.9 AIME Find the answer

Let $r=H_{1}$ be the answer to this problem. Given that $r$ is a nonzero real number, what is the value of $r^{4}+4 r^{3}+6 r^{2}+4 r ?$

Your answer

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

Solution

Since $H_{1}$ is the answer, we know $r^{4}+4 r^{3}+6 r^{2}+4 r=r \Rightarrow(r+1)^{4}=r+1$ . Either $r+1=0$ , or $(r+1)^{3}=1 \Rightarrow r=0$ . Since $r$ is nonzero, $r=-1$ .

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.