From 8c7e52157d34d2796424d7f177df59c72a2a99d5 Mon Sep 17 00:00:00 2001 From: Ashutosh0x Date: Fri, 20 Feb 2026 06:39:24 +0530 Subject: [PATCH] fix: correct sign errors in putnam_like Set_6/A1 grading_scheme.md (fixes #2) Fix incorrect signs in Vieta's formula expansion: - Line 24: a^2+2b -> a^2-2b (since (-a)^2 - 2b = a^2 - 2b) - Line 28: updated final answer to match corrected computation --- .../putnam_like/Set_6/A1/rubrics/grading_scheme.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/eval_hub/putnam_like/putnam_like/Set_6/A1/rubrics/grading_scheme.md b/eval_hub/putnam_like/putnam_like/Set_6/A1/rubrics/grading_scheme.md index 3b9f6a2..165dc5b 100644 --- a/eval_hub/putnam_like/putnam_like/Set_6/A1/rubrics/grading_scheme.md +++ b/eval_hub/putnam_like/putnam_like/Set_6/A1/rubrics/grading_scheme.md @@ -21,8 +21,8 @@ $$ This step is worth 5 points. Doing standard calculations we see that: * $-C = (\alpha\beta\gamma)^2=c^2$, -* $-A = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\alpha\gamma) = a^2+2b$, +* $-A = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\alpha\gamma) = a^2-2b$, * $B = (\alpha\beta+\beta\gamma+\alpha\gamma)^2-2\alpha\beta\gamma(\alpha+\beta+\gamma) = b^2-2ac$. This step is worth 1 point. -Answer: $x^3-(a^2+2b)x^2 + (b^2-2ac)x -c^2 = 0$. \ No newline at end of file +Answer: $x^3-(a^2-2b)x^2 + (b^2-2ac)x -c^2 = 0$. \ No newline at end of file