1983 AIME 第 3 题

先试着解答 1983 AIME 第 3 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1983 AIME 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

3.

求下列方程所有实根的乘积:x2+18x+30=2x2+18x+45? \begin{gathered} x^2+18x+30\\ {}=2\sqrt{x^2+18x+45} \end{gathered}\text{?}
What is the product of the real roots of the equation x2+18x+30=2x2+18x+45? \begin{gathered} x^2+18x+30\\ {}=2\sqrt{x^2+18x+45}? \end{gathered}
答案:20
知识点:根式换元法韦达定理
小提示:
根号外的表达式比被开方数少 1515
The expression outside the radical is 1515 less than the radicand
大提示:
令 t=x2+18x+45t=\sqrt{x^2+18x+45},并解 t2−15=2tt^2-15=2t
Set t=x2+18x+45t=\sqrt{x^2+18x+45} and solve t2−15=2tt^2-15=2t
解答:
令 t=x2+18x+45t=\sqrt{x^2+18x+45},则 t≥0t\geq0。原方程化为 t2−15=2t, t^2-15=2t\text{,}即 (t−5)(t+3)=0(t-5)(t+3)=0。因此 t=5t=5,并且 x2+18x+45=25, x^2+18x+45=25\text{,}所以实根满足 x2+18x+20=0x^2+18x+20=0。由韦达定理,所有实根的乘积为 2020。
Set t=x2+18x+45,t=\sqrt{x^2+18x+45}, so t≥0.t\geq0. The equation becomes t2−15=2t, t^2-15=2t, or (t−5)(t+3)=0.(t-5)(t+3)=0. Thus t=5,t=5, and x2+18x+45=25, x^2+18x+45=25, so the real roots satisfy x2+18x+20=0.x^2+18x+20=0. By Vieta’s formulas, their product is 20.20.
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