Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2026 June Paper 12 Q1

A Level / CIE / P1

CIE 9709 2026 June Paper 12 Paper · Question 1

题目

Problem

Solve the equation

3x2−8−16x2=0.3x^2-8-\frac{16}{x^2}=0.
(3)
题目中文翻译

解方程

3x2−8−16x2=0.3x^2-8-\frac{16}{x^2}=0.
(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

展开

原方程要求 x≠0x\ne0。乘以 x2x^2 后,将 x2x^2 当作一个未知数来解。

答题过程

展开

For x≠0x\ne0, the equation becomes 3x4−8x2−16=03x^4-8x^2-16=0. Putting u=x2u=x^2 gives

3u2−8u−16=(3u+4)(u−4)=0.3u^2-8u-16=(3u+4)(u-4)=0.

The negative value u=−4/3u=-4/3 is impossible. Thus x2=4x^2=4, and the solutions are x=±2x=\pm2.