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FP1 2023 May Q2

A Level / Edexcel / FP1

IAL 2023 May Paper · Question 2

题目

Problem

2. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.

Given that x=2+3ix = 2 + 3\mathrm{i} is a root of the equation

 2x4−8x3+29x2 −12x+39=0.\begin{align*} &\,2x^4-8x^3+29x^2\\ &\,\hspace{2pt}-12x+39=0. \end{align*}

(a) write down another complex root of this equation.

(1)

(b) Use algebra to determine the other 2 roots of the equation.

(4)

(c) Show all 4 roots on a single Argand diagram.

(2)
题目中文翻译
  1. 在本题中必须展示所有解题步骤。依赖计算器技术的解答不可接受。

已知 x=2+3ix = 2 + 3\mathrm{i} 是方程

 2x4−8x3+29x2 −12x+39=0.\begin{align*} &\,2x^4-8x^3+29x^2\\ &\,\hspace{2pt}-12x+39=0. \end{align*}

的一个根,

(a) 写出该方程的另一个复根。

(1)

(b) 用代数方法确定方程的另外两个根。

(4)

(c) 在同一幅 Argand 图上画出所有 4 个根。

(2)

解答

(a)

解法一

思路

展开

多项式的系数全为实数,因此非实复根必成共轭对出现。把 2+3i2+3\mathrm{i} 的虚部变号即可。

答题过程

展开

Since the polynomial has real coefficients, complex roots occur in conjugate pairs. Therefore another root is

2−3i.\boxed{2-3\mathrm{i}}.

(b)

解法一

思路

展开

先把共轭复根对应的两个一次因式相乘,得到实系数二次因式。再用多项式除法从原四次式中除去该因式,最后解余下的二次方程,并保持精确形式。

答题过程

展开

The two known roots give a quadratic factor F(x)F(x), where

 F(x)= (x−2−3i) ×(x−2+3i)= (x−2)2−(3i)2= (x−2)2+9= x2−4x+4+9= x2−4x+13.\begin{align*} &\,F(x)\\ =&\,\big(x-2-3\mathrm{i}\big)\\ &\,\hspace{2pt}\times\big(x-2+3\mathrm{i}\big)\\ =&\,(x-2)^2-(3\mathrm{i})^2\\ =&\,(x-2)^2+9\\ =&\,x^2-4x+4+9\\ =&\,x^2-4x+13. \end{align*}

Let P(x)P(x) denote the polynomial in the question. To divide P(x)P(x) by F(x)F(x), the first term of the quotient is 2x22x^2:

 2x2F(x)= 2x4−8x3+26x2.\begin{align*} &\,2x^2F(x)\\ =&\,2x^4-8x^3+26x^2. \end{align*}

Subtracting gives

 P(x)−2x2F(x)= (29−26)x2−12x+39= 3x2−12x+39= 3(x2−4x+13)= 3F(x).\begin{align*} &\,P(x)-2x^2F(x)\\ =&\,(29-26)x^2-12x+39\\ =&\,3x^2-12x+39\\ =&\,3(x^2-4x+13)\\ =&\,3F(x). \end{align*}

The next term of the quotient is therefore 33, and the remainder is zero. Thus

P(x)= 2x2F(x)+3F(x)= F(x)(2x2+3).\begin{align*} P(x)=&\,2x^2F(x)+3F(x)\\ =&\,F(x)(2x^2+3). \end{align*}

That is,

P(x)= (x2−4x+13) ×(2x2+3).\begin{align*} P(x)=&\,(x^2-4x+13)\\ &\,\hspace{2pt}\times(2x^2+3). \end{align*}

Hence the other roots satisfy

2x2+3=0.2x^2+3=0.

Thus

x2=−32,x^2=-\frac32,

so the other two roots are

x=62i,\boxed{x=\frac{\sqrt6}{2}\mathrm{i}}, x=−62i.\boxed{x=-\frac{\sqrt6}{2}\mathrm{i}}.

解法二

思路

展开

也可以先用两根的和与积构造二次因式,再设另一个二次因式的待定系数。展开并比较系数,可以不写多项式长除法而得到相同的剩余因式。

答题过程

展开

The sum and product of the two known roots are

(2+3i)+(2−3i)=4(2+3\mathrm{i})+(2-3\mathrm{i})=4

and

 (2+3i)(2−3i)= 22−(3i)2= 4+9= 13.\begin{align*} &\,(2+3\mathrm{i})(2-3\mathrm{i})\\ =&\,2^2-(3\mathrm{i})^2\\ =&\,4+9\\ =&\,13. \end{align*}

Therefore their quadratic factor is

x2−4x+13.x^2-4x+13.

Let P(x)P(x) denote the polynomial in the question, and write

P(x)= (x2−4x+13) ×(2x2+px+q).\begin{align*} P(x)=&\,(x^2-4x+13)\\ &\,\hspace{2pt}\times(2x^2+px+q). \end{align*}

Expanding the product,

 P(x)= 2x4+px3+qx2 −8x3−4px2−4qx +26x2+13px+13q= 2x4+(p−8)x3 +(q−4p+26)x2 +(−4q+13p)x +13q.\begin{align*} &\,P(x)\\ =&\,2x^4+px^3+qx^2\\ &\,\hspace{2pt}-8x^3-4px^2-4qx\\ &\,\hspace{4pt}+26x^2+13px+13q\\ =&\,2x^4+(p-8)x^3\\ &\,\hspace{2pt}+(q-4p+26)x^2\\ &\,\hspace{4pt}+(-4q+13p)x\\ &\,\hspace{6pt}+13q. \end{align*}

Comparing the coefficients of x3x^3 and the constant terms gives

p−8=−8p-8=-8

and

13q=39.13q=39.

Hence p=0p=0 and q=3q=3. The remaining coefficients also agree, since

q−4p+26=29q-4p+26=29

and

−4q+13p=−12.-4q+13p=-12.

Thus the remaining factor is 2x2+32x^2+3. Therefore,

x=62i,\boxed{x=\frac{\sqrt6}{2}\mathrm{i}}, x=−62i.\boxed{x=-\frac{\sqrt6}{2}\mathrm{i}}.

(c)

解法一

思路

展开

在 Argand 图中,横坐标是实部,纵坐标是虚部。两根 2±3i2\pm3\mathrm{i} 位于直线 Re⁡(z)=2\operatorname{Re}(z)=2 上;另两根为纯虚数,位于虚轴上,并且关于实轴成对称分布。

答题过程

展开

The four roots are represented by the points

(2,3),(2,3), (2,−3),(2,-3), (0,62),\Big(0,\frac{\sqrt6}{2}\Big), (0,−62).\Big(0,-\frac{\sqrt6}{2}\Big).