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IAL 2023 May FP1 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 + 3i is a root of the equation

2x4−8x3+29x2−12x+39=02x^4 - 8x^3 + 29x^2 - 12x + 39 = 0

(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。完全依赖计算器技术的解答不可接受。

已知 x=2+3ix = 2 + 3i 是方程

2x4−8x3+29x2−12x+39=02x^4 - 8x^3 + 29x^2 - 12x + 39 = 0

的一个根，

(a) 写出该方程的另一个复根。

(b) 用代数方法确定方程的另外两个根。

(c) 在同一幅 Argand 图上画出所有 4 个根。

解答

(a)

解法一

思路

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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

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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

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The two known roots give the factor

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

Dividing the polynomial by this factor,

 2x4−8x3+29x2−12x+39= 2x2(x2−4x+13) +(3x2−12x+39)= 2x2(x2−4x+13) +3(x2−4x+13)= (x2−4x+13)(2x2+3).\begin{align*} &\,2x^4-8x^3+29x^2-12x+39\\ =&\,2x^2(x^2-4x+13)\\ &\,\hspace{2pt}+(3x^2-12x+39)\\ =&\,2x^2(x^2-4x+13)\\ &\,\hspace{2pt}+3(x^2-4x+13)\\ =&\,(x^2-4x+13)(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=62iandx=−62i.\boxed{x=\frac{\sqrt6}{2}\mathrm{i} \quad\text{and}\quad x=-\frac{\sqrt6}{2}\mathrm{i}}.

解法二

思路

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官方评分资料也允许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

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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)=13.(2+3\mathrm{i})(2-3\mathrm{i})=13.

Therefore their quadratic factor is

x2−4x+13.x^2-4x+13.

Let

 2x4−8x3+29x2−12x+39= (x2−4x+13)(2x2+px+q).\begin{align*} &\,2x^4-8x^3+29x^2-12x+39\\ =&\,(x^2-4x+13)(2x^2+px+q). \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=62iandx=−62i.\boxed{x=\frac{\sqrt6}{2}\mathrm{i} \quad\text{and}\quad x=-\frac{\sqrt6}{2}\mathrm{i}}.

(c)

解法一

思路

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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),(0,62),(0,−62).(2,3),\quad (2,-3),\quad \left(0,\frac{\sqrt6}{2}\right),\quad \left(0,-\frac{\sqrt6}{2}\right).