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FP1 2022 Jan Q4

A Level / Edexcel / FP1

IAL 2022 Jan Paper · Question 4

题目

Problem

The equation

 x4+Ax3+Bx2 +Cx+225=0\begin{align*} &\,x^4+Ax^3+Bx^2\\ &\,\hspace{2pt}+Cx+225=0 \end{align*}

where AA, BB and CC are real constants, has

  • a complex root 4+3i4 + 3\mathrm{i}
  • a repeated positive real root

(a) Write down the other complex root of this equation.

(1)

(b) Hence determine a quadratic factor of

 x4+Ax3+Bx2 +Cx+225\begin{align*} &\,x^4+Ax^3+Bx^2\\ &\,\hspace{2pt}+Cx+225 \end{align*}
(2)

(c) Deduce the real root of the equation.

(2)

(d) Hence determine the value of each of the constants AA, BB and CC

(3)
题目中文翻译

方程

 x4+Ax3+Bx2 +Cx+225=0\begin{align*} &\,x^4+Ax^3+Bx^2\\ &\,\hspace{2pt}+Cx+225=0 \end{align*}

其中 AA、BB、CC 为实常数,具有

  • 一个复数根 4+3i4 + 3\mathrm{i}
  • 一个重正实根

(a) 直接写出该方程的另一个复数根。

(1)

(b) 由此确定

 x4+Ax3+Bx2 +Cx+225\begin{align*} &\,x^4+Ax^3+Bx^2\\ &\,\hspace{2pt}+Cx+225 \end{align*}

的一个二次因式。

(2)

(c) 推导该方程的实根。

(2)

(d) 由此确定常数 AA、BB、CC 的值。

(3)

解答

(a)

解法一

思路

展开

实系数多项式的非实复数根成共轭对出现,因此把 4+3i4+3\mathrm{i} 取共轭即可。

答题过程

展开

Since the polynomial has real coefficients, non-real roots occur in conjugate pairs. Hence the other complex root is

4−3i.\boxed{4-3\mathrm{i}}.

(b)

解法一

思路

展开

将一对共轭复根分别写成一次因式并相乘。利用平方差可快速消去虚部,得到实系数二次因式。

答题过程

展开

The quadratic factor formed from the two complex roots is

 [x−(4+3i)] ×[x−(4−3i)]= [(x−4)−3i] ×[(x−4)+3i]= (x−4)2−(3i)2= (x−4)2−9(−1)= (x−4)2+9= x2−8x+16+9= x2−8x+25.\begin{align*} &\,\big[x-(4+3\mathrm{i})\big]\\ &\,\hspace{2pt}\times\big[x-(4-3\mathrm{i})\big]\\ =&\,\big[(x-4)-3\mathrm{i}\big]\\ &\,\hspace{2pt}\times\big[(x-4)+3\mathrm{i}\big]\\ =&\,(x-4)^2-(3\mathrm{i})^2\\ =&\,(x-4)^2-9(-1)\\ =&\,(x-4)^2+9\\ =&\,x^2-8x+16+9\\ =&\,\boxed{x^2-8x+25}. \end{align*}

同样也可先求两根的和与积:

 (4+3i)+(4−3i)= 8, (4+3i)(4−3i)= 42−(3i)2= 16+9= 25.\begin{align*} &\,(4+3\mathrm{i})+(4-3\mathrm{i})\\ =&\,8,\\[4mm] &\,(4+3\mathrm{i})(4-3\mathrm{i})\\ =&\,4^2-(3\mathrm{i})^2\\ =&\,16+9\\ =&\,25. \end{align*}

因此,首项系数为 11 的二次因式为

 x2−(8)x+25= x2−8x+25.\begin{align*} &\,x^2-(8)x+25\\ =&\,x^2-8x+25. \end{align*}

解法二

思路

展开

题目只要求一个二次因式,也可使用重正实根。设该根为 rr,由四个根的乘积求出 r2r^2,利用 r>0r>0 得到 rr,再写出重根对应的二次因式。

答题过程

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The product of the two complex roots is

 (4+3i)(4−3i)= 42+32= 25.\begin{align*} &\,(4+3\mathrm{i})(4-3\mathrm{i})\\ =&\,4^2+3^2\\ =&\,25. \end{align*}

Let the repeated positive real root be rr. Since the product of all four roots is 225225,

25r2= 225,r2= 22525,r2= 9.\begin{align*} 25r^2=&\,225,\\ r^2=&\,\frac{225}{25},\\ r^2=&\,9. \end{align*}

As r>0r>0, r=3r=3. A quadratic factor is therefore

 (x−3)2= x2−3x−3x+9= x2−6x+9.\begin{align*} &\,(x-3)^2\\ =&\,x^2-3x-3x+9\\ =&\,\boxed{x^2-6x+9}. \end{align*}

(c)

解法一

思路

展开

设重正实根为 rr。四个根的乘积等于常数项 225225;共轭复根的乘积为 2525,所以可求得 r2r^2,再利用“正实根”确定符号。

答题过程

展开

Let the repeated positive real root be rr. The product of all four roots is 225225, while

 (4+3i)(4−3i)= 42−(3i)2= 16+9= 25.\begin{align*} &\,(4+3\mathrm{i})(4-3\mathrm{i})\\ =&\,4^2-(3\mathrm{i})^2\\ =&\,16+9\\ =&\,25. \end{align*}

Therefore,

25r2= 225,r2= 22525,r2= 9.\begin{align*} 25r^2=&\,225,\\ r^2=&\,\frac{225}{25},\\ r^2=&\,9. \end{align*}

Since the repeated real root is positive,

r=3.\boxed{r=3}.

(d)

解法一

思路

展开

由 (b) 的复根因式及 (c) 的重根因式构造整个四次多项式,展开后逐项比较 x3,x2,xx^3,x^2,x 的系数。

答题过程

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The factor corresponding to the repeated root is

 (x−3)2= x2−3x−3x+9= x2−6x+9.\begin{align*} &\,(x-3)^2\\ =&\,x^2-3x-3x+9\\ =&\,x^2-6x+9. \end{align*}

Hence

 (x2−8x+25) ×(x2−6x+9)= x4−6x3+9x2 −8x3+48x2−72x +25x2−150x+225= x4+(−6−8)x3 +(9+48+25)x2 +(−72−150)x+225= x4−14x3+82x2 −222x+225.\begin{align*} &\,(x^2-8x+25)\\ &\,\hspace{2pt}\times(x^2-6x+9)\\ =&\,x^4-6x^3+9x^2\\ &\,\hspace{2pt}-8x^3+48x^2-72x\\ &\,\hspace{4pt}+25x^2-150x+225\\ =&\,x^4+(-6-8)x^3\\ &\,\hspace{2pt}+(9+48+25)x^2\\ &\,\hspace{4pt}+(-72-150)x+225\\ =&\,x^4-14x^3+82x^2\\ &\,\hspace{2pt}-222x+225. \end{align*}

Comparing coefficients with

 x4+Ax3+Bx2 +Cx+225,\begin{align*} &\,x^4+Ax^3+Bx^2\\ &\,\hspace{2pt}+Cx+225, \end{align*}

gives

A=−14,B=82,C=−222.\begin{align*} \boxed{A=-14},\\ \boxed{B=82},\\ \boxed{C=-222}. \end{align*}

解法二

思路

展开

由 (a)、(c) 已知四个根。首项系数为 11 的四次多项式中,AA 是四根和的相反数,BB 是所有两根乘积的和,CC 是所有三根乘积和的相反数。两个实根虽然同为 33,但计算时必须按两个根分别计入。

答题过程

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Write

u= 4+3i,v= 4−3i.\begin{align*} u=&\,4+3\mathrm{i},\\ v=&\,4-3\mathrm{i}. \end{align*}

The four roots are uu, vv, 33 and 33, with u+v=8u+v=8 and uv=25uv=25.

Using the sum of the four roots,

A= −(u+v+3+3)= −(8+6)= −14.\begin{align*} A =&\,-(u+v+3+3)\\ =&\,-(8+6)\\ =&\,\boxed{-14}. \end{align*}

Using the sum of all six pairwise products,

B= uv+3u+3u +3v+3v+3×3= uv+6(u+v)+9= 25+6×8+9= 25+48+9= 82.\begin{align*} B =&\,uv+3u+3u\\ &\,\hspace{2pt}+3v+3v+3\times3\\ =&\,uv+6(u+v)+9\\ =&\,25+6\times8+9\\ =&\,25+48+9\\ =&\,\boxed{82}. \end{align*}

Using the sum of all four triple products,

−C= 3uv+3uv +9u+9v= 6uv+9(u+v)= 6×25+9×8= 150+72= 222.\begin{align*} -C =&\,3uv+3uv\\ &\,\hspace{2pt}+9u+9v\\ =&\,6uv+9(u+v)\\ =&\,6\times25+9\times8\\ =&\,150+72\\ =&\,222. \end{align*}

Therefore, C=−222\boxed{C=-222}.