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IAL 2022 Jan FP1 Q4

A Level / Edexcel / FP1

IAL 2022 Jan Paper · Question 4

题目

Problem

The equation

x4+Ax3+Bx2+Cx+225=0x^4 + Ax^3 + Bx^2 + Cx + 225 = 0

where AA, BB and CC are real constants, has

  • a complex root 4+3i4 + 3i
  • 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+225x^4 + Ax^3 + Bx^2 + Cx + 225

(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=0x^4 + Ax^3 + Bx^2 + Cx + 225 = 0

其中 AABBCC 为实常数,具有

  • 一个复数根 4+3i4 + 3i
  • 一个重正实根

(a) 直接写出该方程的另一个复数根。

(b) 由此确定 x4+Ax3+Bx2+Cx+225x^4 + Ax^3 + Bx^2 + Cx + 225 的一个二次因式。

(c) 推导该方程的实根。

(d) 由此确定常数 AABBCC 的值。

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+3i 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

43i.\boxed{4-3i}.

(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(43i)]=[(x4)3i][(x4)+3i]=(x4)2+9=x28x+25.\begin{align*} &\,[x-(4+3i)][x-(4-3i)]\\ =&\,[(x-4)-3i][(x-4)+3i]\\ =&\,(x-4)^2+9\\ =&\,\boxed{x^2-8x+25}. \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)(43i)=42+32=25.(4+3i)(4-3i)=4^2+3^2=25.

Therefore,

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

展开

The factor corresponding to the repeated root is

(x3)2=x26x+9.(x-3)^2=x^2-6x+9.

Hence

(x28x+25)(x26x+9)=x414x3+82x2222x+225.\begin{align*} &\,(x^2-8x+25)(x^2-6x+9)\\ =&\,x^4-14x^3+82x^2-222x+225. \end{align*}

Comparing coefficients with

x4+Ax3+Bx2+Cx+225,x^4+Ax^3+Bx^2+Cx+225,

gives

A=14,B=82,C=222.\boxed{A=-14,\qquad B=82,\qquad C=-222}.