题目
Problem
Solve the quadratic equation ( 2 + i ) w 2 + 4 w + 2 − i = 0 (2+\mathrm{i})w^2+4w+2-\mathrm{i}=0 ( 2 + i ) w 2 + 4 w + 2 − i = 0 x + i y x+\mathrm{i}y x + i y x x x y y y
(5)
题目中文翻译
解一元二次方程 ( 2 + i ) w 2 + 4 w + 2 − i = 0 (2+\mathrm{i})w^2+4w+2-\mathrm{i}=0 ( 2 + i ) w 2 + 4 w + 2 − i = 0 x + i y x+\mathrm{i}y x + i y x x x y y y
(5)
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
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直接使用复系数二次公式，根号内先化成负实数，再分别把两个根的分母实数化。
答题过程
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The quadratic formula gives
w = − 4 ± 16 − 4 ( 2 + i ) ( 2 − i ) 2 ( 2 + i ) = − 4 ± − 4 4 + 2 i = − 4 ± 2 i 4 + 2 i . \begin{align*}
w=&\,\frac{-4\pm\sqrt{16-4(2+\mathrm{i})(2-\mathrm{i})}}
{2(2+\mathrm{i})}\\
=&\,\frac{-4\pm\sqrt{-4}}{4+2\mathrm{i}}
=\frac{-4\pm2\mathrm{i}}{4+2\mathrm{i}}.
\end{align*} w = = 2 ( 2 + i ) − 4 ± 16 − 4 ( 2 + i ) ( 2 − i ) 4 + 2 i − 4 ± − 4 = 4 + 2 i − 4 ± 2 i . For the minus sign, w = ( − 4 − 2 i ) / ( 4 + 2 i ) = − 1 w=(-4-2\mathrm{i})/(4+2\mathrm{i})=-1 w = ( − 4 − 2 i ) / ( 4 + 2 i ) = − 1
For the plus sign, multiply by the conjugate:
w = ( − 4 + 2 i ) ( 4 − 2 i ) ( 4 + 2 i ) ( 4 − 2 i ) = − 12 + 16 i 20 = − 3 5 + 4 5 i . \begin{align*}
w=&\,\frac{(-4+2\mathrm{i})(4-2\mathrm{i})}
{(4+2\mathrm{i})(4-2\mathrm{i})}\\
=&\,\frac{-12+16\mathrm{i}}{20}
=-\frac35+\frac45\mathrm{i}.
\end{align*} w = = ( 4 + 2 i ) ( 4 − 2 i ) ( − 4 + 2 i ) ( 4 − 2 i ) 20 − 12 + 16 i = − 5 3 + 5 4 i . Hence
w = − 1 , w = − 3 5 + 4 5 i . \boxed{w=-1,\qquad w=-\frac35+\frac45\mathrm{i}}. w = − 1 , w = − 5 3 + 5 4 i .
解法二
思路
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先乘最高次项系数的共轭，使首项系数成为实数，再观察并验证因式分解。
答题过程
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Multiplying the equation by 2 − i 2-\mathrm{i} 2 − i
5 w 2 + 4 ( 2 − i ) w + ( 2 − i ) 2 = 0. 5w^2+4(2-\mathrm{i})w+(2-\mathrm{i})^2=0. 5 w 2 + 4 ( 2 − i ) w + ( 2 − i ) 2 = 0. Since ( 2 − i ) 2 = 3 − 4 i (2-\mathrm{i})^2=3-4\mathrm{i} ( 2 − i ) 2 = 3 − 4 i
5 w 2 + ( 8 − 4 i ) w + 3 − 4 i = 0. 5w^2+(8-4\mathrm{i})w+3-4\mathrm{i}=0. 5 w 2 + ( 8 − 4 i ) w + 3 − 4 i = 0. This factors as
( w + 1 ) ( 5 w + 3 − 4 i ) = 0. (w+1)(5w+3-4\mathrm{i})=0. ( w + 1 ) ( 5 w + 3 − 4 i ) = 0. Therefore
w = − 1 , w = − 3 5 + 4 5 i . \boxed{w=-1,\qquad w=-\frac35+\frac45\mathrm{i}}. w = − 1 , w = − 5 3 + 5 4 i .
解法三
思路
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令 w = x + y i w=x+y\mathrm{i} w = x + y i y y y
答题过程
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Put w = x + y i w=x+y\mathrm{i} w = x + y i x , y x,y x , y
2 ( x 2 − y 2 ) − 2 x y + 4 x + 2 = 0 (1) 2(x^2-y^2)-2xy+4x+2=0 \tag{1} 2 ( x 2 − y 2 ) − 2 x y + 4 x + 2 = 0 ( 1 ) and
x 2 − y 2 + 4 x y + 4 y − 1 = 0. (2) x^2-y^2+4xy+4y-1=0. \tag{2} x 2 − y 2 + 4 x y + 4 y − 1 = 0. ( 2 ) Subtracting twice (2) from (1),
− 10 x y + 4 x − 8 y + 4 = 0 , -10xy+4x-8y+4=0, − 10 x y + 4 x − 8 y + 4 = 0 , so
( 5 x + 4 ) y = 2 ( x + 1 ) . (5x+4)y=2(x+1). ( 5 x + 4 ) y = 2 ( x + 1 ) . The value 5 x + 4 = 0 5x+4=0 5 x + 4 = 0 y = 2 ( x + 1 ) / ( 5 x + 4 ) y=2(x+1)/(5x+4) y = 2 ( x + 1 ) / ( 5 x + 4 )
Equation (1), divided by 2 2 2 ( x + 1 ) 2 − y 2 − x y = 0 (x+1)^2-y^2-xy=0 ( x + 1 ) 2 − y 2 − x y = 0 ( 5 x + 4 ) 2 (5x+4)^2 ( 5 x + 4 ) 2
0 = ( x + 1 ) 2 ( 5 x + 4 ) 2 − 4 ( x + 1 ) 2 − 2 x ( x + 1 ) ( 5 x + 4 ) = ( x + 1 ) ( 5 x + 3 ) ( 5 x 2 + 8 x + 4 ) . \begin{align*}
0=&\,(x+1)^2(5x+4)^2-4(x+1)^2\\
&\,\hspace{2pt}-2x(x+1)(5x+4)\\
=&\,(x+1)(5x+3)(5x^2+8x+4).
\end{align*} 0 = = ( x + 1 ) 2 ( 5 x + 4 ) 2 − 4 ( x + 1 ) 2 − 2 x ( x + 1 ) ( 5 x + 4 ) ( x + 1 ) ( 5 x + 3 ) ( 5 x 2 + 8 x + 4 ) . The last quadratic has discriminant 8 2 − 4 ( 5 ) ( 4 ) = − 16 < 0 8^2-4(5)(4)=-16<0 8 2 − 4 ( 5 ) ( 4 ) = − 16 < 0 x = − 1 x=-1 x = − 1 x = − 3 / 5 x=-3/5 x = − 3/5
The corresponding values of y y y 0 0 0 4 / 5 4/5 4/5
w = − 1 , w = − 3 5 + 4 5 i . \boxed{w=-1,\qquad w=-\frac35+\frac45\mathrm{i}}. w = − 1 , w = − 5 3 + 5 4 i . 