题目
Problem
Solve the quadratic equation ( 3 + i ) w 2 − 2 w + 3 − i = 0 (3+\mathrm{i})w^2-2w+3-\mathrm{i}=0 ( 3 + i ) w 2 − 2 w + 3 − 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
解二次方程 ( 3 + i ) w 2 − 2 w + 3 − i = 0 (3+\mathrm{i})w^2-2w+3-\mathrm{i}=0 ( 3 + i ) w 2 − 2 w + 3 − 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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求根公式仍适用于复系数二次方程。先计算判别式，再用共轭把每个根的分母化实。
答题过程
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The quadratic formula gives
w = 2 ± ( − 2 ) 2 − 4 ( 3 + i ) ( 3 − i ) 2 ( 3 + i ) . w=\frac{2\pm\sqrt{(-2)^2-4(3+\mathrm{i})(3-\mathrm{i})}}
{2(3+\mathrm{i})}. w = 2 ( 3 + i ) 2 ± ( − 2 ) 2 − 4 ( 3 + i ) ( 3 − i ) . Since ( 3 + i ) ( 3 − i ) = 9 − i 2 = 10 (3+\mathrm{i})(3-\mathrm{i})=9-\mathrm{i}^2=10 ( 3 + i ) ( 3 − i ) = 9 − i 2 = 10 4 − 40 = − 36 4-40=-36 4 − 40 = − 36
w = 2 ± 6 i 6 + 2 i . w=\frac{2\pm6\mathrm{i}}{6+2\mathrm{i}}. w = 6 + 2 i 2 ± 6 i . For the plus sign,
w = ( 2 + 6 i ) ( 6 − 2 i ) ( 6 + 2 i ) ( 6 − 2 i ) = 24 + 32 i 40 = 3 5 + 4 5 i . \begin{align*}
w=&\,\frac{(2+6\mathrm{i})(6-2\mathrm{i})}{(6+2\mathrm{i})(6-2\mathrm{i})}\\[4mm]
=&\,\frac{24+32\mathrm{i}}{40}
=\frac35+\frac45\mathrm{i}.
\end{align*} w = = ( 6 + 2 i ) ( 6 − 2 i ) ( 2 + 6 i ) ( 6 − 2 i ) 40 24 + 32 i = 5 3 + 5 4 i . For the minus sign,
w = ( 2 − 6 i ) ( 6 − 2 i ) 40 = − 40 i 40 = − i . w=\frac{(2-6\mathrm{i})(6-2\mathrm{i})}{40}
=\frac{-40\mathrm{i}}{40}=-\mathrm{i}. w = 40 ( 2 − 6 i ) ( 6 − 2 i ) = 40 − 40 i = − i . The roots are
w = 3 5 + 4 5 i or w = − i . \boxed{w=\frac35+\frac45\mathrm{i}
\quad\text{or}\quad w=-\mathrm{i}}. w = 5 3 + 5 4 i or w = − i .
解法二
思路
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也可先把方程乘以 3 − i 3-\mathrm{i} 3 − i
答题过程
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Multiplying the equation by 3 − i 3-\mathrm{i} 3 − i
10 w 2 − 2 ( 3 − i ) w + ( 3 − i ) 2 = 0. 10w^2-2(3-\mathrm{i})w+(3-\mathrm{i})^2=0. 10 w 2 − 2 ( 3 − i ) w + ( 3 − i ) 2 = 0. Put v = w / ( 3 − i ) v=w/(3-\mathrm{i}) v = w / ( 3 − i ) 3 − i ≠ 0 3-\mathrm{i}\ne0 3 − i = 0 ( 3 − i ) 2 (3-\mathrm{i})^2 ( 3 − i ) 2
10 v 2 − 2 v + 1 = 0. 10v^2-2v+1=0. 10 v 2 − 2 v + 1 = 0. Thus
v = 2 ± 4 − 40 20 = 1 ± 3 i 10 . v=\frac{2\pm\sqrt{4-40}}{20}
=\frac{1\pm3\mathrm{i}}{10}. v = 20 2 ± 4 − 40 = 10 1 ± 3 i . Therefore
w = ( 3 − i ) ( 1 ± 3 i ) 10 . w=\frac{(3-\mathrm{i})(1\pm3\mathrm{i})}{10}. w = 10 ( 3 − i ) ( 1 ± 3 i ) . The two products give
w = 6 + 8 i 10 = 3 5 + 4 5 i , or w = − 10 i 10 = − i . \begin{align*}
w=&\,\frac{6+8\mathrm{i}}{10}
=\boxed{\frac35+\frac45\mathrm{i}},\\[4mm]
\text{or}\quad w=&\,\frac{-10\mathrm{i}}{10}=\boxed{-\mathrm{i}}.
\end{align*} w = or w = 10 6 + 8 i = 5 3 + 5 4 i , 10 − 10 i = − i . 