填充2怎麼算阿...
我研究了一個小時了QAQ
計算第2.3請參考
我覺得我第二題不完整= =
2.
若\( a+b i \)為根則\( a-bi \),\( 1-a-bi \),\( 1-a+bi \),\( \displaystyle \frac{a+bi}{a^2+b^2} \),\( \displaystyle \frac{a-bi}{a^2+b^2} \)為根
k為根則\( 1-k \),\( \displaystyle \frac{1}{k} \)為根，\( \displaystyle k,1-k,\frac{1}{k} \)只有至多2個相同。
\( \Rightarrow \)3實根2虛根
虛根部份：\( a=1-a \)則\( \displaystyle a^2+b^2=1 \Rightarrow a=\frac{1}{2},b=\frac{\sqrt{3}}{2} \)
虛根\( \displaystyle \frac{1}{2}\pm \frac{\sqrt{3}}{2}i \Rightarrow (x^2-x+1)=0 \)
實根部份：若\( \displaystyle k=\frac{1}{k},k=1 \Rightarrow 1,0,1 \)為根\( \Rightarrow \frac{1}{0} \)為根(不合)
若\( \displaystyle k=1-k,k=\frac{1}{2} \Rightarrow \frac{1}{2},\frac{1}{2},2 \)為根\( \displaystyle \Rightarrow \frac{1}{2},2,-1 \)為根
\( \displaystyle f(x)=(x^2-x+1)(x-\frac{1}{2})(x-2)(x+1) \)
3.
\( f(x)=ax^2+bx-a^2=(x-\alpha)(x-\beta) \)
\( \displaystyle \Rightarrow \alpha+\beta=\frac{-b}{a} \)，\( \alpha \beta=-a \Rightarrow b=\alpha \beta(\alpha+\beta)=\alpha^2 \beta+\alpha \beta^2 \)
又\( a>0 \)故\( \alpha,\beta \)異號，設\( \beta>0 \),\( \alpha<0 \)
\( |\ \alpha |\ + |\ \beta |\ =2 \Rightarrow \beta- \alpha =2 \)，\( \alpha=\beta-2 \)代入b
\( b=(\beta-2)^2 \beta+(\beta-2) \beta^2=2 \beta^3-6 \beta^2+4 \beta \)
\( \displaystyle b=6 \beta^2-12 \beta+4=0 \Rightarrow 3\beta^3-6 \beta+2=0 \Rightarrow \beta=\frac{6 \pm \sqrt{36-24}}{6}=1\pm \frac{\sqrt{3}}{3} \)
\( \displaystyle (\alpha,\beta)=(-1+\frac{\sqrt{3}}{3},1+\frac{\sqrt{3}}{3}) \) or \( (-1-\frac{\sqrt{3}}{3},1-\frac{\sqrt{3}}{3}) \)代回b
得b最大值\( \displaystyle \frac{4}{9}\sqrt{3} \) 最小值\( \displaystyle -\frac{4}{9}\sqrt{3} \)
