數與座標系/7 一元二次方程式-範例

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  • 範例17
    1. 求解一元二次方程式：\left( 2-\sqrt{3} \right)x^{2}-\sqrt{3}x+\left( \sqrt{3}+1 \right)=0
    2. 求解一元二次方程式：z^{2}-4iz-12-6i=0\,
  • 範例18
    • 已知：\left\{ \begin{align} & k\in R,x\in \text{Imaginary Number} \\ & x^{2}-kx+2=0 \\ & x^{2}-2kx+5k+6=0 \\ \end{align} \right.，求解：k\,之範圍。


  • 範例19
    1. 已知：\left\{ \begin{align} & x\in N \\ & x^{2}+\left( m-12 \right)x+\left( m-1 \right)=0 \\ \end{align} \right.，求解：m\,
    2. 已知：\left\{ \begin{align} & m,k,x\in Q,m\ne 3 \\ & x^{2}-4mx+4x+3m^{2}-2m+4k=0 \\ \end{align} \right.，求解：

範例17

  1. 求解一元二次方程式：\left( 2-\sqrt{3} \right)x^{2}-\sqrt{3}x+\left( \sqrt{3}+1 \right)=0
  2. 求解一元二次方程式：z^{2}-4iz-12-6i=0\,

答案

  1. x=\sqrt{3}+1,2+\sqrt{3}
  2. z=3+3i,-3+i\,

詳解

1

  1. \Delta =b^{2}-4ac=\left( -\sqrt{3} \right)^{2}-4\times \left( 2-\sqrt{3} \right)\left( \sqrt{3}+1 \right)
    =\sqrt{3}^{2}-4\times \left( \sqrt{3}-1 \right)=\sqrt{3}^{2}-4\sqrt{3}+4=\sqrt{3}^{2}-2\times 2\times \sqrt{3}+2^{2}
    =\left( \sqrt{3}-2 \right)^{2}
  2. x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{\sqrt{3}\pm \sqrt{\Delta }}{2\left( 2-\sqrt{3} \right)}=\frac{\sqrt{3}\pm \left( \sqrt{3}-2 \right)}{4-2\sqrt{3}}
  3. x=\frac{\sqrt{3}+\left( \sqrt{3}-2 \right)}{4-2\sqrt{3}}=\frac{\sqrt{3}+\left( \sqrt{3}-2 \right)}{\left( \sqrt{3}-1 \right)^{2}}=\frac{2\sqrt{3}-2}{\left( \sqrt{3}-1 \right)^{2}}=\frac{2\left( \sqrt{3}-1 \right)}{\left( \sqrt{3}-1 \right)^{2}}=\frac{2}{\sqrt{3}-1}
    {\color{Red}=\frac{2}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}=\frac{2\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}-1 \right)\left( \sqrt{3}+1 \right)}=\frac{2\left( \sqrt{3}+1 \right)}{2}}=\sqrt{3}+1\#
  4. x=\frac{\sqrt{3}-\left( \sqrt{3}-2 \right)}{4-2\sqrt{3}}=\frac{2}{4-2\sqrt{3}}=\frac{1}{2-\sqrt{3}}{\color{Red}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{1}}=2+\sqrt{3}\#

2

  1. z^{2}-4iz-12-6i=0\,
  2. \Delta =b^{2}-4ac=\left( -4i \right)^{2}-4\left( -12-6i \right)=-16+48+24i=32+24i
    {\color{Red}=36+24i-4=6^{2}+2\times 6\times \left( 2i \right)+\left( 2i \right)^{2}}=\left( 6+2i \right)^{2}
  3. z=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm \sqrt{\Delta }}{2a}=\frac{4i\pm \left( 6+2i \right)}{2}=2i\pm \left( 3+i \right)
  4. \left\{ \begin{align} & z=2i+\left( 3+i \right)=3+3i\# \\ & z=2i-\left( 3+i \right)=-3+i\# \\ \end{align} \right.

範例18

  • 已知：\left\{ \begin{align} & k\in R,x\in \text{Imaginary Number} \\ & x^{2}-kx+2=0 \\ & x^{2}-2kx+5k+6=0 \\ \end{align} \right.，求解：k\,之範圍。

答案

  • -1<k<2\sqrt{2}

詳解

  1. \Delta _{1}=b_{1}^{2}-4a_{1}c_{1}=\left( -k \right)^{2}-4\times 1\times 2=k^{2}-8<0
    \Rightarrow k^{2}<8\Rightarrow \left| k \right|<\sqrt{8}\Rightarrow -2\sqrt{2}<k<2\sqrt{2}
  2. \Delta _{2}=b_{2}^{2}-4a_{2}c_{2}=\left( -2k \right)^{2}-4\times 1\times \left( 5k+6 \right)=4k^{2}-4\times \left( 5k+6 \right)<0
    \Rightarrow k^{2}-\left( 5k+6 \right)=k^{2}-5k-6=\left( k-6 \right)\left( k+1 \right)<0\Rightarrow -1<\left| k \right|<6
  3. \left\{ \begin{align} & -2\sqrt{2}<k<2\sqrt{2} \\ & -1<\left| k \right|<6 \\ \end{align} \right.\Rightarrow -1<k<2\sqrt{2}\#

類題1

已知

求解

答案

詳解

範例19

  1. 已知：\left\{ \begin{align} & x\in N \\ & x^{2}+\left( m-12 \right)x+\left( m-1 \right)=0 \\ \end{align} \right.，求解：m\,
  2. 已知：\left\{ \begin{align} & m,k,x\in Q,m\ne 3 \\ & x^{2}-4mx+4x+3m^{2}-2m+4k=0 \\ \end{align} \right.，求解：k\,

答案

  1. m=6,or,7\,
  2. k=-\frac{5}{4}

詳解

1

  1. 已知：\left\{ \begin{align} & x\in N \\ & x^{2}+\left( m-12 \right)x+\left( m-1 \right)=0 \\ \end{align} \right.
  2. 設兩根為 \alpha ,\beta \in N，則：
  3. \left\{ \begin{align} & \alpha +\beta =-\left( m-12 \right) \\ & \alpha \beta =m-1 \\ \end{align} \right.\Rightarrow \alpha \beta +\alpha +\beta =11
    \Rightarrow \alpha \beta +\alpha +\beta +1=11+1\Rightarrow \left( \alpha +1 \right)\left( \beta +1 \right)=12
  4. \left\{ \begin{align} & \alpha ,\beta \in N \\ & \left( \alpha +1 \right)\left( \beta +1 \right)=12 \\ \end{align} \right.\Rightarrow \begin{matrix} \alpha +1= & 2, & 3, & 4, & 6 \\ \beta +1= & 6, & 4, & 3, & 2 \\ \end{matrix}
    \Rightarrow \begin{matrix} \alpha = & 1, & 2, & 3, & 5 \\ \beta = & 5, & 3, & 2, & 1 \\ \end{matrix}\Rightarrow \alpha \beta =5,or,6
  5. \because \alpha \beta =m-1\therefore m=\alpha \beta +1=6,or,7\#

2

  1. \left\{ \begin{align} & m,k,x\in Q,m\ne 3 \\ & x^{2}-4mx+4x+3m^{2}-2m+4k=0 \\ \end{align} \right.
  2. \left\{ \begin{align} & \Delta =b^{2}-4ac=\left( 4-4m \right)^{2}-4\times \left( 3m^{2}-2m+4k \right) \\ & =\left[ 4\left( 1-m \right) \right]^{2}-4\times \left( 3m^{2}-2m+4k \right) \\ & =4\times 4\left( m^{2}-2m+1 \right)-4\times \left( 3m^{2}-2m+4k \right) \\ & =4\times \left( 4m^{2}-8m+4-3m^{2}+2m-4k \right) \\ & =4\times \left( m^{2}-6m+4-4k \right) \\ \end{align} \right.
  3. 因為 x\, 之根為有理數，故判別式必為有理數之完全平方，即：
    m^{2}-6m+4-4k=m^{2}-2\times 3\times m+3^{2}=\left( m-3 \right)^{2}
    \Rightarrow 4-4k=3^{2}\Rightarrow k=-\frac{9-4}{4}\Rightarrow k=-\frac{5}{4}\#

類題1

已知

求解

答案

詳解

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