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# Find the solutions x of the equation 2ix^2 + x +3i = 0

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Find the solutions x of the equation 2ix^2 + x +3i = 0

Nov 8, 2017

#1
+21819
+2

Find the solutions x of the equation 2ix^2 + x +3i = 0

$$\begin{array}{|rcll|} \hline \mathbf{ax^2+bx+c} &\mathbf{=}& \mathbf{0} \\ \mathbf{x} &\mathbf{=}& \mathbf{{-b \pm \sqrt{b^2-4ac} \over 2a}} \\\\ 2ix^2 + x +3i &=& 0 \quad & | \quad a = 2i \quad b = 1 \quad c = 3i \\ x &=& \dfrac{-1\pm \sqrt{1^2-4\cdot(2i) \cdot (3i)} } {2\cdot 2i } \\ &=& \dfrac{-1\pm \sqrt{1-24i^2} } {4i } \quad & | \quad i^2 = -1 \\ &=& \dfrac{-1\pm \sqrt{1+24} } {4i } \\ &=& \dfrac{-1\pm \sqrt{25} } {4i } \\ &=& \dfrac{-1\pm 5 } {4i } \cdot \dfrac{i}{i} \\ &=& \dfrac{ (-1\pm 5)i } {4i^2 } \quad & | \quad i^2 = -1 \\ &=& \dfrac{ (-1\pm 5)i } {-4} \\ \\ x_1 &=& \dfrac{ (-1+ 5)i } {-4} \\ &=& \dfrac{ 4i } {-4} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{-i} \\\\ x_2 &=& \dfrac{ (-1- 5)i } {-4} \\ &=& \dfrac{ -6i } {-4} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{ \dfrac{3} {2}i } \\ \hline \end{array}$$

Nov 8, 2017

#1
+21819
+2
$$\begin{array}{|rcll|} \hline \mathbf{ax^2+bx+c} &\mathbf{=}& \mathbf{0} \\ \mathbf{x} &\mathbf{=}& \mathbf{{-b \pm \sqrt{b^2-4ac} \over 2a}} \\\\ 2ix^2 + x +3i &=& 0 \quad & | \quad a = 2i \quad b = 1 \quad c = 3i \\ x &=& \dfrac{-1\pm \sqrt{1^2-4\cdot(2i) \cdot (3i)} } {2\cdot 2i } \\ &=& \dfrac{-1\pm \sqrt{1-24i^2} } {4i } \quad & | \quad i^2 = -1 \\ &=& \dfrac{-1\pm \sqrt{1+24} } {4i } \\ &=& \dfrac{-1\pm \sqrt{25} } {4i } \\ &=& \dfrac{-1\pm 5 } {4i } \cdot \dfrac{i}{i} \\ &=& \dfrac{ (-1\pm 5)i } {4i^2 } \quad & | \quad i^2 = -1 \\ &=& \dfrac{ (-1\pm 5)i } {-4} \\ \\ x_1 &=& \dfrac{ (-1+ 5)i } {-4} \\ &=& \dfrac{ 4i } {-4} \\ \mathbf{x_1} &\mathbf{=}& \mathbf{-i} \\\\ x_2 &=& \dfrac{ (-1- 5)i } {-4} \\ &=& \dfrac{ -6i } {-4} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{ \dfrac{3} {2}i } \\ \hline \end{array}$$