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== 公式 ==
壹元二次方程''x''<sup>2</sup>+''px''+''q''=0, p≠0,q≠0有李煌解：

<math>x_{1}=\frac{q}{-p{sin^{2}\left(\frac{\arcsin\left(\frac{2}{\pm\sqrt\frac{p^2}{q}}\right)}{2}\right)}}</math>

<math>x_{2}=\frac{q}{-p{cos^{2}\left(\frac{\arcsin\left(\frac{2}{\pm\sqrt\frac{p^2}{q}}\right)}{2}\right)}}</math>

==李煌出的习题==

代数方程<math>x^2+px=q,p>0,q>0</math>之解为

<math>x_1{={\frac {\sqrt{2}q} {\sqrt{p^2+2q+\sqrt{p^4+4p^2q}}}}}</math>

<math>x_2{={\frac {-\sqrt{2}q} {\sqrt{p^2+2q-\sqrt{p^4+4p^2q}}}}}</math>

代数方程<math>x^2+px=q,p>0,q<0</math>之解为

<math>x_1{={\frac {\sqrt{2}q} {\sqrt{p^2+2q+\sqrt{p^4+4p^2q}}}}}</math>

<math>x_2{={\frac {\sqrt{2}q} {\sqrt{p^2+2q-\sqrt{p^4+4p^2q}}}}}</math>

代数方程<math>x^2+px=q,p<0,q>0</math>之解为

<math>x_1{={\frac {-\sqrt{2}q} {\sqrt{p^2+2q+\sqrt{p^4+4p^2q}}}}}</math>

<math>x_2{={\frac {\sqrt{2}q} {\sqrt{p^2+2q-\sqrt{p^4+4p^2q}}}}}</math>

代数方程<math>x^2+px=q,p<0,q<0</math>之解为

<math>x_1{={\frac {-\sqrt{2}q} {\sqrt{p^2+2q+\sqrt{p^4+4p^2q}}}}}</math>

<math>x_2{={\frac {-\sqrt{2}q} {\sqrt{p^2+2q-\sqrt{p^4+4p^2q}}}}}</math>

== 來源 ==
* 《南昌理工學院學報》.李煌

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