# Quod Erat Demonstrandum

## 2010/11/05

### 找二次方程

Filed under: NSS — johnmayhk @ 12:05 下午

$x = \frac{1}{y}$，代入

$ax^2 + bx + c = 0$，得

$a(\frac{1}{y})^2 + b(\frac{1}{y}) + c = 0$

$a + by + cy^2 = 0$

$cx^2 + bx + a = 0$

(*) $\left \{ \begin{array}{ll} a(\alpha)^2 + b(\alpha) + c = 0\\ a(\beta)^2 + b(\beta) + c = 0\end{array}\right.$

$\left \{ \begin{array}{ll} y_1 = \frac{1}{\alpha}\\ y_2 = \frac{1}{\beta}\end{array}\right.$

$\left \{ \begin{array}{ll} \alpha = \frac{1}{y_1}\\ \beta = \frac{1}{y_2}\end{array}\right.$

$\left \{ \begin{array}{ll} a(\frac{1}{y_1})^2 + b(\frac{1}{y_1}) + c = 0\\ a(\frac{1}{y_2})^2 + b(\frac{1}{y_2}) + c = 0\end{array}\right.$

$\left \{ \begin{array}{ll} a + b(y_1) + c(y_1)^2 = 0\\ a + b(y_2) + c(y_2)^2 = 0\end{array}\right.$

（注：隨便「換件衫」，得：「$\frac{1}{\alpha}, \frac{1}{\beta}$ 就是 $a + bx + cx^2 = 0$ 的根。」）

::::::::::

$y = x^2$，代入 $ax^2 + bx + c = 0$，得

$ay + bx + c = 0$
$bx = -ay - c$
$b^2x^2 = a^2y^2 + 2acy + c^2$
$b^2y = a^2y^2 + 2acy + c^2$
$a^2y^2 + (2ac - b^2)y + c^2 = 0$

$a^2x^2 + (2ac - b^2)x + c^2 = 0$

::::::::::

$x^2 - (\alpha^3 + \beta^3)x + (\alpha^3 \beta^3) = 0$