公式解:
$x=-\dfrac {b\pm \sqrt {b^{2}-4a}c} {2a}$推導過程:
設一方程式, $ax^{2}+bx+c=0$
求解
- 既然是求根的話,先將 $x^{2}$ 項的係數化為 1 → $x^{2}+\dfrac {b} {a}x+\dfrac {c} {a}=0$
- 為了完整的拆解 $x^{2}$ ,湊出一個完全平方式
→ $\left( x+\dfrac {b} {2a}\right) ^{2}-\left( \dfrac {b} {2a}\right) ^{2}+\dfrac {c} {a}=0$ - 移項
→ $()^{2}=\dfrac {b^{2}-4ac} {\left( 2a\right) ^{2}}$ - 左式先取絕對值,右式開根號
→ $\left\| ()\right\| =\dfrac {\sqrt {b^{2}-4ac}} {2a}$
最後,左式把絕對值拿掉,右式加上正負號,再將 b/2a 移項後,得證
=> $x=-\dfrac {b\pm \sqrt {b^{2}-4a}c} {2a}$
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參考資料
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回覆刪除裡面有很多可以輸入數學公式的方案