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# 求根公式证明复习

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

$a{x}^{2}+bx+c=0$

## 推导

### 第一步：配方法

$\begin{array}{rlr}a{x}^{2}+bx+c& =0& \left(1\right)\\ \\ a{x}^{2}+bx& =-c& \left(2\right)\\ \\ {x}^{2}+\frac{b}{a}x& =-\frac{c}{a}& \left(3\right)\\ \\ {x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}& =\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}& \left(4\right)\\ \\ {\left(x+\frac{b}{2a}\right)}^{2}& =\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}& \left(5\right)\end{array}$

## 第二步：代数！代数！代数！

$\begin{array}{rlr}{\left(x+\frac{b}{2a}\right)}^{2}& =\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}& \left(5\right)\\ \\ {\left(x+\frac{b}{2a}\right)}^{2}& =\frac{{b}^{2}}{4{a}^{2}}-\frac{4ac}{4{a}^{2}}& \left(6\right)\\ \\ {\left(x+\frac{b}{2a}\right)}^{2}& =\frac{{b}^{2}-4ac}{4{a}^{2}}& \left(7\right)\\ \\ x+\frac{b}{2a}& =±\frac{\sqrt{{b}^{2}-4ac}}{\sqrt{4{a}^{2}}}& \left(8\right)\\ \\ x+\frac{b}{2a}& =±\frac{\sqrt{{b}^{2}-4ac}}{2a}& \left(9\right)\\ \\ x& =-\frac{b}{2a}±\frac{\sqrt{{b}^{2}-4ac}}{2a}& \left(10\right)\\ \\ x& =\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}& \left(11\right)\end{array}$