WEBVTT

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在这段视频中，我将向你们展示一种技巧

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叫做配方法。

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它的奇妙之处在于它适用于任何二次方程，

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它实际上是

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二次方程的基。

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在下个或者在那之后的视频，

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我会用配方法来证明这个二次公式。

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但在此之前，我们需要理解。

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它到底是怎么回事。

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这是建立在上一集视频基础上的，

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我们用完全平方

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求解了二次方程。

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已知二次方程x² - 4x

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等于5。

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我在这里放这么大的空间是有原因的。

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在上一集视频中，我们知道

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如果左边是完全平方，

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就可以很直接地求解出来。

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你看，配方就是把二次方程变成完全平方，

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对它进行工程处理，

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两边加减

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使它变成完全平方。

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我们该怎么做呢?

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为了使左边是完全平方的，

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这里必须有某个数。

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这里一定有某个数，

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如果我有这个数的平方，

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我用2乘以这个数，我得到-4。

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记住这一点，

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我想通过几个例子你就会明白。

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我想要x²- 4x

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加上某项等于x - a²。

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我们还不知道a是多少，

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但我们知道一些东西。

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当我平方时，这个等于

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x²- 2a + a²。

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如果你看这里的模式，它必须是 --

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抱歉，x² - 2ax -- 这里必须是2ax。

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这里应该是a的平方。

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所以这个数，a是-4的一半，

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a是-2，对吧

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因为2乘以a等于-4。

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a是-2，如果a是-2，a的平方是多少?

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那么a的平方就是+4。

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现在你们可能觉得这些很复杂，

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但是我给你们展示的是基本原理。

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你只需要看这里的系数，

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好，系数的一半是多少?

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系数的一半是-2。

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所以我们可以说a等于-2，同样的，

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然后平方。

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对a平方，得到正4。

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所以这里加上4。

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加4。

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从我们做过的第一个方程，

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你应该知道

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不能只在方程的一边运算。

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不能只在等式的一边加上4。

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如果x²- 4x = 5，那么当我加4时，

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它就不再等于5了。

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这将会是5 + 4。

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我们在左边加了4，

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因为我们想让它是一个完全平方。

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但是如果你在左边加上某项，

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你必须在右边加上它。

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现在，我们遇到了一个

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和上个视频中做的问题一样的问题。

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左边是什么？

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我把整个式子重写一下。

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我们现在得到x² - 4x + 4 = 9。

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我们所做的就是方程两边同时加上4。

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但是我们为了左边变成完全平方，

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而加上了4。

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这是什么?

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什么数乘以它本身等于4？

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而当我把它自身相加等于-2?

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我们已经回答了这个问题。

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它是- 2。

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我们得到(x - 2)(x - 2) = 9。

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或者我们可以跳过这一步

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写成(x - 2)² = 9。

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然后两边同时开根号，

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就得到x - 2 = ±3。

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两边同时加上2，就得到x = 2 ± 3。

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这告诉我们x可以等于2 + 3，也就是5。

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或者x可以等于2 - 3，也就是-1。

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我们完成了。

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现在我想说明一下。

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你可以完成这个方程而不用配方法。

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我们可以从

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x²- 4x = 5开始。

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我们可以两边同时减去5，

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得到x² - 4x - 5 = 0。

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你可以这样想，

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如果我有一个-5乘以1，那么它们的乘积是-5，

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它们的和是-4。

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所以我可以说这是

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（x - 5)(x + 1) = 0。

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然后我们说x等于5

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或者x等于-1。

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在这种情况下，

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实际上可能有更快的解题方法。

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但是配方法的奇妙之处

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就在于它总是有效的。

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它总是有效的，不管系数是多少，

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不管问题有多复杂。

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我来证明一下。

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我们来做一个

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传统上会很麻烦的问题，

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如果我们试着做分解，特别是用分组

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就像这样。

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假设我们现在有

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10x² - 30x - 8 = 0。

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现在，从一开始，

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我们可以两边除以2。

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这确实简化了一点。

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我们两边同时除以2。

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如果把所有数都除以2，会得到什么?

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我们得到5x² - 15x - 4 = 0。

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但是同样的，现在系数前面有一个疯狂的5，

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我们必须通过分组来解决它。

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这是一个相当痛苦的过程。

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但现在我们可以直接完成这个配方，

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为了完成这个，

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我现在要除以5得到1的系数。

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你们会看到为什么

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这和我们传统的做法不同。

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如果我把这整个除以5，

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我可以从一开始就除以10，

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但我想先做这个，只是为了让你们知道

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题目并没有告诉我们多少。

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所有项都除以5。

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如果每项都除以5，

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就得到x² - 3x - 4/5 = 0。

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你可能会问，

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为什么我们要用分组来分解呢?

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