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Consider the following quadratic equation: x2 = 9. If asked to solve it, we would naturally take the square root of 9 and end up with 3 and -3. But what if simple square root methods won't do? What if the equation includes x raised to the first power and cannot be easily factored?
考虑下面的二次方程： x 2 = 9 。 如果要求解决它，我们自然会取9的平方根，最后得到3和-3 。 但是，如果简单的平方根方法不起作用怎么办？ 如果方程式包含x升至一阶幂且不易分解的情况怎么办？
Fortunately, there is a method for completing the square. As as result, a quadratic equation can be solved by taking the square root. Let's explore this step by step together.
幸运的是，有一种完成平方的方法。 结果，可以通过取平方根来求解二次方程。 让我们一起逐步探索这一点。
Say we are given the following equation:
假设我们得到以下方程式：
Let's simplify our equation. First, separate the terms that include variables from the constant terms. Next, subtract x from 13x (result is 12x) and subtract 7 from 6 (result is -1).
让我们简化方程式。 首先，将包含变量的项与常数项分开。 接下来，从13x减去x (结果为12x )，然后从6减去7 (结果为-1 )。
The method of completing the square works a lot easier when the coefficient of x2 equals 1. The coefficient in our case equals 4. Dividing 4 into each member results in x2 + 3x = - 1/4.
当x 2的系数等于时，完成平方的方法容易得多 1 。 在我们的情况下，系数等于4 。 划分 每个成员4个结果为x 2 + 3x =-1/4 。
First we need to find the constant term of our complete square. The coefficient of x, which equals 3 is divided by 2 and squared, giving us 9/4.
首先，我们需要找到完整平方的常数项。 x的系数，等于 3除以2并平方，得到9/4 。
Then we add and subtract 9/4 as shown above. Doing so does not affect our equation (9/4 - 9/4 = 0), but gives us an expression for the complete square x2 + 3x + 9/4.
然后我们如上所示加减9/4 。 这样做不会影响我们的方程( 9/4-9/4 = 0 )，但会为我们提供完整平方x 2 + 3x + 9/4的表达式。
Let's now remember a more general (x + a)2 = x2 + 2ax + a2 and use it in the current example. Substituting our numbers gives us: x2 + 3x + 9/4 = x2 + 2*(3/2)*x + (3/2)2 = (x + 3/2)2.
现在让我们记住一个更一般的(x + a) 2 = x 2 + 2ax + a 2并在当前示例中使用它。 代入数字得到： x 2 + 3x + 9/4 = x 2 + 2 *(3/2)* x +(3/2) 2 = (x + 3/2) 2 。
Finally, taking the square root from both sides gives us √(x + 3/2)2 = ±√2. Or simply x + 3/2 = ±√2. We conclude this by solving for x: X1= √2 - 3/2 and X2 = - √2 - 3/2.
最后，从两边取平方根即可得出√(x + 3/2) 2 =±√2 。 或者简单地 x + 3/2 =±√2 。 我们通过求解x得出结论： X 1 =√2-3/2 和X 2 =-√2-3-2/2 。
Simplify by separating the terms with variables from constant terms. Then perform subtraction and addition on both sides of the equation.
通过将带有变量的术语与恒定项分开来简化。 然后在方程的两边进行减法和加法。
Here, the coefficient of X2 already equals 1, so no further action needed.
在此， X 2的系数已经等于1 ，因此不需要其他操作。
As in previous example, we find the constant term of our complete square. The coefficient of x, which equals -8 is divided by 2 and squared, giving us 16.
与前面的示例一样，我们找到了完整平方的常数项。 x的系数，等于 -8除以2并平方，得到16 。
We add and subtract 16 and can see that x2 - 8x + 16 gives us a complete square.
我们加减16，可以看出，X 2 - 8X + 16为我们提供了一个完整的正方形。
Since the constant term -8 is with the minus sign, we use this general form: (x - a)2 = x2 - 2ax + a2. Using our numbers gives us: x2 - 8x + 16 = x2 - 2*(4)*x + (4)2 = (x - 4)2.
由于常数项-8与减号，我们使用这种一般形式为：(X - A)2 = X 2 - 2AX + 2。 使用我们的数字给了我们：X 2 - 8X + 16 = X 2 - 2 *(4)* X +(4)2 =(X - 4)2。
Finally, taking the square root from both sides gives us √(x - 4)2 = ±√11. Or simply x - 4 = ±√11. We conclude this by solving for x: X1 = 4 + √11 and X2 = 4 - √11
最后，从两边取平方根即可得出√(x-4) 2 =±√11 。 或者简单地 x-4 =±√11 。 我们通过求解x得出结论： X 1 = 4 +√11 并且X 2 = 4-√11
And there you have it!
在那里，您拥有了！
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