课程： 5.5 简化平方根

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选择节简化平方根

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简化平方根
The length of the two legs of a right triangle are $\begin{align*}2\sqrt{5}\end{align*}$ and $\begin{align*}3\sqrt{4}\end{align*}$ . What is the length of the triangle's hypotenuse?
::右三角形两腿的长度是25和34 三角形的下限长度是多少?

Simplifying Square Roots
::简化平方根

Before we can solve a quadratic equation using square roots, we need to review how to simplify , add, subtract, and multiply them. Recall that the square root is a number that, when multiplied by itself, produces another number. 4 is the square root of 16, for example. -4 is also the square root of 16 because $\begin{align*}(-4)^2 = 16\end{align*}$ . The symbol for square root is the radical sign, or $\begin{align*}\sqrt{}\end{align*}$ . The number under the radical is called the radicand . If the square root of an integer is not another integer, it is an irrational number .
::在我们用正方根解决二次方程之前, 我们需要审查如何简化、添加、减、乘。回顾正方根是一个数字, 当它本身乘以后, 就会产生另一个数字。例如。 4 是 16 的平方根。 4 也是 16 的平方根, 因为 (- 4) 2=16 。平方根的符号是激进符号 , 或者。基底下的数字叫做 radicand 。如果整数的平方根不是另一个整数, 它是一个不合理的数字。

Let's find $\begin{align*}\sqrt{50}\end{align*}$ .
::找五十块吧

using a calculator.
::使用计算器。

To plug the square root into your graphing calculator, typically there is a $\begin{align*}\sqrt{}\end{align*}$ or SQRT button. Depending on your model, you may have to enter 50 before or after the square root button. Either way, your answer should be $\begin{align*}\sqrt{50}=7.071067811865 \ldots\end{align*}$ In general, we will round to the hundredths place, so 7.07 is sufficient.
::要将平方根插入您的图形计算器, 通常需要一个或 SQRT 按钮。取决于您的模型, 您可能需要在平方根按钮之前或之后输入50 个。无论哪种方式, 您的回答应该是 50= 7. 071067811865 。一般来说, 我们将四舍五入到第一百个位置, 所以 7. 07 就足够了。

by simplifying the square root.
::通过简化平方根。

To simplify the square root, the square numbers must be “pulled out.” Look for factors of 50 that are square numbers: 4, 9, 16, 25... 25 is a factor of 50, so break the factors apart.
::为了简化平方根,平方数字必须“拉出来 ” 。寻找正方数字的50个因素:4,9,16,25... 25是50个因素,所以打破因素。

$\begin{align*}\sqrt{50}=\sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5 \sqrt{2}\end{align*}$ . This is the most accurate answer.
::50=252=252=52 这是最准确的答案

Radical Rules
::激进激进规则

1. $\begin{align*}\sqrt{ab} = \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\end{align*}$ Any two radicals can be multiplied together.
::1. ab=ab=ab

2. $\begin{align*}x\sqrt{a} \pm y \sqrt{a} = x \pm y \sqrt{a}\end{align*}$ The radicands must be the same in order to add or subtract.
::2. Xaya=xya 弧度必须相同才能增减。

3. $\begin{align*}\left(\sqrt{a}\right)^2 = \sqrt{a}^2 = a\end{align*}$ The square and square root cancel each other out.
::3. (a)2=a2=a 平方根和平方根相互勾销。

Now, let's simplify the following expressions.
::现在,让我们简化以下表达式。

$\begin{align*}\sqrt{45}+\sqrt{80}-2\sqrt{5}\end{align*}$

At first glance, it does not look like we can simplify this. But, we can simplify each radical by pulling out the perfect squares.
::乍一看,我们似乎无法简化这一点。但是,我们可以通过拉出完美的方形来简化每个激进的方形。

$\begin{aligned} \sqrt{45} & = \sqrt{9 \cdot 5} = 3 \sqrt{5} \\ \sqrt{80} & = \sqrt{16 \cdot 5} = 4 \sqrt{5} \end{aligned}$
$\begin{align*}\sqrt{45} &= \sqrt{9 \cdot 5} = 3 \sqrt{5}\\ \sqrt{80} &= \sqrt{16 \cdot 5} = 4 \sqrt{5}\end{align*}$

Rewriting our expression, we have: $\begin{align*}3\sqrt{5}+4\sqrt{5}-2\sqrt{5}\end{align*}$ and all the radicands are the same. Using the Order of Operations, our answer is $\begin{align*}5 \sqrt{5}\end{align*}$ .
::重写我们的表达方式,我们有:35+45-25,所有射线都是相同的。使用行动命令,我们的答复是55。

$\begin{align*}2\sqrt{35} \cdot 4 \sqrt{7}\end{align*}$

Multiply across.
::乘以横跨。

$\begin{aligned} 2 \sqrt{35} \cdot 4 \sqrt{7} = 2 \cdot 4 \sqrt{35 \cdot 7} = 8 \sqrt{245} \end{aligned}$
$\begin{align*}2\sqrt{35} \cdot 4 \sqrt{7} = 2 \cdot 4 \sqrt{35 \cdot 7} = 8 \sqrt{245}\end{align*}$

Now, simplify the radical. $\begin{align*}8\sqrt{245} = 8 \sqrt{49 \cdot 5} = 8 \cdot 7 \sqrt{5} = 56\sqrt{5}\end{align*}$
::现在,简化底线8245=8495=875=565

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the length of the triangle's hypotenuse.
::早些时候,有人要求你找到三角形的下垂长度

We must use the Pythagorean Theorem, which states that the square of one leg of a right triangle plus the square of the other leg equals the square of the hypotenuse.
::我们必须使用Pythagorean Theorem, 它说右三角的一条腿加上另一条腿的正方形等于下方的正方形。

So we are looking for c such that $\begin{align*}(2\sqrt{5})^2 + (3\sqrt{4})^2 = c^2\end{align*}$ .
::因此,我们正在寻找(25)2+(34)2=c2的c。

Simplifying, we get $\begin{align*}4 \cdot 5 + 9 \cdot 4 = c^2\end{align*}$ , or $\begin{align*}20 + 36 = c^2\end{align*}$ .
::简化后,我们得到 45+94=c2,或20+36=c2。

Therefore, $\begin{align*}c^2 = 56\end{align*}$ , so to find c , we must take the square root of 56.
::因此,C2=56,为了找到C, 我们必须以56为平方根。

$\begin{align*}\sqrt{56} = \sqrt{4 \cdot 14} = 2\sqrt{14}\end{align*}$ .

Therefore, $\begin{align*}c = 2\sqrt{14}\end{align*}$ .
::因此,C=214。

Example 2
::例2

Simplify the following radical.
::简化以下激进主义。

$\begin{align*}\sqrt{150}\end{align*}$

Pull out all the square numbers.
::把所有的平方数字都拿出来

$\begin{aligned} \sqrt{150} = \sqrt{25 \cdot 6} = 5 \sqrt{6} \end{aligned}$
$\begin{align*}\sqrt{150}=\sqrt{25 \cdot 6}= 5 \sqrt{6}\end{align*}$

Alternate Method : Write out the of 150.
::替代方法:写出150个。

$\begin{aligned} \sqrt{150} = \sqrt{2 \cdot 3 \cdot 5 \cdot 5} \end{aligned}$
$\begin{align*}\sqrt{150} = \sqrt{2 \cdot 3 \cdot 5 \cdot 5}\end{align*}$

Now, pull out any number that has a pair. Write it once in front of the radical and multiply together what is left over under the radical.
::现在,拉出任何有一对的号码。在激进分子面前写一次, 并把在激进分子之下留下的东西一起乘在一起。

$\begin{aligned} \sqrt{150} = \sqrt{2 \cdot 3 \cdot 5 \cdot 5} = 5 \sqrt{6} \end{aligned}$
$\begin{align*}\sqrt{150} = \sqrt{2 \cdot 3 \cdot {\color{red}5 \cdot 5}}={\color{red}5}\sqrt{6}\end{align*}$

Example 3
::例3

Simplify the following radical.
::简化以下激进主义。

$\begin{align*}2\sqrt{3}-\sqrt{6}+\sqrt{96}\end{align*}$

Simplify $\begin{align*}\sqrt{96}\end{align*}$ to see if anything can be combined. We will use the alternate method above.
::简化 96 , 以确定是否有东西可以合并。我们将使用上面的替代方法。

$\begin{aligned} \sqrt{96} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3} = 2 \cdot 2 \sqrt{6} = 4 \sqrt{6} \end{aligned}$
$\begin{align*}\sqrt{96}= \sqrt{{\color{red}2 \cdot 2} \cdot {\color{blue}2 \cdot 2} \cdot 2 \cdot 3} = {\color{red}2} \cdot {\color{blue}2} \sqrt{6} = 4 \sqrt{6}\end{align*}$

Rewrite the expression: $\begin{align*}2 \sqrt{3} - \sqrt{6} + 4\sqrt{6} = 2\sqrt{3}+3\sqrt{6}\end{align*}$ . This is fully simplified. $\begin{align*}\sqrt{3}\end{align*}$ and $\begin{align*}\sqrt{6}\end{align*}$ cannot be combined because they do not have the same value under the radical.
::重写表达式: 23-6+46=23+36。完全简化。 3 和 6 不能合并, 因为它们在基底下没有相同的值。

Example 4
::例4

Simplify the following radical.
::简化以下激进主义。

$\begin{align*}\sqrt{8} \cdot \sqrt{20}\end{align*}$

This problem can be done two different ways.
::这一问题可以采取两种不同的方式。

First Method : Multiply radicals, then simplify the answer.
::第一种方法:乘以基,然后简化答案。

$\begin{aligned} \sqrt{8} \cdot \sqrt{20} = \sqrt{160} = \sqrt{16 \cdot 10} = 4 \sqrt{10} \end{aligned}$
$\begin{align*}\sqrt{8} \cdot \sqrt{20} = \sqrt{160} = \sqrt{16 \cdot 10} = 4\sqrt{10}\end{align*}$

Second Method : Simplify radicals, then multiply.
::第二种方法: 简化基体, 然后乘以。

$\begin{aligned} \sqrt{8} \cdot \sqrt{20} = (\sqrt{4 \cdot 2}) \cdot (\sqrt{4 \cdot 5}) = 2 \sqrt{2} \cdot 2 \sqrt{5} = 2 \cdot 2 \sqrt{2 \cdot 5} = 4 \sqrt{10} \end{aligned}$
$\begin{align*}\sqrt{8} \cdot \sqrt{20} = \left(\sqrt{4 \cdot 2}\right) \cdot \left(\sqrt{4 \cdot 5}\right)= 2 \sqrt{2} \cdot 2 \sqrt{5}=2 \cdot 2 \sqrt{2 \cdot 5} = 4 \sqrt{10}\end{align*}$

Depending on the complexity of the problem, either method will work. Pick whichever method you prefer.
::视问题的复杂性而定,两种方法都有效。选择你喜欢哪种方法。

Review
::回顾

Find the square root of each number by using the calculator. Round your answer to the nearest hundredth.
::使用计算器来查找每个数字的平方根。将您的答案四舍五入到最接近的百位。

56

12

92

Simplify the following radicals. If it cannot be simplified further, write " cannot be simplified ".
::简化以下基体。如果无法进一步简化, 请写“ 不能简化 ” 。

$\begin{align*}\sqrt{18}\end{align*}$

$\begin{align*}\sqrt{75}\end{align*}$

$\begin{align*}\sqrt{605}\end{align*}$

$\begin{align*}\sqrt{48}\end{align*}$

$\begin{align*}\sqrt{50} \cdot \sqrt{2}\end{align*}$

$\begin{align*}4\sqrt{3} \cdot \sqrt{21}\end{align*}$

$\begin{align*}\sqrt{6} \cdot \sqrt{20}\end{align*}$

$\begin{align*}\left(4\sqrt{5}\right)^2\end{align*}$

$\begin{align*}\sqrt{24} \cdot \sqrt{27}\end{align*}$

$\begin{align*}\sqrt{16} + 2\sqrt{8}\end{align*}$

$\begin{align*}\sqrt{28} + \sqrt{7}\end{align*}$

$\begin{align*}-8 \sqrt{3} - \sqrt{12}\end{align*}$

$\begin{align*}\sqrt{72} - \sqrt{50}\end{align*}$

$\begin{align*}\sqrt{6} +7 \sqrt{6} - \sqrt{54}\end{align*}$

$\begin{align*}8 \sqrt{10} - \sqrt{90}+7\sqrt{5}\end{align*}$

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

简化平方根

Simplifying Square Roots ::简化平方根

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Simplifying Square Roots
::简化平方根

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)