10.2 评价nth根

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  A professional baseball diamond is 8,100 square feet, and a little league baseball diamond is 3,600 square feet. What is the difference between the distances from home plate to 1st base between the fields?
  ::职业棒球钻石是8,100平方英尺,而一个小联盟棒球钻石是3,600平方英尺。 从家用车牌到田间第一基地之间的距离有什么区别?

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  We can find the length of one side of the diamond by using roots, which we will discuss in this section.
  ::我们可以通过利用根源找到钻石的一方长度,我们将在本节中讨论这个问题。

  

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  Principal n ^th Roots
  ::主要 nth 根根

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  Before we consider roots other than the square root , let's discuss the vocabulary we use to describe roots.
  ::在我们考虑除平方根以外的根之前, 让我们来讨论我们用来描述根的词汇。

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     Principal n ^th Roots
  ::主要 nth 根根

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  For   $\begin{array}{r}n\ge 2,\end{array}$  where  $\begin{array}{r}n\end{array}$  is an integer , the principal root of   $\begin{array}{r}\sqrt[n]{a}\end{array}$   is   $\begin{array}{r}b,\end{array}$  where   $\begin{array}{r}a=b^n\end{array}$ .
  ::n2, 其中n为整数, a的主要根为 b, 其中 a=bn 。

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       If $\begin{array}{r}n\end{array}$  is even, $\begin{array}{r}a\ge 0\end{array}$  and  $\begin{array}{r}b\ge 0\end{array}$ .
  ::如果 n 是偶数, a0 和 b0 。

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       If $\begin{array}{r}n\end{array}$  is odd, $\begin{array}{r}a\end{array}$  is a real number and so is $\begin{array}{r}b\end{array}$ .
  ::如果 n 是 奇数, a 是 实际数字, b 也是 。

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  For even values of n , to indicate the other root—that is, the other solution of  $\begin{array}{r}a=b^n\end{array}$ —we write the root as  $\begin{array}{r}-\sqrt[n]{a}\end{array}$ .  
  ::对n的值来说,表示另一个根——即a=bn的另一种解决办法——我们将根写成an。

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  Vocabulary
  ::词汇表

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  $\begin{array}{r}\sqrt[n]{a}\end{array}$
  ::a , a , a , a , a , a

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  $\begin{array}{r}\sqrt{}\end{array}$   This symbol is called the  radical .
  ::这个符号叫做激进符号。

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  $\begin{array}{r}a\end{array}$   This symbol is called the  radicand .
  ::a 这个符号称为射线。

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  $\begin{array}{r}n\end{array}$  This symbol is called the  index.
  ::n 此符号称为索引。

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       When  $\begin{array}{r}n=2\end{array}$ , which we tend not to  write explicitly, the root is called the square root .
  ::当 n=2 时, 根被称为平方根, 而我们往往不明确写入 n=2 。

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       When $\begin{array}{r}n=3\end{array}$ , the root is called the cube root .  
  ::n=3 时, 根被称为立方根 。

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       For all other values of n , we call the roots the n ^th root. 
  ::对于n的所有其他价值, 我们称根为nth根。

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  Evaluating n ^th Roots
  ::评估 nth 根

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  Let's use our knowledge of square roots to determine how to find roots with a different index.
  ::让我们利用我们对正根的知识 来决定如何用不同的指数找到根。

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  Example 1
  ::例1

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  A professional baseball diamond is 8,100 square feet, and a little league baseball diamond is 3,600 square feet ¹ . What is the difference between the distances from home plate to 1st base between the fields?
  ::专业棒球钻石是8,100平方英尺, 小型棒球钻石是3,600平方英尺。 从家用车牌到第一基地的距离, 在田地之间有什么区别?

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  Solution:  A baseball diamond is a square. To find the area of a square, we square the length of a side. To find the length of one side, we can thus take the square root of the area.
  ::解决方案: 棒球钻石是方形。 要找到方形区域, 我们将方形平开。 要找到一方的长度, 我们就可以选择方形平方形平方形平方形平方形平方形平方形平方形平方形平方形平方形平方形平方形平方形平方形。

  $$\begin{array}{rlr}\sqrt{8,100}=90& & \sqrt{3,600}=60\end{array}$$

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  Notice that $\begin{array}{r}90\cdot 90=8,100,\end{array}$  or the product of two 90s is 8,100. Likewise, $\begin{array}{r}60\cdot 60=3,600\end{array}$  or the product of two 60s is 3,600. We will use the idea that the index is the number of times we multiply the root by itself for roots with a higher index.
  ::请注意,90-90=8,100,或两个90年代的产物为8,100。同样,60-60=3,600,或两个60年代的产物为3,600。我们将使用这个概念,即指数是用更高的指数将根根本身乘以根数的倍数。

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  To answer the question, the distance between home plate and 1st base on a little league field is 60 feet, and the distance between home plate and 1st base on a professional field is 90 feet. There is a difference of  $\begin{array}{r}90-60=30\end{array}$  feet.
  ::要回答这个问题,家用车牌和一垒基之间的距离是60英尺,家用车牌和一垒基之间的距离是90英尺,相差90-60=30英尺。

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     Evaluating n ^th Roots
  ::评估 nth 根

  To evaluate  n ^th roots, find a number that multiplied by itself the number of times in the index is the radicand. It is helpful to find the prime factorization of the number to do this. 

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  This video by CK-12 demonstrates how to find real n ^th  roots of $\begin{array}{r}a\end{array}$  (where n is an integer).    
  ::CK-12的这段影片展示了如何找到一个(n是整数的)真实的 nth 根。

   

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  Example 2
  ::例2

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  Find $\begin{array}{r}\sqrt[6]{729}\end{array}$ .
  ::查找7296号

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  Solution: To find  the 6th root of 729, we need to find the prime factorization. 
  ::解决方案:为了找到729的第六根, 我们需要找到质因子化。

  $$\begin{array}{r}729=3\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3=3^6\end{array}$$

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  Therefore , $\begin{array}{r}\sqrt[6]{729}=\sqrt[6]{3^6}=3\end{array}$ . We say that 3 is the 6th root of 729.
  ::因此,7296=366=3. 我们说3是729的第6根根

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  This video by CK-12 demonstrates how to find n ^th  roots of $\begin{array}{r}a\end{array}$  (where n is an even integer and $\begin{array}{r}n>1\end{array}$  ).    
  ::CK-12的这段视频展示了如何找到一个(n是偶整数和n>1的) nth根。

  
   

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  Example 3
  ::例3

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  Evaluate :
  ::评价:

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  a.  $\begin{array}{r}\sqrt[5]{{32}^3}\end{array}$
  ::a. 3235

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  b.  $\begin{array}{r}\sqrt[3]{\text{-}216}\end{array}$
  ::b. -2163

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  Solution:
  ::解决方案 :

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  a. Find the prime factorization of 32, and then we will have three copies of those values due to the exponent . 
  ::a. 找出32的基数乘数,然后我们得到3份这些数值的副本,作为指数。

  $$\begin{array}{rl}32& =2\cdot 2\cdot 2\cdot 2\cdot 2=2^5\\ {32}^3& =2^5\cdot 2^5\cdot 2^5=2^{15}\end{array}$$

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  We need to then divide that into five groups as we do below. 
  ::然后,我们需要像下面那样,将这分为五组。

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  $$\begin{array}{rl}{2}^{15}& =2^3\cdot 2^3\cdot 2^3\cdot 2^3\cdot 2^3\\ \\ \sqrt[5]{{32}^3}& =\sqrt[5]{(2^3{)}^5}=\sqrt[5]{8^5}=8\end{array}$$

  As we will see in an upcoming section, we can apply the root 1st and then the exponent. 
  ::215=232323232323232335=(2335=855=855=8A,我们可以看到,在即将到来的一节中,我们可以应用根1,然后应用引号。

  $$\begin{array}{r}(\sqrt[5]{32}{)}^3=2^3=8\end{array}$$

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  b.  We need to find the prime factorization of 216. 
  ::b. 我们需要找到216个主要因素。

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  $$\begin{array}{rl}216& =4\cdot 54\\ & =2\cdot 2\cdot 2\cdot 27\\ & =2\cdot 2\cdot 2\cdot 3\cdot 9\\ & =2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 3\\ & =2^3\cdot 3^3\\ & ={(2\cdot 3)}^3\end{array}$$

  The cube root of $\begin{array}{r}\text{-}216\end{array}$  is then 
  ::216=454=2222227=22222}39=2222223}33=2333=(23)3 3 立方根为 -216。

  $$\begin{array}{r}\sqrt[3]{\text{-}216}=\sqrt[3]{(\text{-}1{)}^3(2\cdot 3{)}^3}=\sqrt[3]{(\text{-}2\cdot 3{)}^3}=\sqrt[3]{(\text{-}6{)}^3}=\text{-}6\end{array}$$

    

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     Properties of Radicals
  ::激进党的属性

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  $$\begin{array}{rl}\sqrt[n]{ab}& =\sqrt[n]{a}\cdot \sqrt[n]{b}\\ \\ \sqrt[n]{\frac{a}{b}}& =\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\\ \\ (\sqrt[n]{a}{)}^b& =\sqrt[n]{a^b}\\ \\ \sqrt[m]{\sqrt[n]{a}}& =\sqrt[mn]{a}\end{array}$$

  
  ::abn=anbnabn=anbn(an)b=abnanm=amn (an)b=abnanm=amn)

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  Note that all these properties are related to properties of exponents, as we will see later in this chapter.
  ::请注意,所有这些财产都与借方财产有关,我们将在本章下文中看到这一点。

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  Example 4
  ::例4

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  Simplify:
  ::简化 :

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  a.  $\begin{array}{r}\sqrt[4]{625z^8}\end{array}$
  ::a. 625z84

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  b.  $\begin{array}{r}\sqrt[4]{64}\end{array}$
  ::b. 644

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  c. $\begin{array}{r}\sqrt[5]{9,216}\end{array}$
  ::c. 9 2165

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  Solution:
  ::解决方案 :

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  a. First, you can separate this expression  into two separate  roots by the 1st property :  $\begin{array}{r}\sqrt[4]{625}\cdot \sqrt[4]{z^8}\end{array}$ . Now  find  each root.
  ::a. 首先,您可以通过第一个属性6254z84将这个表达式分为两个独立的根。现在找到每个根。

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  $$\begin{array}{r}\sqrt[4]{625}\cdot \sqrt[4]{z^8}=\sqrt[4]{5^4}\cdot \sqrt[4]{(z^2{)}^4}=5z^2\end{array}$$

  
  ::6254Z84=544(z2,44=5z2)

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  b. To simplify the 4th root of a number, write the prime factorization of 64:  $\begin{array}{r}64=2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^6\end{array}$ . 
  ::b. 为简化数字的第四根,写入64:64=22222=26。

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  Notice that there are six 2's in 64. Using the 1st property, we can separate the radical with a group of four 2's and a group of the remaining 2's. 
  ::注意64分有6 2分 使用第1个财产 我们可以把激进分子和4 2分和其余2分分开

  $$\begin{array}{r}\sqrt[4]{64}=\sqrt[4]{2^6}=\sqrt[4]{2^4}\cdot \sqrt[4]{2^2}=2\sqrt[4]{4}\end{array}$$

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  c. Find the prime factorization of  9,216, and place factors into groups of five.
  ::c. 确定9 216的初级因数,并将各种因素分为五组。

  $$\begin{array}{rl}\sqrt[5]{9216}& =\sqrt[5]{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3}\\ & =\sqrt[5]{2^5\cdot 2^5\cdot 3^2}\\ & =2\cdot 2\sqrt[5]{3^2}\\ & =4\sqrt[5]{9}\end{array}$$

   

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  by Mathispower4u demonstrates how to simplify radicals.     
  ::Mathispower4u 演示如何简化激进。

    

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  Example 5
  ::例5

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  Simplify :
  ::简化 :

   

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  a. $\begin{array}{r}\sqrt[4]{\frac{160}{81}}\end{array}$
  ::a. 160814

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  b.  $\begin{array}{r}\sqrt[3]{\frac{54x^3}{125y^5}}\end{array}$
  ::b. 54x3125y53

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  Solution:
  ::解决方案 :

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  a . Using the 2nd property, separate the root to the numerator and denominator.
  ::a. 使用第二个属性,将根与分子和分母分开。

  $$\begin{array}{r}\sqrt[4]{\frac{160}{81}}=\frac{\sqrt[4]{160}}{\sqrt[4]{81}}=\frac{\sqrt[4]{16}\cdot \sqrt[4]{10}}{\sqrt[4]{81}}=\frac{2\sqrt[4]{10}}{3}\end{array}$$

   

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  b. Again, using the 2nd property , we can separate the numerator and denominator.
  ::b. 同样,使用第二个属性,我们可以分离分子和分母。

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  $$\begin{array}{r}\sqrt[3]{\frac{54x^3}{125y^5}}=\frac{\sqrt[3]{54x^3}}{\sqrt[3]{125y^5}}=\frac{\sqrt[3]{2\cdot 3\cdot 3\cdot 3\cdot x^3}}{\sqrt[3]{5\cdot 5\cdot 5\cdot y^3\cdot y^2}}=\frac{3x\sqrt[3]{2}}{5y\sqrt[3]{y^2}}\end{array}$$

  
  ::54x3125y53 = 54x33125y53 = 2}3}3}3}3}3×3}3}3}3}3}3}3}3}3}3}5}5}5}5}}}y}3}23=3×23}3×23}23

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  Example 6
  ::例6

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  Simplify :
  ::简化 :

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  a.  $\begin{array}{r}(\sqrt[5]{7}{)}^6\end{array}$
  ::a. (75)6

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  b.  $\begin{array}{r}\sqrt[3]{\sqrt{64x^6}}\end{array}$
  ::b. 64x63

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  Solution:  
  ::解决方案 :

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  a. Using the 3rd property, we have
  ::a. 使用第三财产,我们拥有

  $$\begin{array}{r}(\sqrt[5]{7}{)}^6=\sqrt[5]{7^6}=\sqrt[5]{7^5\cdot 7}=7\sqrt[5]{7}.\end{array}$$

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  b. We can multiply the indexes to create one index according to the 4th property.  
  

  $$\begin{array}{r}\sqrt[3]{\sqrt{64x^6}}=\sqrt[3\cdot 2]{64x^6}=\sqrt[6]{2^6\cdot x^6}=2x\end{array}$$

  
  ::b. 我们可以根据第四属性乘以指数来创建一个索引。 64x63=64x63}2=26x66=2x

   

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  Feature: To Mars and Back
  ::地物:到火星和后退

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  by Deirdre Mundy
  ::由Deirdre Mundy 编辑

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  What would a group of astronauts need to travel to Mars and back? At the very least, they would need a ship with enough fuel to escape Earth's orbit, travel to Mars, and then escape Mars's orbit so they could return home. NASA wants to send astronauts to Mars by 2035. Will a manned mission to the Red Planet be possible? It all comes down to the type of fuel that is used.
  ::一群宇航员需要前往火星和返回火星?他们至少需要一艘具有足够燃料的船舶,以逃离地球轨道,前往火星,然后逃离火星轨道,以便返回本国。NASA希望到2035年派遣宇航员到火星。有人造访红行星是可能的吗?这一切都取决于所使用的燃料类型。

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  Leaving Orbit
  ::离开轨道

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  Escape velocity is the speed a spaceship needs to reach to break free from a planet's gravitational attraction. The equation for escape velocity is $\begin{array}{r}v=\sqrt{\frac{2GM}{R}},\end{array}$  where $\begin{array}{r}G\end{array}$ is the universal gravitational constant , $\begin{array}{r}M\end{array}$ is the mass of the planet, and $\begin{array}{r}R\end{array}$ is the radius of the planet. The more massive a planet, the harder it is to break out of its orbit. Escaping orbit requires a lot of power. If a rocket were to carry enough fuel to escape two different orbits, it would have to be quite large and expensive.
  ::逃跑速度是宇宙飞船从地球引力吸引力中解脱出来所需的速度。 逃跑速度的方程式是 v=2GMR, G是宇宙引力常数, M是行星的质量, R是行星的半径。 行星越大, 越难脱离轨道。 脱轨轨道需要很大的动力。 如果火箭携带足够的燃料以逃离两个不同的轨道, 它必须非常庞大和昂贵。

  

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  Some scientists believe fusion power could provide the answer. Nuclear fusion, a process in which the nuclei of atoms collide, generates a lot of energy from little fuel. Nuclear-powered spaceships could travel to Mars and back, completing the entire journey in about six months. The only problem is that we are not yet sure how to build a ship that can be powered by nuclear fusion.
  ::一些科学家认为核聚变能可以提供答案。 核聚变是一个原子核相撞的过程,它从少量燃料中产生大量能量。 核动力飞船可以往返火星,在大约六个月内完成整个旅程。 唯一的问题是我们还不确定如何建造一艘核聚变能的船舶。

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  British scientists have proposed sending two rockets to Mars. The 1st rocket would carry a robot capable of mining Mars for fuel, and the 2nd would carry the astronauts. When the astronauts arrive on Mars, the fuel for their trip home would be waiting at their destination. Right now, several nations and private companies are all working to send people to Mars. Each group has different ideas about how to make the trip. But perhaps in a just few decades, there will be human beings walking and living on another world.
  ::英国科学家们已经提议向火星发射两枚火箭。 第一枚火箭将携带一个能够开采火星燃料的机器人,第二枚火箭将携带宇航员。 当宇航员抵达火星时,他们回家的燃料将在目的地等待。现在,若干国家和私营公司都在努力将人们送入火星。每个团体对于如何进行这次旅行都有不同的想法。但也许在短短几十年内,就会有人类在另一个世界行走和生活。

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  Summary
  ::摘要

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  • To evaluate n ^th roots, find a number that multiplied by itself the number of times in the index is the radicand. It is helpful to find the prime factorization of the number to do this. 
    ::要评估 nth 根, 需要找到一个数字, 该数字本身会乘以指数中的次数是 radicand 。 找到数字的基因来做此操作很有帮助 。
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  • We can use the properties of radicals to separate radicals of  products or quotients, and we can take a root 1st and then apply a power or vice versa.
    ::我们可以利用激进分子的特性来分离产品或商数的激进分子,我们可以从头开始,然后运用一种力量,反之亦然。

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  Review
  ::回顾

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  Evaluate or simplify the following radical expressions:
  ::评价或简化以下激进表达式:

  1. $\begin{array}{r}\sqrt[3]{81}\end{array}$

  2. $\begin{array}{r}\sqrt[4]{625}\end{array}$

  3. $\begin{array}{r}\sqrt{9^5}\end{array}$

  4. $\begin{array}{r}\sqrt[5]{128}\end{array}$

  5.  $\begin{array}{r}(\sqrt{\frac{25}{8}}{)}^4\end{array}$

  6. $\begin{array}{r}(\sqrt[6]{64}{)}^5\end{array}$

  7. $\begin{array}{r}(\sqrt[3]{\frac{8}{81}}{)}^2\end{array}$

  8. $\begin{array}{r}\sqrt[4]{\frac{243}{16}}\end{array}$

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  9. $\begin{array}{r}\sqrt[3]{24x^5}\end{array}$
  ::9. 24x53

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  10. $\begin{array}{r}\sqrt[4]{48x^7{y}^{13}}\end{array}$
  ::10. 48x7y134

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  11. $\begin{array}{r}\sqrt[5]{\frac{160x^8}{y^7}}\end{array}$
  ::11. 160x88y75

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  12. $\begin{array}{r}(\sqrt[3]{1000x^6}{)}^2\end{array}$
  :1000x63)2

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  13. $\begin{array}{r}\sqrt[4]{\frac{162x^5}{y^3{z}^{10}}}\end{array}$
  ::13. 162x5y3z104

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  14. $\begin{array}{r}(\sqrt{40x^3{y}^4}{)}^3\end{array}$
  ::14. (40x3y4)3

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  Explore More
  ::探索更多

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  1.  The volume of a cube is found to be $\begin{array}{r}343z^6\end{array}$ . What is the length of each side of the cube?
  ::1. 立方体的体积为343z6. 立方体每一侧的长度是多少?

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  2. Is $\begin{array}{r}\sqrt[6]{a}\cdot \sqrt{a}\end{array}$   the same as $\begin{array}{r}\sqrt[6]{a^2}\end{array}$ ? Explain why or why not.
  ::2. a6a是否与a26相同?解释原因或原因。

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  3. In a thunderstorm, the wind velocity in meters per second can be described by the function $\begin{array}{r}v(p)=5.7\sqrt{998-p},\end{array}$  where $\begin{array}{r}p\end{array}$  is the air pressure in millibars. What is the air pressure of a thunderstorm in which the wind velocity is 49.3 meters per second? Round your answer to the nearest tenth of a millibar.
  ::3. 在雷暴中,以每秒米计的风速可以用函数 v(p)=5.7998-p来描述,p是毫巴的气压。风速为每秒49.3米的雷暴的气压是多少?您回答的答案最接近毫巴的十分之一。

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  4. The population of Apple Island is slowly increasing, and can be calculated by $\begin{array}{r}P(t)=88.2\sqrt{1+0.13t},\end{array}$  where $\begin{array}{r}P\end{array}$  is the population (measured by millions) and $\begin{array}{r}t\end{array}$  the time (measured by years). How many years will it take the population of the island to reach 121 million? Round your answer to one decimal point.
  ::4. 苹果岛的人口正在缓慢增加,可以按P(t)=88.21+0.13t计算,P是人口(以百万计)和时间(以年计),岛屿人口需要多少年才能达到1.21亿?

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  5. The supply of computers is surely increasing in recent years. A survey shows that this phenomenon can be calculated by the function $\begin{array}{r}S(t)=33.2\sqrt{2+2.9t},\end{array}$  where  $\begin{array}{r}S\end{array}$  represents the supply (measured by millions) and $\begin{array}{r}t\end{array}$  the time (measured by years). How many years will it take for the supply of computers to reach 132.2 million? Round your answer to one decimal point.
  ::5. 近年来,计算机的供应量肯定在增加,一项调查显示,这种现象可以用S(t)=33.22+2.9t函数来计算,S函数代表供应量(以百万计),而S函数代表时间(以年计),而计算机供应量需要多少年才能达到1.322亿?

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  Answers for Review and Explore More Problems
  ::回顾和探讨更多问题的答复

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  Please see the Appendix.
  ::请参看附录。

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  PLIX
  ::PLIX

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  Try these interactives that reinforce the concepts explored in this section:
  ::尝试这些强化本节所探讨概念的交互作用 :

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  References
  ::参考参考资料

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  1. "Baseball Field Dimensio ns | Youth, College, High School, Pro and Softball," posted on December 30, 2007,
  ::1. 2007年12月30日张贴的“棒球球场层面 青年、学院、高中、职业和软球”,