节: 根和合理化 | 2.5 根和合理化-interactive

章节大纲

The Purpose of this Lesson

In this lesson, you will encounter irrational solutions to problems. Sometimes those solutions will involve in the denominator of fractions. You will learn to rationalize denominators in those cases.
::在这个教训中,你会遇到不合理的解决问题的办法。有时这些办法会涉及到分数分母。你会学会在这些情况下理顺分母。

Introduction: Irrational Roots
::导言:不合理根根

Many of the scenarios you've explored so far in this book can create problems whose solutions are irrational roots.
::你在这本书中至今所探讨的许多情况都会产生一些问题, 这些问题的解决方案是非理性的。

Activity 1: Roots and Powers of Fractions
::活动1:分数的根源和权力

Interactive
::交互式互动

The interactive below allows you to create right triangles with different leg lengths. Find two (remember that these are sets of three numbers that satisfy the Pythagorean Theorem : $\begin{align*}a^2+b^2=c^2\end{align*}$ ). Create a triangle with integer leg lengths whose hypotenuse is irrational. Find the hypotenuse and write it in the form of a square root , not a decimal approximation.
::下面的互动允许您创建右三角, 其腿长度不同。查找 2 个( 记住, 它们是符合 Pytagoren 理论的三组数 : a2+b2=c2 ) 。创建一个带有整数腿长度的三角形, 其下限不合理。查找下限, 以平方根的形式写, 而不是小数近似值。

Some of the triangles created with this interactive are Pythagorean Triples. To find the hypotenuse algebraically, substitute and solve. For example, if the two legs are 3 and 4, then solve for the hypotenuse as follows:
::通过此互动创建的一些三角形是 Pythagorean Triples 。要找到低温代数、替代和解析。例如, 如果双腿是 3 和 4, 那么对下限的解答如下 :

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ 3^{2} + 4^{2} = c^{2} & The Pythagorean Theorem. \\ c^{2} = 25 & Simplifying. \\ c = \pm \sqrt{25} & If c squared is 25, c is the square root of 25. \\ c = \pm 5 & Either positive or negative 5 makes the original equation true. \\ c = 5 & Take the positive, since triangles can't have negative lengths. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 3^2+4^2=c^2 & \text{The Pythagorean Theorem.}\\ c^2=25 & \text{Simplifying.}\\ c=\pm \sqrt{25} & \text{If c squared is 25, c is the square root of 25.}\\ c=\pm 5 & \text{Either positive or negative 5 makes the original equation true.}\\ c=5 & \text{Take the positive, since triangles can't have negative lengths.}\\ \end{array}\end{align*}$
::赤道Explation32+42=c2 Pythagorean Theorem.c2=25 简化. c25if cquand is 25, c是25c+5或负5的平方根,使原始方程式成为正数。 c=5 采取正数, 因为三角形不能有负长度。

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ 3^{2} + 4^{2} = c^{2} & The Pythagorean Theorem. \\ c^{2} = 25 & Simplifying. \\ c = \pm \sqrt{25} & If c squared is 25, c is the square root of 25. \\ c = \pm 5 & Either positive or negative 5 makes the original equation true. \\ c = 5 & Take the positive, since triangles can't have negative lengths. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 3^2+4^2=c^2 & \text{The Pythagorean Theorem.}\\ c^2=25 & \text{Simplifying.}\\ c=\pm \sqrt{25} & \text{If c squared is 25, c is the square root of 25.}\\ c=\pm 5 & \text{Either positive or negative 5 makes the original equation true.}\\ c=5 & \text{Take the positive, since triangles can't have negative lengths.}\\ \end{array}\end{align*}$
::赤道Explation32+42=c2 Pythagorean Theorem.c2=25 简化. c25if cquand is 25, c是25c+5或负5的平方根,使原始方程式成为正数。 c=5 采取正数, 因为三角形不能有负长度。

Other leg lengths will result in a hypotenuse that is irrational. For example:
::其他腿的长度将导致不合理的低耗。例如 :

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ 1^{2} + 2^{2} = c^{2} & The Pythagorean Theorem. \\ c^{2} = 5 & Simplifying. \\ c = \pm \sqrt{5} & If c squared is 5, c is the square root of 5. \\ c = \pm \sqrt{5} & Either positive or negative makes the original equation true. \\ c = \sqrt{5} & Only the positive value makes sense. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 1^2+2^2=c^2 & \text{The Pythagorean Theorem.}\\ c^2=5 & \text{Simplifying.}\\ c=\pm \sqrt{5} & \text{If c squared is 5, c is the square root of 5.}\\ c=\pm \sqrt{5} & \text{Either positive or negative makes the original equation true.}\\ c=\sqrt5 & \text{Only the positive value makes sense.}\\ \end{array}\end{align*}$
::QquationExplation12+22=c2 Pythagorean Theorem.c2=5Simplication.c5if c squand is 5,c是5.c5Ether正或负的平方根,使原始方程式变为正数。c=5 只有正值才有意义。

Example 1-1
::例1-1

Find the hypotenuse of a right triangle with leg lengths of 0.5 and 1.
::找到右三角形的下限,右三角形的腿长为0.5和1。

Solution:

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ (0.5)^{2} + (1)^{2} = c^{2} & The Pythagorean Theorem. \\ {(\frac{1}{2})}^{2} + 1^{2} & Writing decimals as fractions. \\ \frac{1}{4} + 1 = c^{2} & Squaring. \\ \frac{5}{4} = c^{2} & Adding. \\ c = \sqrt{\frac{5}{4}} & Square rooting. (Only positive makes sense.) \\ c = \frac{\sqrt{5}}{\sqrt{4}} & The root of a fraction is the fraction of the roots. \\ c = \frac{\sqrt{5}}{2} & Simplifying. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline (0.5)^2+(1)^2=c^2 & \text{The Pythagorean Theorem.}\\ \left(\frac12 \right)^2+1^2 & \text{Writing decimals as fractions.}\\ \frac14 + 1 =c^2 & \text{Squaring.}\\ \frac54 = c^2 & \text{Adding.}\\ c=\sqrt{\frac54} & \text{Square rooting. (Only positive makes sense.)}\\ c=\frac{\sqrt5}{\sqrt4} & \text{The root of a fraction is the fraction of the roots.}\\ c=\frac{\sqrt5}{2} & \text{Simplifying.}\\ \end{array} \end{align*}$
::解析度( 0.5) 2+(1)2=c2 Pythagorean Theorem. (12) 2+ 12Writing 小数小数作为分数 14+1=c2Squaling. 54=c2Adding.c=54Square rooting. (只有阳性才有意义。) c=54 分数的根数是根数的分数。 c=52Simplication。

The last example finished with a simplification that may be new to you: $\begin{align*}\sqrt{\frac54}=\frac{\sqrt5}{2}.\end{align*}$ In the following examples and problems, explore the reasoning behind this simplification. The exploration begins by looking at powers of products.
::最后一个例子结尾的简化对你们来说可能是新的:54=52。在下面的例子和问题中,探索简化背后的推理。探索从研究产品的力量开始。

Activity 2: Powers of Products
::活动2:产品的权力

Example 2-1
::例2-1

Simplify $\begin{align*}(2\cdot3)^2\end{align*}$ by two different methods.
::以两种不同的方法简化 (2Q3) 2 。

Solution:   In the first method, perform the multiplication in " data-term="Parentheses" role="term" tabindex="0"> parentheses first:
::解决办法:在第一种方法中,首先在括号内执行乘法:

$\begin{aligned} \begin{matrix} (2 \cdot 3)^{2} \\ 6^{2} \\ 36 \end{matrix} \end{aligned}$
$\begin{align*}\begin{array}{c} (2\cdot3)^2\\ 6^2\\ 36 \end{array}\end{align*}$

In the second method, begin by writing out the multiplication implied by the exponent :
::在第二种方法中,首先写出引言人暗示的乘法:

$\begin{aligned} \begin{array}{ll} Expression & Explanation \\ (2 \cdot 3)^{2} & Given expression. \\ (2 \cdot 3) (2 \cdot 3) & The definition of the power of 2. \\ 2 \cdot 3 \cdot 2 \cdot 3 & Because all quantities are multiplied, parentheses not needed. \\ 2 \cdot 2 \cdot 3 \cdot 3 & Commutative property of multiplication. \\ 2^{2} \cdot 3^{2} & Definition power of 2. \\ 4 \cdot 9 & Simplifying. \\ 36 & Simplifying. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Expression} & \text{Explanation}\\ \hline (2\cdot 3)^2 & \text{Given expression.}\\ (2\cdot3)(2\cdot 3)& \text{The definition of the power of 2.}\\ 2\cdot3\cdot2\cdot3 & \text{Because all quantities are multiplied, parentheses not needed.}\\ 2\cdot2\cdot3\cdot3 & \text{Commutative property of multiplication.}\\ 2^2 \cdot 3^2 & \text{Definition power of 2.}\\ 4\cdot9& \text{Simplifying.}\\ 36 & \text{Simplifying.} \end{array}\end{align*}$
::表达式Explation(2)3,2(3)3 表达式。(2)3(2)3) 2.23/23的功率定义,因为所有数量都是乘以,括号不需要。 2223}3 乘以的商品属性 22(32) 定义功率 2.49 简化36。

Work it Out
::工作出来

Generalize your results from the last example by performing the same process to simplify $\begin{align*}(a\cdot b)^2.\end{align*}$ (Given $\begin{align*}a \text{ and } b\end{align*}$ are integers.) Explain each step.
::通过执行相同的简化过程(ab) 2. (给的a和b为整数。) 解释每个步骤。

Perform the same process to simplify $\begin{align*}\left(\frac{a}{b}\right)^2 .\end{align*}$ (These variables are integers and $\begin{align*}b\ne0.\end{align*}$ ) Explain each step.
::执行相同的简化进程 (ab) 2. (这些变量为整数和 b0.) 解释每个步骤。

Do you think these results will hold for powers of 3, 4, and more? Experiment. Justify your conclusions.
::你认为这些结果能维持3,4和3的权力吗?试验。请解释你的结论。

Powers of Products or Fractions of Integers
::产品权或整数分数

The power of a product is the product of the powers:
$\begin{aligned} (a b)^{n} = a^{n} b^{n} \end{aligned}$
$\begin{align*}(ab)^n=a^nb^n\end{align*}$
::产品的力量是以下权力的产物ab)n=anbn
The power of a quotient is the quotient of the powers:
$\begin{aligned} {(\frac{a}{b})}^{n} = (\frac{a^{n}}{b^{n}}) \end{aligned}$
$\begin{align*}\left(\frac{a}{b}\right)^n = \left(\frac{a^n}{b^n}\right)\end{align*}$

Example 2-2
::例2-2

Simplify each of the following numerical expressions. What do you observe? Can you explain why?

$\begin{aligned} \begin{array}{cc} First Expression & Second Expression \\ \sqrt{4 \cdot 9} & \sqrt{4} \cdot \sqrt{9} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{c|c} \text{First Expression} & \text{Second Expression}\\ \hline \sqrt{4\cdot9} & \sqrt4 \cdot \sqrt9\\ \end{array}\end{align*}$
::简化以下每个数字表达式。您观察了什么 ? 您可以解释原因吗 ? 第一个表达式二

Solution:
::解决方案 :

$\begin{aligned} \begin{array}{cc} First Expression & Second Expression \\ \sqrt{4 \cdot 9} & \sqrt{4} \cdot \sqrt{9} \\ \sqrt{36} & 2 \cdot 3 \\ 6 & 6 \end{array} \end{aligned}$
$\begin{align*}\begin{array}{c|c} \text{First Expression} & \text{Second Expression}\\ \hline \sqrt{4\cdot9} & \sqrt4 \cdot \sqrt9\\ \sqrt{36} & 2\cdot3\\ 6 & 6\\ \end{array}\end{align*}$
::第二表达式49439362366

The two expressions are equivalent. To explain why, remember that square rooting is the same as raising to the power of $\begin{align*}\frac12:\end{align*}$
::两个表达式相同。要解释原因, 请记住, 平方根与 12 的功率相同 :

$\begin{aligned} \sqrt{4 \cdot 9} = (4 \cdot 9)^{\frac{1}{2}} \end{aligned}$
$\begin{align*}\sqrt{4\cdot 9}=(4 \cdot 9)^{\frac12}\end{align*}$

The power of a product is the product of powers, so:
::产品的力量是力量的产物,所以:

$\begin{aligned} (4 \cdot 9)^{\frac{1}{2}} = 4^{\frac{1}{2}} \cdot 9^{\frac{1}{2}} \end{aligned}$
$\begin{align*}(4 \cdot 9)^{\frac12}=4^{\frac12} \cdot 9^{\frac12}\end{align*}$

And finally:
::最后:

$\begin{aligned} 4^{\frac{1}{2}} \cdot 9^{\frac{1}{2}} = \sqrt{4} \cdot \sqrt{9} \end{aligned}$
$\begin{align*}4^{\frac12} \cdot 9^{\frac12}=\sqrt4 \cdot \sqrt9\end{align*}$

Powers and Roots of Products of Integers
::整数产品产品的权力和根源

The root of a product is the product of the roots:
$\begin{aligned} \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \\ \sqrt[n]{a \times b} = \sqrt[n]{a} \times \sqrt[n]{b} \end{aligned}$
$\begin{align*}\sqrt{a\times b}=\sqrt{a}\times\sqrt{b}\\ \sqrt[n]{a\times b}=\sqrt[n]{a}\times \sqrt[n]{b}\\\end{align*}$
::产品的根是根的产物: axb=axbaxbn=anxbn
The root of a quotient is the quotient of the roots:
$\begin{aligned} \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} \\ \sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \end{aligned}$
$\begin{align*}\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\\ \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}\\ \end{align*}$

  Hint: Simplifying Square Roots
::提示: 简化平方根

If possible, factor the radicand (the number under the square root) so that one of the factors is a perfect square . For example:

Activity 3: Exponent Rules with Rational Exponents
::活动3:有理有理有理有理有理有理有理有理有理有理有理有据的规则

Work it Out
::工作出来

Use the method shown in the last example to show that $\begin{align*}\sqrt[3]{a\cdot b}=\sqrt[3]{a}\cdot \sqrt[3]{b}.\end{align*}$
::使用上一个示例中显示的方法显示 a&b3=a3&b3。

A right triangle

Find the hypotenuse for the right triangles with the given legs. Simplify your result as much as possible. An example with given legs of 0.25 and 0.75 is provided for you.
::用给定的腿查找右三角形的下限。尽可能简化结果。给定的腿为0. 25 和 0. 75 的示例提供给您。

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ (0.25)^{2} + (0.75)^{2} = c^{2} & Pythagorean Theorem. \\ {(\frac{1}{4})}^{2} + {(\frac{3}{4})}^{2} = c^{2} & Written as fractions. \\ \frac{1}{16} + \frac{9}{16} = c^{2} & Squaring. \\ \frac{10}{16} = c^{2} & Adding. \\ \frac{5}{8} = c^{2} & Reducing. \\ c = \sqrt{\frac{5}{8}} & Square rooting. (Only positive makes sense.) \\ c = \frac{\sqrt{5}}{\sqrt{8}} & The root of a fraction is the fraction of the roots. \\ c = \frac{\sqrt{5}}{\sqrt{4 \cdot 2}} & Preparing to simplify further by factoring the 8... \\ c = \frac{\sqrt{5}}{\sqrt{4} \cdot \sqrt{2}} & The root of a product is the product of the roots. \\ c = \frac{\sqrt{5}}{2 \cdot \sqrt{2}} & The square root of 4 is 2. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline (0.25)^2+(0.75)^2=c^2 & \text{Pythagorean Theorem.}\\ \left(\frac14 \right)^2+\left(\frac34 \right)^2=c^2 & \text{Written as fractions.}\\ \frac{1}{16}+\frac{9}{16}=c^2 & \text{Squaring.}\\ \frac{10}{16}=c^2 & \text{Adding.}\\ \frac58 = c^2 & \text{Reducing.}\\ c=\sqrt{\frac58} & \text{Square rooting. (Only positive makes sense.)}\\ c=\frac{\sqrt5}{\sqrt8} & \text{The root of a fraction is the fraction of the roots.}\\ c=\frac{\sqrt5}{\sqrt{4\cdot 2}} & \text{Preparing to simplify further by factoring the 8...}\\ c=\frac{\sqrt5}{\sqrt4 \cdot \sqrt2} & \text{The root of a product is the product of the roots.}\\ c=\frac{\sqrt5}{2\cdot \sqrt2} & \text{The square root of 4 is 2.}\\ \end{array} \end{align*}$
$\begin{align*}\quad\begin{array}{ll} \text{a.} & 0.5, 2.5\\ \text{b.} & 0.25, 0.5\\ \text{c.} & \frac13, \frac32\\ \text{d.} & \frac25, \frac32\\ \text{e.} & 3,\frac47\\ \end{array}\end{align*}$
::QquationExplation( 0. 252) 2+( 0. 752) = c2Pythagorean Theorem. (142) +( 342) = c2Wititit 以分数表示。 116+916= c2Squaring.1016= c2Add. 58= c2 降为58Square rooting. c= c2=58Squeting。 (只有正数才有意义。) c c=58=58Square rooting. c=58 root is the roots. c= 522 4的产物。 a. 0.5, 2.5b. 0.25, 0.5c.13, 32d.25, 32e3, 32e3, 47

Example 3-1
::例3-1

Solving for an unknown side of a right triangle with the Pythagorean Theorem resulted in a quadratic equation . Equations like $\begin{align*}c^2=\frac79\end{align*}$   are recognizable as quadratic because you must solve for a variable that is squared.
::与 Pythagorean 理论为右三角形的未知侧面进行溶解,结果产生了二次方程。 C2=79 等方程可被识别为二次方程,因为您必须解决正方形变量。

Imagine a car accelerating at 16 meters per second, per second. Create an equation for the distance the car has traveled as a function of time. Find the distance the car has traveled in 7.5 seconds. Write and solve an equation to determine when the car reaches the 100-meter mark. Write your solution as simplified as possible. Then find a decimal approximation for your solution.
::想象一下车速加速为每秒16米, 每秒每秒。为车行驶的距离建立一个方程式。找到车行驶的距离在7. 5 秒内。写入并解析一个方程式, 以确定汽车何时到达100米的标记。尽可能简化您的解题。然后为您的解题找到小数近似值。

Solution: Recall that the equation for the distance the car has traveled as a function of time is given by the equation:
::解答: 提醒注意车行驶距离的方程式是时间函数, 方程式给出的方程式 :

$\begin{aligned} \begin{array}{l} y = \frac{1}{2} a x^{2} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l} y=\frac12 a x^2\\ \end{array}\end{align*}$
::y=12轴2

Recall that $\begin{align*}a\end{align*}$ here represents the acceleration, $\begin{align*}x\end{align*}$ the time in seconds, and $\begin{align*}y\end{align*}$ the distance in meters. Given an acceleration of 16 meters per second, per second:
::回顾这里有一个代表加速度, x 秒时间, y 米距离。鉴于加速度为每秒16米, 每秒:

$\begin{aligned} \begin{array}{l} y = 8 x^{2} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l} y=8x^2\\ \end{array}\end{align*}$
::y=8x2 y=8x2

$\begin{aligned} \begin{array}{l} y = 8 x^{2} \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l} y=8x^2\\ \end{array}\end{align*}$
::y=8x2 y=8x2

After 7.5 seconds:
::7.5秒后:

$\begin{aligned} \begin{array}{l} y = 8 {(7.5)}^{2} \\ y = 8 {(7 \frac{1}{2})}^{2} \\ y = 8 {(\frac{15}{2})}^{2} \\ y = 8 \cdot \frac{225}{4} \\ y = 2 \cdot 225 \\ y = 450 m \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l} y=8\left(7.5\right) ^2\\ y=8\left(7 \frac12 \right)^2\\ y=8\left(\frac{15}{2} \right)^2\\ y=8\cdot \frac{225}{4}\\ y=2\cdot 225\\ y=450~\text{m} \end{array}\end{align*}$
::y=8( 7.5) 2y=8( 712) 2y=8( 152) 2yy=8( 2254y=2) 225y=450米

$\begin{aligned} \begin{array}{l} y = 8 {(7.5)}^{2} \\ y = 8 {(7 \frac{1}{2})}^{2} \\ y = 8 {(\frac{15}{2})}^{2} \\ y = 8 \cdot \frac{225}{4} \\ y = 2 \cdot 225 \\ y = 450 m \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l} y=8\left(7.5\right) ^2\\ y=8\left(7 \frac12 \right)^2\\ y=8\left(\frac{15}{2} \right)^2\\ y=8\cdot \frac{225}{4}\\ y=2\cdot 225\\ y=450~\text{m} \end{array}\end{align*}$
::y=8( 7.5) 2y=8( 712) 2y=8( 152) 2yy=8( 2254y=2) 225y=450米

Here's the equation to determine when the car reaches the 100-meter mark:
::确定汽车何时到达100米标志的方程式如下:

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ 100 = 8 x^{2} & The distance is 100 meters. The time is unknown. \\ \frac{100}{8} = x^{2} & Dividing by 8 to isolate the variable squared. \\ \frac{25}{2} = x^{2} & Reducing the fraction. \\ x^{2} = \frac{25}{2} & Re-arranging. \\ x = \sqrt{\frac{25}{2}} & Square-rooting. Only positive makes sense. \\ x = \frac{\sqrt{25}}{\sqrt{2}} & The root of a quotient is the quotient of the roots. \\ x = \frac{5}{\sqrt{2}} & Simplifying. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline 100=8x^2\ & \text{The distance is 100 meters. The time is unknown.}\\ \frac{100}{8}=x^2 & \text{Dividing by 8 to isolate the variable squared.}\\ \frac{25}{2}=x^2 & \text{Reducing the fraction.}\\ x^2=\frac{25}{2} & \text{Re-arranging.}\\ x=\sqrt{\frac{25}{2}} & \text{Square-rooting. Only positive makes sense.}\\ x=\frac{\sqrt{25}}{\sqrt2} & \text{The root of a quotient is the quotient of the roots.}\\ x=\frac{5}{\sqrt2} & \text{Simplifying.} \end{array} \end{align*}$
::方位分布 100 = 8x2 距离为 100 公尺。时间为未知 1008 = x2 = x2 乘以 8 以分离变量方位d.252 = x2 。正在减少分数. x2 = 252 re- arrange. x= 252Square root。只有正数才有意义。 x = 252 商数的根部是根部的商数。 x = 52 简化。

You can use a calculator to determine that's about 3.54 seconds.
::您可以使用计算器确定大约3.54秒。

Activity 4: R ationalizing Denominators
::活动4:合理化分计器

In both the last example and the last problem, the simplified result featured an irrational root in the denominator. It is a mathematical convention to rationalize the denominator . This means performing a final arithmetic step so that the denominator is not an irrational root. Rationalizing the denominator makes it easier to compare the many irrational solutions you'll encounter.
::在最后一个例子和最后一个问题中,简化的结果在分母中都有一个非理性的根。这是一个使分母合理化的数学惯例。这意味着要执行最后的算术步骤,使分母不是非理性根。理顺分母可以更容易地比较你将会遇到的许多非理性的解决方法。

Example 4-1
::例4-1

Below, find a fraction with an irrational root in the denominator. What can you multiply the denominator by in order to transform it into an integer? Does multiplying the numerator and denominator by this value create an equivalent fraction? How do you know? Perform this arithmetic. How is the result different from the original?

$\begin{aligned} \frac{3 \sqrt{5}}{\sqrt{7}} \end{aligned}$
$\begin{align*}\frac{3\sqrt5}{\sqrt7}\end{align*}$
::在下面, 在分母中找到一个有非理性根的分数。您可以通过什么将分母乘以来将其转换成整数 ? 通过这个值乘以数字和分母会产生一个相等的分数吗? 你怎么知道? 执行此算术。结果与原数有什么不同? 357

Solution:

$\begin{aligned} \begin{array}{ll} Expression & Explanation \\ \frac{3 \sqrt{5}}{\sqrt{7}} & Given Expression \\ \frac{3 \sqrt{5}}{\sqrt{7}} \cdot \frac{\sqrt{7}}{\sqrt{7}} & Expanding the fraction by the square root of 7. \\ \frac{3 \cdot \sqrt{5} \cdot \sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} & How we multiply fractions. \\ \frac{3 \cdot \sqrt{35}}{\sqrt{7} \sqrt{7}} & The product of roots is the root of the product. \\ \frac{3 \sqrt{35}}{7} & By definition of a square root, the root of 7 times the root of 7 is 7. \end{array} \end{aligned}$
$\begin{align*}\begin{array}{l|l} \text{Expression} & \text{Explanation}\\ \hline \frac{3\sqrt5}{\sqrt7} & \text{Given Expression}\\ \frac{3\sqrt5}{\sqrt7}\cdot \frac{\sqrt7}{\sqrt7} & \text{Expanding the fraction by the square root of 7.}\\ \frac{3\cdot \sqrt5 \cdot \sqrt7}{\sqrt7 \cdot \sqrt7} & \text{How we multiply fractions.}\\ \frac{3\cdot \sqrt{35}}{\sqrt{7}\sqrt7} & \text{The product of roots is the root of the product.}\\ \frac{3\sqrt{35}}{7}& \text{By definition of a square root, the root of 7 times the root of 7 is 7.}\\ \end{array}\end{align*}$
::溶液: ExplainExplainment 357 Explainment 357 Expression 357-7777 将分数除以平方根7.3 577-7。我们如何乘以分数 33577 根的产物是产物的根。 3357B 根根根定义,7根根根的根数是7。

Work it Out
::工作出来

A rocket is accelerating at 24 meters per second, per second. Create an equation that gives the distance the rocket has traveled as a function of time. Determine how far the rocket has traveled in 3 and one-third seconds. Write and solve an equation to determine when the rocket passes the 1-kilometer mark. Give a simplified, rationalized answer and a decimal approximation.
::一枚火箭正在以每秒24米的速度加速, 每秒每秒每秒24米的速度加速。创建一个公式, 显示火箭所穿越的距离。确定火箭在三秒和三分之一秒内飞到的距离。写入和解析一个公式, 以确定火箭何时经过一公里的标记。给出一个简化、合理化的答案和一个小数点的近似值。

Describe a general procedure for rationalizing the denominator of the fraction $\begin{align*}\frac{a}{\sqrt{b}}.\end{align*}$ ( $\begin{align*}a\end{align*}$ and $\begin{align*}b\end{align*}$ are positive integers.)
::描述使分数 ab 分数分母合理化的一般程序。 (a和b为正整数。 )

Simplify each of the following. Rationalize denominators when necessary.
::简化以下各点。必要时对分母进行合理化。

$\begin{align*}\quad\begin{array}{ll} \quad \text{a.} & \sqrt{8}\\ \quad \text{b.} & \sqrt{\frac{75}{4}}\\ \quad \text{c.} & \sqrt{\frac{20}{5}}\\ \quad \text{d.} & \frac{\sqrt{49}}{\sqrt7}\\ \quad \text{e.} & \sqrt{12}\cdot\sqrt{6}\\ \quad \text{f.} & \sqrt{\frac{72}{75}}\cdot \sqrt{\frac{200}{147}}\\ \quad \text{g.} & \sqrt{300} \cdot \sqrt{363}\\ \quad \text{h.} & \frac{1}{\sqrt{12}}\\ \end{array}\end{align*}$
::a. a. b. 754 c. 205d. 497e. 12.6f. 7275-200147g. 300-363h.112

Summary
::摘要

The power or root of a product of integers is the product of the powers or roots.
::整数产物的力量或根是力量或根的产物。

The power or root of a quotient of integers is the quotient of the powers or roots.
::整数商数的功率或根值是功率或根数的商数。

Rationalize the denominator by expanding the fraction. (Multiply the numerator and denominator by the irrational root in the denominator.)
::通过扩展分数来理顺分母。 (由分数中非理性根根部分的分子和分母组成。)

PLIX Interactive
::PLIX 交互式互动

根和合理化

章节大纲

Introduction: Irrational Roots
::导言:不合理根根

Activity 1: Roots and Powers of Fractions
::活动1:分数的根源和权力

Interactive
::交互式互动

Example 1-1
::例1-1

Activity 2: Powers of Products
::活动2:产品的权力

Example 2-1
::例2-1

Work it Out
::工作出来

Powers of Products or Fractions of Integers
::产品权或整数分数

Example 2-2
::例2-2

Powers and Roots of Products of Integers
::整数产品产品的权力和根源

Hint: Simplifying Square Roots
::提示: 简化平方根

Activity 3: Exponent Rules with Rational Exponents
::活动3:有理有理有理有理有理有理有理有理有理有理有理有据的规则

Work it Out
::工作出来

Activity 4: R ationalizing Denominators
::活动4:合理化分计器

Example 4-1
::例4-1

Work it Out
::工作出来

Summary
::摘要

PLIX Interactive
::PLIX 交互式互动

章节大纲

Introduction: Irrational Roots ::导言:不合理根根

Activity 1: Roots and Powers of Fractions ::活动1:分数的根源和权力

Interactive ::交互式互动

Example 1-1 ::例1-1

Activity 2: Powers of Products ::活动2:产品的权力

Example 2-1 ::例2-1

Work it Out ::工作出来

Powers of Products or Fractions of Integers ::产品权或整数分数

Example 2-2 ::例2-2

Powers and Roots of Products of Integers ::整数产品产品的权力和根源

Hint: Simplifying Square Roots ::提示: 简化平方根

Activity 3: Exponent Rules with Rational Exponents ::活动3:有理有理有理有理有理有理有理有理有理有理有理有据的规则

Work it Out ::工作出来

Activity 4: R ationalizing Denominators ::活动4:合理化分计器

Example 4-1 ::例4-1

Work it Out ::工作出来

Summary ::摘要

PLIX Interactive ::PLIX 交互式互动

Introduction: Irrational Roots
::导言:不合理根根

Activity 1: Roots and Powers of Fractions
::活动1:分数的根源和权力

Interactive
::交互式互动

Example 1-1
::例1-1

Activity 2: Powers of Products
::活动2:产品的权力

Example 2-1
::例2-1

Work it Out
::工作出来

Powers of Products or Fractions of Integers
::产品权或整数分数

Example 2-2
::例2-2

Powers and Roots of Products of Integers
::整数产品产品的权力和根源

Hint: Simplifying Square Roots
::提示: 简化平方根

Activity 3: Exponent Rules with Rational Exponents
::活动3:有理有理有理有理有理有理有理有理有理有理有理有据的规则

Work it Out
::工作出来

Activity 4: R ationalizing Denominators
::活动4:合理化分计器

Example 4-1
::例4-1

Work it Out
::工作出来

Summary
::摘要

PLIX Interactive
::PLIX 交互式互动