节: 解决简单的激进方程式 | 7.7 解决简单激进方程式

章节大纲

The legs of a right triangle measure 3 and $\begin{aligned} 2 \sqrt{x} \end{aligned}$ . The hypotenuse measures 5. What is the length of the leg with the unknown value?
::右三角的腿 3 和 2x 。下限测量 5 。腿的长度是多少, 值未知 ?

Solving Radical Equations
::解决激进等号

Solving radical equations are very similar to solving other types of equations. The objective is to get $\begin{aligned} x \end{aligned}$ by itself. However, now there are radicals within the equations. Recall that the opposite of the square root of something is to square it.
::解决激进方程式与解决其他方程式非常相似。目标是自己获得 x 。但是, 现在方程式中存在激进方程式。回顾某方方程式的正方根对面是正方方形。

Let's determine if $\begin{aligned} x = 5 \end{aligned}$ is the solution to $\begin{aligned} \sqrt{2 x + 15} = 8 \end{aligned}$ .
::让我们确定 x=5 是否是 2x+15=8 的解决方案。

Plug in 5 for $\begin{aligned} x \end{aligned}$ to see if the equation holds true. If it does, then 5 is the solution.
::x 的插件在 5 中, x 看看方程式是否正确。如果是, 5 是解决方案。

$\begin{aligned} \sqrt{2 (5) + 15} & = 8 \\ \sqrt{10 + 15} & = 9 \\ \sqrt{25} & \neq 8 \end{aligned}$

We know that $\begin{aligned} \sqrt{25} = 5 \end{aligned}$ , so $\begin{aligned} x = 5 \end{aligned}$ is not the solution.
::我们知道25=5,所以x=5不是解决办法。

Now, let's solve the following equations for x.
::现在,让我们解决x的以下方程式。

$\begin{aligned} \sqrt{2 x - 5} + 7 = 16 \end{aligned}$
::2x-5+7=16

To solve for $\begin{aligned} x \end{aligned}$ , we need to isolate the radical. Subtract 7 from both sides.
::要解决x,我们需要隔离激进分子,从两边减7

$\begin{aligned} \sqrt{2 x - 5} + 7 & = 16 \\ \sqrt{2 x - 5} & = 9 \end{aligned}$

::2x-5+7=162x-5=9

Now, we can square both sides to eliminate the radical. Only square both sides when the radical is alone on one side of the equals sign.
::现在,我们可以让双方平分来消灭激进分子。只有激进分子独自站在平等标志的一方时,双方才能平分。

$\begin{aligned} {\sqrt{2 x - 5}}^{2} & = 9^{2} \\ 2 x - 5 & = 81 \\ 2 x & = 86 \\ x & = 43 \end{aligned}$

::2 - 52=922x-5=812x=86x=43

Check: $\begin{aligned} \sqrt{2 (43) - 5} + 7 = \sqrt{86 - 5} + 7 = \sqrt{81} + 7 = 9 + 7 = 16 \end{aligned}$
::检查:2(43)-5+7=86-5+7=81+7=7=9+7=16

ALWAYS check your answers when solving radical equations. Sometimes, you will solve an equation, get a solution, and then plug it back in and it will not work. These types of solutions are called extraneous solutions and are not actually considered solutions to the equation.
::ALWAS 在解答激进方程式时检查您的答案。有时, 您会解答方程式, 找到一个解决方案, 然后插入它, 它不会起作用。这些类型的解决方案被称为不相干解决方案, 而实际上不被认为是公式的解决方案。

$\begin{aligned} 3 \sqrt[3]{x - 8} - 2 = - 14 \end{aligned}$
::3-83-214

Again, isolate the radical first. Add 2 to both sides and divide by 3.
::再次将激进先锋孤立开来。双方加2,分3,分2。

$\begin{aligned} 3 \sqrt[3]{x - 8} - 2 & = - 14 \\ 3 \sqrt[3]{x - 8} & = - 12 \\ \sqrt[3]{x - 8} & = - 4 \end{aligned}$

::3 - 83 - 2 143x - 83 12x - 83 4

Now, cube both sides to eliminate the radical.
::现在立方两边消灭激进分子

$\begin{aligned} {\sqrt[3]{x - 8}}^{3} & = (- 4)^{3} \\ x - 8 & = - 64 \\ x & = - 56 \end{aligned}$

::x- 833=(- 4)3x-864x56

Check: $\begin{aligned} 3 \sqrt[3]{- 56 - 8} - 2 = 3 \sqrt[3]{- 64} - 2 = 3 \cdot - 4 - 2 = - 12 - 2 = - 14 \end{aligned}$
::检查: 3-56- 83-2=3- 643-2=3- 4-2}12-214

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the length of the leg with the unknown value.
::早些时候,有人要求你用未知的值来寻找腿的长度。

Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the leg with the unknown.
::使用 Pythagorean 理论并解析 x , 然后用未知值替换此值, 以未知值解析腿。

$\begin{aligned} 3^{2} + (2 \sqrt{x})^{2}) = 5^{2} \\ 9 + 4 x = 25 \\ 4 x = 16 \\ x = 4 \end{aligned}$

::32+(2x)2=529+4x=254x=16x=4

Now substitute this value into the leg with the unknown.
::现在将此值替换为未知值。

$\begin{aligned} 2 \sqrt{4} = 2 \cdot 2 = 4 \end{aligned}$

Therefore the leg with the unknown has a length of 4.
::因此,未知的腿长为4。

Example 2
::例2

Solve for x: $\begin{aligned} \sqrt{x + 5} = 6 \end{aligned}$ . Check your answer.
::解决 x: x+5=6. 请检查您的答案。

The radical is already isolated here. Square both sides and solve for $\begin{aligned} x \end{aligned}$ .
::激进分子已经在这里被隔离了将两侧平开解决 x 。

$\begin{aligned} {\sqrt{x + 5}}^{2} & = 6^{2} \\ x + 5 & = 36 \\ x & = 31 \end{aligned}$

::x+52=62x+5=36x=31

Check: $\begin{aligned} \sqrt{31 + 5} = \sqrt{36} = 6 \end{aligned}$
::检查: 31+5=36=6

Example 3
::例3

Solve for x: $\begin{aligned} 5 \sqrt{2 x - 1} + 1 = 26 \end{aligned}$ . Check your answer.
::x: 52x- 1+1=26。请检查您的答案。

Isolate the radical by subtracting 1 and then dividing by 5.
::将激进分子隔离,减去1,然后除以5。

$\begin{aligned} 5 \sqrt{2 x - 1} + 1 & = 26 \\ 5 \sqrt{2 x - 1} & = 25 \\ \sqrt{2 x - 1} & = 5 \end{aligned}$

::52x-1+1=2652x-1=252x-1=5

Square both sides and continue to solve for $\begin{aligned} x \end{aligned}$ .
::两边的广场继续解决 x 。

$\begin{aligned} {\sqrt{2 x - 1}}^{2} & = 5^{2} \\ 2 x - 1 & = 25 \\ 2 x & = 26 \\ x & = 13 \end{aligned}$

::2 - 12=522x-1=252x=26x=13

Check: $\begin{aligned} 5 \sqrt{2 (13) - 1} + 1 = 5 \sqrt{26 - 1} = 5 \sqrt{25} + 1 = 5 \cdot 5 + 1 = 25 + 1 = 26 \end{aligned}$
::检查: 52( 13) - 1+1=126 - 1=525+1=525=1=5_5=5=5=5=5=5=5=5=5+5=1=25+1=26

Example 4
::例4

Solve for x: $\begin{aligned} \sqrt[4]{3 x + 11} - 2 = 3 \end{aligned}$ . Check your answer.
::x: 3x+114-2=3。请检查您的答案。

In this problem, we have a fourth root . That means, once we isolate the radical, we must raise both sides to the fourth power to eliminate it.
::在这个问题上,我们有第四根根根,这意味着一旦我们孤立了激进分子,我们必须将双方推向第四势力,以消灭它。

$\begin{aligned} \sqrt[4]{3 x + 11} - 2 & = 3 \\ {\sqrt[4]{3 x - 11}}^{4} & = 5^{4} \\ 3 x - 11 & = 625 \\ 3 x & = 636 \\ x & = 212 \end{aligned}$

::3x+114-2=33x-1144=543x-111=6253x=636x=212

Check: $\begin{aligned} \sqrt[4]{3 (212) + 11} - 2 = \sqrt[4]{636 - 11} - 2 = \sqrt[4]{625} - 2 = 5 - 2 = 3 \end{aligned}$
::检查:3(212)+114-2=636-114-2=6254-2=5-2=3

Review
::回顾

Determine if the given values of x are solutions to the radical equations below.
::确定 x 的给定值是否是以下基方程的解决方案。

$\begin{aligned} \sqrt{x - 3} = 7; x = 32 \end{aligned}$
::x-3=7;x=32

$\begin{aligned} \sqrt[3]{6 + x} = 3; x = 21 \end{aligned}$
::6+x3=3;x=21

$\begin{aligned} \sqrt[4]{2 x + 3} - 11 = - 9; x = 6 \end{aligned}$
::2x+34- 11\\\\\\\\\\\9;x=6

Solve the equations and check your answers.
::解决方程式检查你的答案

$\begin{aligned} \sqrt{x + 5} = 6 \end{aligned}$
::x+5=6 x+5=6

$\begin{aligned} 2 - \sqrt{x + 1} = 0 \end{aligned}$
::2 - x+1=0

$\begin{aligned} 4 \sqrt{5 - x} = 12 \end{aligned}$
::45-x=12

$\begin{aligned} \sqrt{x + 9} + 7 = 11 \end{aligned}$
::x+9+7=11 x9+7=11

$\begin{aligned} \frac{1}{2} \sqrt[3]{x - 2} = 1 \end{aligned}$
::12 - 23=1

$\begin{aligned} \sqrt[3]{x + 3} + 5 = 9 \end{aligned}$
::x+33+5=9

$\begin{aligned} 5 \sqrt{15 - x} + 2 = 17 \end{aligned}$
::515-x+2=17

$\begin{aligned} - 5 = \sqrt[5]{x - 5} - 7 \end{aligned}$
::- 5=x-55-7

$\begin{aligned} \sqrt[4]{x - 6} + 10 = 13 \end{aligned}$
::x- 64+10=13

$\begin{aligned} \frac{8}{5} \sqrt[3]{x + 5} = 8 \end{aligned}$
::85x+53=8

$\begin{aligned} 3 \sqrt{x + 7} - 2 = 25 \end{aligned}$
::3x+7-2=25

$\begin{aligned} \sqrt[4]{235 + x} + 9 = 14 \end{aligned}$
::235+x4+9=14

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

Solving Radical Equations ::解决激进等号

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)

Solving Radical Equations
::解决激进等号

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)