Course: 11.7 双方与激进分子的均等 | The Way To Learn

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与两边的激进分子的对等
Equations with Radicals on Both Sides
::与两边的激进分子的对等

Often equations have more than one radical expression . The strategy in this case is to start by isolating the most complicated radical expression and raise the equation to the appropriate power. We then repeat the process until all radical signs are eliminated.
::方程式通常有一个以上的激进表达方式。本案的策略是先孤立最复杂的激进表达方式,然后将方程式提升到适当的力量。然后我们重复这一过程,直到所有激进迹象都消失为止。

Finding the Real Roots of an Equation
::寻找赤道的真正根

Find the real roots of the equation $\begin{aligned} \sqrt{2 x + 1} - \sqrt{x - 3} = 2 \end{aligned}$ .
::查找公式 2x+1 - x-3=2 的真正根。

$\begin{aligned} Isolate one of the radical expressions: & \sqrt{2 x + 1} & = 2 + \sqrt{x - 3} \\ Square both sides: & {(\sqrt{2 x + 1})}^{2} & = {(2 + \sqrt{x - 3})}^{2} \\ Eliminate parentheses: & 2 x + 1 & = 4 + 4 \sqrt{x - 3} + x - 3 \\ Simplify: & x & = 4 \sqrt{x - 3} \\ Square both sides of the equation: & x^{2} & = {(4 \sqrt{x - 3})}^{2} \\ Eliminate parentheses: & x^{2} & = 16 (x - 3) \\ Simplify: & x^{2} & = 16 x - 48 \\ Move all terms to one side of the equation: & x^{2} - 16 x + 48 & = 0 \\ Factor: & (x - 12) (x - 4) & = 0 \\ Solve: & x & = 12 or x = 4 \end{aligned}$

::分离一个基表达式 : 2x+1=2+1=2+2+2+2+x-3Square 双侧 : (2x+1) 2= (2++x-3) 2= (2++x-3) 2 括号 : 2x+1=4+4x-3+x-3+x-3-3 简化 :x=4x-3Square 方程式的两侧: x2=(4x-3) 2ELimate 括号: x2=16(x-3) 简单化: x2=16x=16x-48=0Fator: (x-12) (x-4) =0Solve:x=12 或x=4

Check: $\begin{aligned} \sqrt{2 (12) + 1} - \sqrt{12 - 3} = \sqrt{25} - \sqrt{9} = 5 - 3 = 2. \end{aligned}$ The solution checks out.
::检查:2(12)+1-12-3=25-9=5-3=2.

$\begin{aligned} \sqrt{2 (4) + 1} - \sqrt{4 - 3} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2 \end{aligned}$ The solution checks out.
::2(4)+1-4-3=9-1=3-1=2 验证解决方案。

The equation has two solutions: $\begin{aligned} x = 12 \end{aligned}$ and $\begin{aligned} x = 4 \end{aligned}$ .
::方程式有两个解决方案: x=12 和 x=4。

Identify Extraneous Solutions to Radical Equations
::识别极端等异异异的外部解决方案

We saw in Example 3 that some of the solutions that we find by solving radical equations do not check out when we substitute (or “plug in”) those solutions back into the original radical equation. These are called extraneous solutions. It is very important to check the answers we obtain by plugging them back into the original equation, so we can tell which of them are real solutions.
::我们从例3中看到,我们通过解决激进方程式找到的一些解决方案,当我们将这些解决方案替换(或“插入 ” ) 回到原始的激进方程式时,无法检查这些解决方案。这些解决方案被称为不相干解决方案。将这些解决方案塞回原始方程式,检查我们获得的答案非常重要,这样我们才能发现其中哪一个是真正的解决方案。

Solving for Unknown Values
::解决未知值

Find the real solutions of the equation $\begin{aligned} \sqrt{x - 3} - \sqrt{x} = 1 \end{aligned}$ .
::找到方程式 x-3- x=1 的真正解决方案。

$\begin{aligned} Isolate one of the radical expressions: & \sqrt{x - 3} & = \sqrt{x} + 1 \\ Square both sides: & {(\sqrt{x - 3})}^{2} & = {(\sqrt{x} + 1)}^{2} \\ Remove parenthesis: & x - 3 & = {(\sqrt{x})}^{2} + 2 \sqrt{x} + 1 \\ Simplify: & x - 3 & = x + 2 \sqrt{x} + 1 \\ Now isolate the remaining radical: & - 4 & = 2 \sqrt{x} \\ Divide all terms by 2: & - 2 & = \sqrt{x} \\ Square both sides: & x & = 4 \end{aligned}$

::分离一个基表达式 : x-3=x+1Square 双侧 : (x- 3) 2= (x+1) 2, 重新移动括号 : x-3= (x) 2+2x+1 简化 : x-3=x+2x+1 现在将剩余基表达式 : - 4=2xDivide 全部条件除以 2: - 2=xSquare 双侧: x=4

Check: $\begin{aligned} \sqrt{4 - 3} - \sqrt{4} = \sqrt{1} - 2 = 1 - 2 = - 1 \end{aligned}$ The solution does not check out.
::检查: 4 - 3 - 4=1 - 2=1 - 2 =1

The equation has no real solutions. $\begin{aligned} x = 4 \end{aligned}$ is an extraneous solution .
::等式没有真正的解决方案。 x=4 是一个无关的解决方案。

Applications using Special Case of Radical Equations
::使用激进等同特别案例的申请

Radical equations often appear in problems involving areas and volumes of objects.
::激进方程式经常出现在涉及物体面积和数量的问题中。

Real-World Application: Garden Fencing
::真实世界应用程序:花园围栏

Anita’s square vegetable garden is 21 square feet larger than Fred’s square vegetable garden. Anita and Fred decide to pool their money together and buy the same kind of fencing for their gardens. If they need 84 feet of fencing, what is the size of each garden?
::安妮塔的平方菜园面积比弗雷德的平方菜园大21平方英尺。安妮塔和弗雷德决定把他们的钱集中起来,为他们花园购买同样的围栏。如果他们需要84英尺的围栏,那么每个花园的大小是多少?

Make a sketch:
::绘制一张草图:

Define variables: Let Fred’s area be $\begin{aligned} x \end{aligned}$ ; then Anita’s area is $\begin{aligned} x + 21 \end{aligned}$ .
::定义变量:让 Fred 的面积为 x; 然后 Anita 的面积为 x+21 。

Find an equation:
::查找方程式 :

Side length of Fred’s garden is $\begin{aligned} \sqrt{x} \end{aligned}$
::Fred 花园的侧侧长为 x

Side length of Anita’s garden is $\begin{aligned} \sqrt{x + 21} \end{aligned}$
::安妮塔花园的侧长为x+21

The amount of fencing is equal to the combined perimeters of the two squares:
::围栏的面积等于两个广场的合并周界:

$\begin{aligned} 4 \sqrt{x} + 4 \sqrt{x + 21} = 84 \end{aligned}$

::4x+4x+21=84

Solve the equation:
::解决方程式:

$\begin{aligned} Divide all terms by 4: & \sqrt{x} + \sqrt{x + 21} & = 21 \\ Isolate one of the radical expressions: & \sqrt{x + 21} & = 21 - \sqrt{x} \\ Square both sides: & {(\sqrt{x + 21})}^{2} & = {(21 - \sqrt{x})}^{2} \\ Eliminate parentheses: & x + 21 & = 441 - 42 \sqrt{x} + x \\ Isolate the radical expression: & 42 \sqrt{x} & = 420 \\ Divide both sides by 42: & \sqrt{x} & = 10 \\ Square both sides: & x & = 100 f t^{2} \end{aligned}$

::将所有条件除以 4:x+x+x+21=21:21 孤立一个基表达式:x+21=21-x方形两侧x+212=(21-x)2 =(21-x) 括号:x+21=441 - 42x+x孤立两个边的基表达式:42x=420Divide 以42:x=10Square两侧:x=100英尺

Check: $\begin{aligned} 4 \sqrt{100} + 4 \sqrt{100 + 21} = 40 + 44 = 84 \end{aligned}$ . The solution checks out.
::检查: 4100+4100+21=40+44=84。解决方案检查完毕。

Fred’s garden is $\begin{aligned} 10 f t \times 10 f t = 100 f t^{2} \end{aligned}$ and Anita’s garden is $\begin{aligned} 11 f t \times 11 f t = 121 f t^{2} \end{aligned}$ .
::Fred的花园面积为10英尺×10英尺=100英尺2,Anita的花园面积为11英尺×11英尺=121英尺2。

Example
::示例示例示例示例

Example 1
::例1

Find the real solutions of the equation $\begin{aligned} \sqrt{9 - x} = 3 + \sqrt{2 x} \end{aligned}$ .
::找到公式 9 - x=3+2x 的真正解决方案。

$\begin{aligned} Isolate one of the radical expressions: & \sqrt{9 - x} & = 3 + \sqrt{2 x} \\ Square both sides: & (\sqrt{9 - x})^{2} & = (3 + \sqrt{2 x})^{2} \\ Remove parenthesis: & 9 - x & = 9 + 6 \sqrt{2 x} + 2 x \\ Simplify: & x & = - 2 \sqrt{2 x} \\ Square both sides: & x^{2} & = 4 (2 x) \\ Simplify: & x^{2} & = 8 x \\ Set one side equal to zero: & x^{2} - 8 x & = 0 \\ Factor: & x (x - 8) & = 0 \\ Use the zero product principle: & x & = 0 or x = 8 \end{aligned}$

::分离一个基表达式 : 9 - x= 3+2xSquare 双侧 : (9 - x) 2= (3+2x) 2 重新删除括号 : 9 - x= 9+ 62x+2xSimplication: x\\ 22xSquare 双侧 : x2= 4 (2x) 简单化: x2= 8xSet 一面等于 0: x2 - 8x= 0Facor:x(x- 8) =0 使用零产品原理: x=0 或 x=8

Check: First check $\begin{aligned} x = 0 \end{aligned}$ :
::选中: 第一检查 x=0 :

$\begin{aligned} Start with the original equation: & \sqrt{9 - x} & = 3 + \sqrt{2 x} \\ Substitute in the x-value: & \sqrt{9 - 8} & = 3 + \sqrt{2 (8)} \\ Simplify: & \sqrt{1} & = 3 + \sqrt{16)} \\ Take the square root: & 1 & = 3 + 4 = 7 \end{aligned}$

::在 x- value: 9 - 8= 3+2(8) 中以原始方程式开始: 9- x=3+2xSubtitude: 9- x=3+2=3+2/2; 简化: 1=3+16 将平方根: 1=3+4=7 选择平方根:1=3+4=7

The solution does not check out.
::解决方案不检查。

Then check $\begin{aligned} x = 8 \end{aligned}$ :
::然后检查 x=8 :

$\begin{aligned} Start with the original equation: & \sqrt{9 - x} & = 3 + \sqrt{2 x} \\ Substitute in the x-value: & \sqrt{9 - 0} & = 3 + \sqrt{2 (0)} \\ Simplify: & \sqrt{9} & = 3 + \sqrt{0)} \\ Take the square root: & 3 & = 3 + 0 = 3 \end{aligned}$

::在 x- value: 9- 0=3+2(0) 中以初始方程式开始: 9- x=3+2xSubtitude: 9- x=3+2; 9- 0 简化: 9=3+0) 选择平方根: 3=3+0=3

The solution checks out.
::解决方案已经检查了。

The equation has one real solution. $\begin{aligned} x = 8 \end{aligned}$ is an extraneous solution.
::方程式有一个真正的解决方案。x=8是一个不相干解决方案。

Review
::回顾

Find the solution to each of the following radical equations. Identify extraneous solutions.
::找到以下每个极端方程式的解决方案。找出不相干方程式。

$\begin{aligned} \sqrt{x} = \sqrt{x - 9} + 1 \end{aligned}$
::x=x-9+1x=x-9+1

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

$\begin{aligned} 5 \sqrt{x} = \sqrt{x + 12} + 6 \end{aligned}$
::5x=xx+12+6

$\begin{aligned} \sqrt{10 - 5 x} + \sqrt{1 - x} = 7 \end{aligned}$
::10-5x+1-x=7

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

$\begin{aligned} \sqrt{2 x + 5} - 3 \sqrt{2 x - 3} = \sqrt{2 - x} \end{aligned}$
::2x+5-32x-3=2-x

$\begin{aligned} 3 \sqrt{x} - 9 = \sqrt{2 x - 14} \end{aligned}$
::3x-9=2x-14

$\begin{aligned} \sqrt{x + 7} = \sqrt{x + 4} + 1 \end{aligned}$
::x+7=x+4+1 x7=x+4+1

$\begin{aligned} \sqrt{4 x} = \sqrt{3 - 2 x} - 1 \end{aligned}$
::4x=3-2x-1

$\begin{aligned} \sqrt{2 x + 5} + \sqrt{3 - x} = 10 \end{aligned}$
::2x+5+3-x=10

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。

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