Course: 7.8 解决双方有变数的激进等式

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解决双方有变量的激进方程式
The legs of a right triangle measure 12 and $\begin{align*}\sqrt {x + 1}\end{align*}$ . The hypotenuse measures $\begin{align*}\sqrt{7x + 1}\end{align*}$ . What are the lengths of the sides with the unknown values?
::右三角的腿度量 12 和 x+1 。下限度量 7x+1 。边的长度与未知值的长度是多少 ?

Solving Radical Equations with Variables
::用变量解决极端等式

In this concept, we will continue solving radical equations. Here, we will address variables and radicals on both sides of the equation .
::在这个概念中,我们将继续解决激进方程式问题,在这里,我们将处理方程式两侧的变数和激进方程式问题。

Let's solve the following radical equations for x.
::让我们解决x的以下基方程。

$\begin{align*}\sqrt{4x+1}-x=-1\end{align*}$
::4x+1 -x1

Now we have an $\begin{align*}x\end{align*}$ that is not under the radical. We will still isolate the radical.
::现在,我们有一个不是在激进下的x。我们仍会孤立激进。

$\begin{aligned} \sqrt{4 x + 1} - x & = - 1 \\ \sqrt{4 x - 1} & = x - 1 \end{aligned}$
$\begin{align*}\sqrt{4x+1}-x&=-1\\ \sqrt{4x-1}&=x-1\end{align*}$
::4x+1-x=1=x-1

Now, we can square both sides. Be careful when squaring $\begin{align*}x-1\end{align*}$ , the answer is not $\begin{align*}x^2-1\end{align*}$ .
::现在,我们可以对齐两边。当对齐 x - 1 时要小心, 答案不是 x2 - 1 。

$\begin{aligned} {\sqrt{4 x + 1}}^{2} & = (x - 1)^{2} \\ 4 x + 1 & = x^{2} - 2 x + 1 \end{aligned}$
$\begin{align*}\sqrt{4x+1}^2&=(x-1)^2\\ 4x+1&=x^2-2x+1\end{align*}$
::4x+12=(x- 1) 24x+1=x2-2x+1

This problem is now a quadratic. To , we must either factor , when possible, or use the Quadratic Formula . Combine like terms and set one side equal to zero.
::这个问题现在是一个四面形的问题。对来说,我们必须在可能的时候考虑因素,或者使用“四面形公式 ” 。将类似条件合并,将一面设置为零。

$\begin{aligned} 4 x + 1 & = x^{2} - 2 x + 1 \\ 0 & = x^{2} - 6 x \\ 0 & = x (x - 6) \\ x & = 0 o r 6 \end{aligned}$
$\begin{align*}4x+1&=x^2-2x+1\\ 0&=x^2-6x\\ 0&=x(x-6)\\ x&=0 \ or \ 6\end{align*}$
::4x+1=x2-2x+10=x2-6x0=x(x-6)xx=0或6

Check both solutions: $\begin{align*}\sqrt{4 \left(0\right)+1}-1=\sqrt{0+1}-1=1-1=0 \neq -1\end{align*}$ . 0 is an extraneous solution .
::检查两个解决方案 : 4 (0)+1 - 1=0+1 - 1=1 - 1=1 - 1=0 *1. 0是一个不相干解决方案。

$\begin{align*}\sqrt{4 \left(6 \right)+1}-6=\sqrt{24+1}-6=5-6=-1\end{align*}$

Therefore , 6 is the only solution.
::因此,6是唯一的解决办法。

$\begin{align*}\sqrt{8x-11}-\sqrt{3x+19}=0\end{align*}$
::8-11-3x+19=0

In this problem, you need to isolate both radicals. To do this, subtract the second radical from both sides. Then, square both sides to eliminate the variable .
::在此问题上, 您需要将两个激进分子隔离开来。要做到这一点, 请从双方中减去第二个激进分子。然后, 将双方平方来消除变量。

$\begin{aligned} \sqrt{8 x - 11} - \sqrt{3 x + 19} & = 0 \\ {\sqrt{8 x - 11}}^{2} & = {\sqrt{3 x + 19}}^{2} \\ 8 x - 11 & = 3 x + 19 \\ 5 x & = 30 \\ x & = 6 \end{aligned}$
$\begin{align*}\sqrt{8x-11}-\sqrt{3x+19}&=0 \\ \sqrt{8x-11}^2&=\sqrt{3x+19}^2 \\ 8x-11&=3x+19 \\ 5x&=30 \\ x&=6\end{align*}$
::8-11-3x+19=08x-112=3x+1928x-11=3x+1995x=30x=6

Check: $\begin{align*}\sqrt{8 \left(6 \right)-11}-\sqrt{3 \left(6 \right)+19}=\sqrt{48-11}-\sqrt{18+19}=\sqrt{37}-\sqrt{37}=0\end{align*}$
::检查时间: 8(6)-11-3(6)+19=48-11-18+19=37-37=0

$\begin{align*}\sqrt[4]{4x+1}=x\end{align*}$
::4x+14=x 4x+14=x

The radical is isolated. To eliminate it, we must raise both sides to the fourth power.
::激进分子是孤立的,要消灭它,我们必须将双方提升到第四势力。

$\begin{aligned} {\sqrt[4]{2 x^{2} - 1}}^{4} & = x^{4} \\ 2 x^{2} - 1 & = x^{4} \\ 0 & = x^{4} - 2 x^{2} + 1 \\ 0 & = (x^{2} - 1) (x^{2} - 1) \\ 0 & = (x - 1) (x + 1) (x - 1) (x + 1) \\ x & = 1 o r - 1 \end{aligned}$
$\begin{align*}\sqrt[4]{2x^2-1}^4&=x^4 \\ 2x^2-1&=x^4 \\ 0&=x^4-2x^2+1 \\ 0&=(x^2-1)(x^2-1)\\ 0&=(x-1)(x+1)(x-1)(x+1)\\ x&=1 \ or \ -1\end{align*}$
::2x2 - 144=x42x2 - 1=x40=x4 - 2x2+10=(x2 - 1)(x2 - 1)(x2 - 1)(x2 - 1)0=(x - 1)(x+1)(x - 1)(x - 1)(x+1)(x+1)x1=1或-1

Check: $\begin{align*}\sqrt[4]{2(1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1 \end{align*}$
::查询: 2(1)2- 14=2- 14=14=14=1

$\begin{align*}\sqrt[4]{2(-1)^2-1}=\sqrt[4]{2-1}=\sqrt[4]{1}=1\end{align*}$

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the lengths of the sides with the unknown values.
::早些时候,你被要求找到双方的长度与未知的价值观。

Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the sides with unknowns.
::使用 Pythagorean 定理并解析 x , 然后替换该值, 以解析有未知点的侧面。

$\begin{aligned} 12^{2} + (\sqrt{x + 1})^{2}) = (\sqrt{7 x + 1})^{2} \\ 144 + x + 1 = 7 x + 1 \\ 144 = 6 x \\ x = 24 \end{aligned}$
$\begin{align*}12^2 +(\sqrt {x + 1})^2) = (\sqrt {7x + 1})^2 \\ 144 + x + 1 = 7x + 1\\ 144 = 6x\\ x = 24\end{align*}$
::122+(x+1)2=(7x+1)2144+x+1=7x+1144=6xx=24

Now substitute this value into the sides with the unknowns.
::现在用未知来将这个值替换为两面。

$\begin{align*} \sqrt{x + 1} = \sqrt {24 + 1} = 5\end{align*}$ and
::x+1=24+1=5和

$\begin{align*} \sqrt{7x + 1} = \sqrt {[7(24)] + 1} = \sqrt {169} = 13\end{align*}$
::7x+1=[7(24)]+1=169=13

Therefore the leg with the unknown measures 5 and the hypotenuse measures 13.
::因此,有未知措施5和下限措施13的腿上有未知措施5和下限措施13。

Solve the following radical equations. Check for extraneous solutions.
::解决以下的激进方程式。检查不相干的解决办法。

Example 2
::例2

$\begin{align*}\sqrt[3]{4x^3-24}=x\end{align*}$
::4x3-2243=x

The radical is isolated. Cube both sides to eliminate the cubed root .
::激进分子是孤立的立方体两边消灭立方根

$\begin{aligned} {\sqrt[3]{4 x^{3} - 24}}^{3} & = x^{3} \\ 4 x^{3} - 24 & = x^{3} \\ - 24 & = - 3 x^{3} \\ 8 & = x^{3} \\ 2 & = x \end{aligned}$
$\begin{align*}\sqrt[3]{4x^3-24}^3&=x^3\\ 4x^3-24&=x^3\\ -24&=-3x^3\\ 8&=x^3\\ 2&=x\end{align*}$
::4x3 - 2433=x34x3 - 24=x3 - 24*3 - 3x38=x32=x

Check: $\begin{align*}\sqrt[3]{4 \left(2 \right)^3-24}=\sqrt[3]{32-24}=\sqrt[3]{8}=2\end{align*}$
::查询:4(2)3-243=32-243=83=2

Example 3
::例3

$\begin{align*}\sqrt{5x-3}=\sqrt{3x+19}\end{align*}$
::5x-3=3x+19

Square both sides to solve for $\begin{align*}x\end{align*}$ .
::双方平方解决 x 。

$\begin{aligned} {\sqrt{5 x - 3}}^{2} & = {\sqrt{3 x + 19}}^{2} \\ 5 x - 3 & = 3 x + 19 \\ 2 x & = 22 \\ x & = 11 \end{aligned}$
$\begin{align*}\sqrt{5x-3}^2&=\sqrt{3x+19}^2\\ 5x-3&=3x+19\\ 2x&=22\\ x&=11\end{align*}$
::5x-32=3x+1925x-3=3x+192x=22x=11

Check:
$\begin{aligned} \sqrt{5 (11) - 3} & = \sqrt{3 (11) + 19} \\ \sqrt{55 - 3} & = \sqrt{33 + 19} \\ \sqrt{52} & = \sqrt{52} \end{aligned}$
$\begin{align*}\sqrt{5 \left(11 \right)-3}&=\sqrt{3 \left(11 \right)+19} \\ \sqrt{55-3}&=\sqrt{33+19} \quad \\ \sqrt{52}&=\sqrt{52}\end{align*}$
::检查: 5( 11)- 3=3( 11)+1955-3=33+1952=52

Example 4
::例4

$\begin{align*}\sqrt{6x-5}-x=-10\end{align*}$
::6 - 5 - x* 10

Add $\begin{align*}x\end{align*}$ to both sides and square to eliminate the radical.
::在两边和广场上加x 来消灭激进分子

$\begin{aligned} {\sqrt{6 x - 5}}^{2} & = (x - 10)^{2} \\ 6 x - 5 & = x^{2} - 20 x + 100 \\ 0 & = x^{2} - 26 x + 105 \\ 0 & = (x - 21) (x - 5) \\ x & = 21 o r 5 \end{aligned}$
$\begin{align*}\sqrt{6x-5}^2&=(x-10)^2\\ 6x-5&=x^2-20x+100\\ 0&=x^2-26x+105\\ 0&=(x-21)(x-5)\\ x&=21 \ or \ 5\end{align*}$
::6-52=(x-(10)26x-5=x2-20x+1000=x2-2-26x+1050=(x-21)(x-55x=21或5)

Check both solutions:
$\begin{aligned} x & = 21 : \sqrt{6 (21) - 5} - 21 = \sqrt{126 - 5} - 21 = \sqrt{121} - 21 = 11 - 21 = - 10 \\ x & = 5 : \sqrt{6 (5) - 5} - 21 = \sqrt{30 - 5} - 21 = \sqrt{25} - 21 = 5 - 21 \neq - 10 \end{aligned}$
$\begin{align*}x&= 21: \sqrt{6 \left(21 \right)-5}-21=\sqrt{126-5}-21=\sqrt{121}-21=11-21=-10 \\ x&= 5: \sqrt{6 \left(5 \right)-5}-21=\sqrt{30-5}-21=\sqrt{25}-21=5-21 \neq -10\end{align*}$
::检查两个解决方案 : x=21: 6(21)- 5- 21= 126-5- 5- 21= 121- 21= 11- 21_ @ 10x= 5: 6(5)- 5- 21= 30- 5- 21= 25- 21= 5- 21\ @ 10

5 is an extraneous solution.
::5是一个不相干的解决办法。

Review
::回顾

Solve the following radical equations. Be sure to check for extraneous solutions.
::解决以下的激进方程式。请务必检查不相干的解决办法。

$\begin{align*}\sqrt{x-3}=x-5\end{align*}$
::x-3=x-5

$\begin{align*}\sqrt{x+3}+15=x-12\end{align*}$
::x+3+15=x- 12

$\begin{align*}\sqrt[4]{3x^2+54}=x\end{align*}$
::3x2+544=x

$\begin{align*}\sqrt{x^2+60}=4\sqrt{x}\end{align*}$
::x2+60=4x

$\begin{align*}\sqrt{x^4+5x^3}=2\sqrt{2x+10}\end{align*}$
::x4+5x3=22x+10

$\begin{align*}x=\sqrt{5x-6}\end{align*}$
::x=5x- 6x=5x-6

$\begin{align*}\sqrt{3x+4}=x-2\end{align*}$
::3x+4=x-2

$\begin{align*}\sqrt{x^3+8x}-\sqrt{9x^2-60}=0\end{align*}$
::x3+8x-9x2-60=0

$\begin{align*}x=\sqrt[3]{4x+4-x^2}\end{align*}$
::x=4x+4-x23 x=4x+4-x23

$\begin{align*}\sqrt[4]{x^3+3}=2\sqrt[4]{x+3}\end{align*}$
::x3+34=2x+34

$\begin{align*}x^2-\sqrt{42x^2+343}=0\end{align*}$
::x2 - 42x2+343=0

$\begin{align*}x\sqrt{x^2-21}=2\sqrt{x^3-25x+25}\end{align*}$
::x2-21-21=2x3-25x+25

For questions 13-15, you will need to use the method illustrated in the example below.
::对于问题13-15,你将需要使用下面示例中说明的方法。

$\begin{aligned} \sqrt{x - 15} & = \sqrt{x} - 3 \\ {(\sqrt{x - 15})}^{2} & = {(\sqrt{x} - 3)}^{2} \\ x - 15 & = x - 6 \sqrt{x} + 9 \\ - 24 & = - 6 \sqrt{x} \\ (4)^{2} & = {(\sqrt{x})}^{2} \\ 16 & = x \end{aligned}$
$\begin{align*}\sqrt{x-15}&=\sqrt{x}-3\\ \left(\sqrt{x-15}\right)^2&=\left(\sqrt{x}-3 \right)^2\\ x-15&=x-6 \sqrt{x}+9\\ -24&=-6 \sqrt{x}\\ (4)^2&=\left(\sqrt{x}\right)^2\\ 16&=x\end{align*}$
::x- 15=x-3(x- 15) 2=(x-3) 2x- 15=x-6x+9- 24=_ 6x(4)2=(x) 216=x

Square both sides
::两边广场

Combine like terms to isolate the remaining radical
::结合用词来隔离剩下的激进分子

Square both sides again to solve
::双方重开广场,再解决

Check: Don't forget to check your answers for extraneous solutions!
::检查: 不要忘了检查您的答案, 以寻找不相干的解决办法 !

$\begin{aligned} \sqrt{16 - 15} & = \sqrt{16} - 3 \\ \sqrt{1} & = 4 - 3 \\ 1 & = 1 \end{aligned}$
$\begin{align*}\sqrt{16-15}&=\sqrt{16}-3 \\ \sqrt{1}&=4-3 \\ 1&=1\end{align*}$

$\begin{align*}\sqrt{x+11}-2=\sqrt{x-21}\end{align*}$
::x+11-2=x-21

$\begin{align*}\sqrt{x-6}=\sqrt{7x}-22\end{align*}$
::x-6=7x-22xx-6=7x-22

$\begin{align*}2+\sqrt{x+5}=\sqrt{4x-7}\end{align*}$
::2+x+5=4x-7

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。

Section outline

General

解决双方有变量的激进方程式

Solving Radical Equations with Variables ::用变量解决极端等式

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 with Variables
::用变量解决极端等式

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)