Section: 解决具有复杂数字解决方案的赤道等量 | 5.10 用复杂数字解决方案解决赤道赤道

Section outline

Miss Harback writes the equation $\begin{aligned} 5 x^{2} + 125 = 0 \end{aligned}$ on the board. She asks the class how many solutions the equation has and what type they are.
::Harback 小姐在棋盘上写了 5x2+125=0 的方程式。她问该类方程式有多少解决方案以及它们是什么类型。

Corrine says the equation has two real solutions. Drushel says the equation has a double root, so only one solution. Farrah says the equation has two imaginary solutions.
::Corrine说方程式有两个真正的解决方案。Drushel说方程式有一个双根,所以只有一个解决方案。Farrah说方程式有两个假想解决方案。

Which one of them is correct?
::哪一个是正确的?

Quadratic Equations with Complex Number Solutions
::具有复杂数字解决方案的赤道等赤道

When you solve a quadratic equation, there will always be two answers. Until now, we thought the answers were always real numbers. In actuality, there are quadratic equations that have imaginary solutions as well. The possible solutions for a quadratic are:
::当您解答二次方程式时, 总是有两个答案。直到现在, 我们一直认为答案总是真实的数字。实际上, 有二次方程式也有想象中的解决方案。对于二次方程式, 可能的解决办法是 :

2 real solutions
::2个实际解决方案

$\begin{aligned} x^{2} - 4 & = 0 \\ x & = - 2, 2 \end{aligned}$

::x2-4=0x%2,2

Double root
::双根双根

$\begin{aligned} x^{2} + 4 x + 4 & = 0 \\ x & = - 2, - 2 \end{aligned}$

::x2+4x+4=0x%2,-2

2 imaginary solutions
::2个想象的解决方案

$\begin{aligned} x^{2} + 4 & = 0 \\ x & = - 2 i, 2 i \end{aligned}$

::x2+4=0x%2i,2i

Let's solve the following quadratic equations.
::让我们解决以下的二次方程。

Solve $\begin{aligned} 3 x^{2} + 27 = 0 \end{aligned}$ .
::解决 3x2+27=0。

First, factor out the GCF.
::首先,考虑一下全球合作框架。

$\begin{aligned} 3 (x^{2} + 9) = 0 \end{aligned}$

::3(x2+9)=0

Now, try to factor $\begin{aligned} x^{2} + 9 \end{aligned}$ . Rewrite the quadratic as $\begin{aligned} x^{2} + 0 x + 9 \end{aligned}$ to help. There are no factors of 9 that add up to 0. Therefore, this is not a factorable quadratic. Let’s solve it using square roots.
::现在, 尝试乘以 x2+9 。重写二次曲线为 x2+0x+9 来帮助您。没有9 乘以 0 的系数。因此, 这不是一个可乘因数的二次曲线。让我们用正方根解决它。

$\begin{aligned} 3 x^{2} + 27 & = 0 \\ 3 x^{2} & = - 27 \\ x^{2} & = - 9 \\ x & = \pm \sqrt{- 9} = \pm 3 i \end{aligned}$

::3x2+27=03x227x29x993i

Quadratic equations with imaginary solutions are never factorable.
::带有假想解决方案的赤道方程式从来就不是可考虑的因素。

Solve $\begin{aligned} (x - 8)^{2} = - 25 \end{aligned}$ .
::解决(x-8)2+25。

Solve using square roots.
::用平方根解决

$\begin{aligned} (x - 8)^{2} & = - 25 \\ x - 8 & = \pm 5 i \\ x & = 8 \pm 5 i \end{aligned}$

:x-8)225x-85ix=85i

Solve $\begin{aligned} 2 (3 x - 5) + 10 = - 30 \end{aligned}$ .
::解决 2( 2 3x- 5) + 10 30 。

Solve using square roots.
::用平方根解决

$\begin{aligned} 2 (3 x - 5)^{2} + 10 & = - 30 \\ 2 (3 x - 5)^{2} & = - 40 \\ (3 x - 5)^{2} & = - 20 \\ 3 x - 5 & = \pm 2 i \sqrt{5} \\ 3 x & = 5 \pm 2 i \sqrt{5} \\ x & = \frac{5}{3} \pm \frac{2 \sqrt{5}}{3} i \end{aligned}$

::2(3x-55)2+10 302(3x-5)2+40(3x-5)2}(3x-5)2}(203x-5)2}(203x-5}2i53x=52i5x=53253i

Examples
::实例

Example 1
::例1

Earlier, you were asked to find which student is correct.
::早些时候,有人要求你找出哪个学生是正确的。

To solve $\begin{aligned} 5 x^{2} + 125 = 0 \end{aligned}$ , we first need to factor out the GCF.
::为了解决5x2+125=0,我们首先需要考虑全球合作框架。

$\begin{aligned} 5 (x^{2} + 25) = 0 \end{aligned}$

::5(x2+25)=0

Now, try to factor $\begin{aligned} x^{2} + 25 \end{aligned}$ . Rewrite the quadratic as $\begin{aligned} x^{2} + 0 x + 25 \end{aligned}$ to help. There are no factors of 25 that add up to 0. Therefore, this is not a factorable quadratic. Let’s solve it using square roots.
::现在, 尝试乘以 x2+25。重写二次方块为 x2+0x+25 来提供帮助。没有25个系数加起来等于 0。因此, 这不是一个可乘因数的二次方块。让我们用正方根解决这个问题。

$\begin{aligned} 5 x^{2} + 125 & = 0 \\ 5 x^{2} & = - 125 \\ x^{2} & = - 25 \\ x & = \pm \sqrt{- 5} = \pm 5 i \end{aligned}$

::5x2+125=05x2125x225x555i

The equation has two roots and both of them are imaginary, so Farrah is correct.
::方程式有两根根两者都是想象出来的所以法拉是对的

Example 2
::例2

Solve $\begin{aligned} 4 (x - 5)^{2} + 49 = 0 \end{aligned}$ .
::4 (x-5)2+49=0 解决(x-5)2+49=0。

$\begin{aligned} 4 (x - 5)^{2} + 49 & = 0 \\ 4 (x - 5)^{2} & = - 49 \\ (x - 5)^{2} & = - \frac{49}{4} \\ x - 5 & = \pm \frac{7}{2} i \\ x & = 5 \pm \frac{7}{2} i \end{aligned}$

::4(x-5-5)2+49=04(x-5)249(x-5)249(x-5)249x-5*5*72ix=572i

Example 3
::例3

Solve $\begin{aligned} - \frac{1}{2} (3 x + 8)^{2} - 16 = 2 \end{aligned}$ .
::解决 - 12( 3x+8) 2 - 16=2。

$\begin{aligned} - \frac{1}{2} (3 x + 8)^{2} - 16 & = 2 \\ - \frac{1}{2} (3 x + 8)^{2} & = 18 \\ (3 x + 8)^{2} & = - 36 \\ 3 x + 8 & = \pm 6 i \\ 3 x & = - 8 \pm 6 i \\ x & = - \frac{8}{3} \pm 2 i \end{aligned}$

::- 12(3x+8)2--16=2-12(3x+8)2=18(3x+8)2363x+8_6i3x_8_6x_8_6x_8_6x_8_6x_8_8_2i

Both of the quadratic equations in Examples 2 and 3 can be solved by using square roots.
::例2和例3中的两个二次方程都可以用平方根来解决。

Review
::回顾

Solve the following quadratic equations.
::解决以下的二次方程。

$\begin{aligned} x^{2} = - 9 \end{aligned}$
::x2% 9

$\begin{aligned} x^{2} + 8 = 3 \end{aligned}$
::x2+8=3

$\begin{aligned} (x + 1)^{2} = - 121 \end{aligned}$
:x+1)2_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

$\begin{aligned} 5 x^{2} + 16 = - 29 \end{aligned}$
::5x2+16=29

$\begin{aligned} 14 - 4 x^{2} = 38 \end{aligned}$
::14-4x2=38

$\begin{aligned} (x - 9)^{2} - 2 = - 82 \end{aligned}$
:x--9)2-282

$\begin{aligned} - 3 (x + 6)^{2} + 1 = 37 \end{aligned}$
::-3(x+6)2+1=37

$\begin{aligned} 4 (x - 5)^{2} - 3 = - 59 \end{aligned}$
::4(x-5-5)2-359

$\begin{aligned} (2 x - 1)^{2} + 5 = - 23 \end{aligned}$
:2x-1)2+523

$\begin{aligned} - (6 x + 5)^{2} = 72 \end{aligned}$
::- (6x+5)2=72

$\begin{aligned} 7 (4 x - 3)^{2} - 15 = - 68 \end{aligned}$
::7(4x-3-3)2-1568

If a quadratic equation has $\begin{aligned} 4 - i \end{aligned}$ as a solution, what must the other solution be?
::如果四方方程式有4-i作为解决办法,其他解决办法又是什么?

If a quadratic equation has $\begin{aligned} 6 + 2 i \end{aligned}$ as a solution, what must the other solution be?
::如果四方方程式有6+2i作为解决办法,其他解决办法又是什么?

Challenge Recall that the factor of a quadratic equation has the form $\begin{aligned} (x \pm m) \end{aligned}$ where $\begin{aligned} m \end{aligned}$ is any number. Find a quadratic equation that has the solution $\begin{aligned} 3 + 2 i \end{aligned}$ .
::挑战回顾二次方程的因数为m为任意数字的表( xm) 。找到一个有 3+2i 解决方案的二次方程。

Find a quadratic equation that has the solution $\begin{aligned} 1 - i \end{aligned}$ .
::找到一个具有1-i解决方案的二次方程。

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

Quadratic Equations with Complex Number Solutions ::具有复杂数字解决方案的赤道等赤道

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

Quadratic Equations with Complex Number Solutions
::具有复杂数字解决方案的赤道等赤道

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)