课程： 5.11 当领导系数等于1时完成广场

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选择节当领导系数等于1时完成广场

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当领导系数等于1时完成广场
The area of a parallelogram is given by the equation $\begin{align*}x^2 + 8x - 5 = 0\end{align*}$ , where x is the length of the base. What is the length of this base?
::x2+8x-5=0 方程式给出了平行图区域, x 是基数的长度。此基数的长度是多少 ?

Completing the Square
::完成广场

is another technique used to . When completing the square, the goal is to make a perfect square trinomial and factor it.
::使用的另一个技术是.在完成广场时,目标是使一个完美的正方形三边形,并将它作为因素。

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

$\begin{aligned} x^{2} - 8 x - 1 = 10 \end{aligned}$
$\begin{align*}x^2-8x-1=10\end{align*}$
::x2-8x- 1=10

Step 1: Write the polynomial so that $\begin{align*}x^2\end{align*}$ and $\begin{align*}x\end{align*}$ are on the left side of the equation and the constants on the right. This is only for organizational purposes, but it really helps. Leave a little space after the $\begin{align*}x-\end{align*}$ term .
::步骤 1: 写入多义, 使 x2 和 x 位于方程的左侧, 以及右边的常数。这只用于组织目的, 但确实有帮助。在 x- 期后留下一点空间。

$\begin{aligned} x^{2} - 8 x = 11 \end{aligned}$
$\begin{align*}x^2-8x=11\end{align*}$
::x2- 8x=11

Step 2: Now, "complete the square." Determine what number would make a perfect square trinomial with $\begin{align*}x^2-8x+c\end{align*}$ . To do this, divide the $\begin{align*}x-\end{align*}$ term by 2 and square that number, or $\begin{align*}\left(\frac{b}{2}\right)^2\end{align*}$ .
::步数 2: 现在, “ 完成广场 ” 。确定哪个数字可以使 X2 - 8x+c 的完全平方三角形成为 X2 - 8x+c 。要做到这一点, 将 x - term 除以 2 和该数字的正方形, 或者 (b2) 2 。

$\begin{aligned} {(\frac{b}{2})}^{2} = {(\frac{8}{2})}^{2} = 4^{2} = 16 \end{aligned}$
$\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{8}{2}\right)^2=4^2=16\end{align*}$
:b2)2=((82)2=42=16)

Step 3: Add this number to both sides in order to keep the equation balanced.
::第3步:在双方增加这一数目,以保持平衡。

$\begin{aligned} x^{2} - 8 x + 16 = 11 + 16 \end{aligned}$
$\begin{align*}x^2-8x {\color{red}+16}=11 {\color{red}+16}\end{align*}$
::x2-8x+16=11+16

Step 4: Factor the left side to the square of a binomial and simplify the right.
::第4步:将左侧乘以二进制方形,并简化右侧。

$\begin{aligned} (x - 4)^{2} = 27 \end{aligned}$
$\begin{align*}(x-4)^2=27\end{align*}$
:x-4)2=27

Step 5: Solve by using square roots.
::步骤5:用平方根解决。

$\begin{aligned} x - 4 & = \pm 3 \sqrt{3} \\ x & = 4 \pm 3 \sqrt{3} \end{aligned}$
$\begin{align*}x-4 &=\pm3\sqrt{3}\\ x &=4 \pm 3 \sqrt{3}\end{align*}$
::x - 433x=433

Completing the square enables you to solve any quadratic equation using square roots. Through this process, we can make an unfactorable quadratic equation solvable, like the one above. It can also be used with quadratic equations that have imaginary solutions.
::完成方形后, 您可以使用平方根解决任何二次方程。通过此进程, 我们可以使一个不可计数的二次方程可以溶解, 如上方方方程式。它也可以用于具有想象解决方案的二次方程。

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

$\begin{aligned} x^{2} + 12 x + 37 = 0 \end{aligned}$
$\begin{align*}x^2+12x+37=0\end{align*}$
::x2+12x+37=0

First, this is not a factorable quadratic equation. Therefore , the only way we know to solve this equation is to complete the square. Follow the steps from the problem above.
::首先,这不是一个可考虑因素的二次方程。因此,我们唯一能解决这个方程的方法就是完成方程。遵循上述问题的步骤。

Step 1: Organize the polynomial, $\begin{align*}x\end{align*}$ 's on the left, constant on the right.
::第1步:组织多面体,x在左侧,在右侧恒定。

$\begin{aligned} x^{2} + 12 x = - 37 \end{aligned}$
$\begin{align*}x^2+12x=-37\end{align*}$
::x2+12x% 37

Step 2: Find $\begin{align*}\left(\frac{b}{2}\right)^2\end{align*}$ and add it to both sides.
::第2步:找出(b2)2 并将其添加到双方。

$\begin{aligned} {(\frac{b}{2})}^{2} = {(\frac{12}{2})}^{2} & = 6^{2} = 36 \\ x^{2} + 12 x + 36 & = - 37 + 36 \end{aligned}$
$\begin{align*}\left(\frac{b}{2}\right)^2 = \left(\frac{12}{2}\right)^2 &=6^2={\color{red}36}\\ x^2+12x+{\color{red}36} &=-37+{\color{red}36}\end{align*}$
:b2)2=(122)2=62=36x2+12x+36+37+36)

Step 3: Factor the left side and solve.
$\begin{aligned} (x + 6)^{2} & = - 1 \\ x + 6 & = \pm i \\ x & = - 6 \pm i \end{aligned}$
$\begin{align*}(x+6)^2 &=-1\\ x+6 &=\pm i\\ x &=-6\pm i\end{align*}$
::步骤 3 : 将左侧和溶解值乘以。 (x+6) 2 @% 1x+6_ix @ 6i

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

$\begin{aligned} x^{2} - 11 x - 15 = 0 \end{aligned}$
$\begin{align*}x^2-11x-15=0\end{align*}$
::x2- 11x- 15=0

This is not a factorable equation. Use completing the square.
::这不是一个可乘方程。使用完成方形。

Step 1: Organize the polynomial, $\begin{align*}x\end{align*}$ ’s on the left, constant on the right.
::第一步 : 组织多面性, x 在左侧, 在右侧常数。

$\begin{aligned} x^{2} - 11 x = 15 \end{aligned}$
$\begin{align*}x^2-11x=15\end{align*}$
::x2- 11x=15

Step 2: Find $\begin{align*}\left(\frac{b}{2}\right)^2\end{align*}$ and add it to both sides.
::第2步:找出(b2)2 并将其添加到双方。

$\begin{aligned} {(\frac{b}{2})}^{2} & = {(\frac{11}{2})}^{2} = \frac{121}{4} \\ x^{2} - 11 x + \frac{121}{4} & = 15 + \frac{121}{4} \end{aligned}$
$\begin{align*}\left(\frac{b}{2}\right)^2 &=\left(\frac{11}{2}\right)^2 = {\color{red}\frac{121}{4}}\\ x^2-11x+{\color{red}\frac{121}{4}} &=15+{\color{red}\frac{121}{4}}\end{align*}$
:b2)2=(112)2=1214x2-11x+1214=15+1214)

Step 3: Factor the left side and solve.
::第三步: 将左侧因素乘以解决。

$\begin{aligned} {(x - \frac{11}{2})}^{2} & = \frac{60}{4} + \frac{121}{4} \\ {(x - \frac{11}{2})}^{2} & = \frac{181}{4} \\ x - \frac{11}{2} & = \pm \frac{\sqrt{181}}{2} \\ x & = \frac{11}{2} \pm \frac{\sqrt{181}}{2} \end{aligned}$
$\begin{align*}\left(x-\frac{11}{2}\right)^2 &=\frac{60}{4}+{\color{red}\frac{121}{4}}\\ \left(x-\frac{11}{2}\right)^2 &=\frac{181}{4}\\ x-\frac{11}{2} &=\pm \frac{\sqrt{181}}{2}\\ x &=\frac{11}{2}\pm \frac{\sqrt{181}}{2}\end{align*}$
:x-112)2=604+1214(x-112)2=1814x-112)2=1814x-112=1812x=1112-1812)

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the length of the base of the parallelogram.
::早些时候,有人要求你找到平行图的底部长度。

We can't factor $\begin{align*}x^2 + 8x - 5 = 0\end{align*}$ , so we must complete the square.
::我们不能乘以 x2+8x-5=0, 所以我们必须完成方块。

Step 1: Write the polynomial so that $\begin{align*}x^2\end{align*}$ and $\begin{align*}x\end{align*}$ are on the left side of the equation and the constants are on the right.
::步骤 1: 写入多数值, 使 x2 和 x 位于方程的左侧, 而常数位于右侧。

$\begin{aligned} x^{2} + 8 x = 5 \end{aligned}$
$\begin{align*}x^2 + 8x = 5\end{align*}$
::x2+8x=5

Step 2: Now, complete the square.
$\begin{aligned} {(\frac{b}{2})}^{2} = {(\frac{8}{2})}^{2} = 4^{2} = 16 \end{aligned}$
$\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{8}{2}\right)^2=4^2=16\end{align*}$
::第2步:现在,完成方形。 (b2) 2=(82)2=42=16

Step 3: Add this number to both sides in order to keep the equation balanced.
::第3步:在双方增加这一数目,以保持平衡。

$\begin{aligned} x^{2} + 8 x + 16 = 5 + 16 \end{aligned}$
$\begin{align*}x^2 + 8x {\color{red} + 16}=5 {\color{red} + 16}\end{align*}$
::x2+8x+16=5+16

Step 4: Factor the left side to the square of a binomial and simplify the right.
::第4步:将左侧乘以二进制方形,并简化右侧。

$\begin{aligned} (x + 4)^{2} = 21 \end{aligned}$
$\begin{align*}(x + 4)^2=21\end{align*}$
:x+4)2=21

Step 5: Solve by using square roots.
::步骤5:用平方根解决。

$\begin{aligned} x + 4 & = \pm \sqrt{21} \\ x & = - 4 \pm \sqrt{21} \end{aligned}$
$\begin{align*}x + 4 &=\pm\sqrt{21}\\ x &= -4 \pm \sqrt{21}\end{align*}$
::x+421x421

However, because x is the length of the parallelogram's base, it must be a positive value. Only $\begin{align*}-4 + \sqrt{21}\end{align*}$ results in a positive value. Therefore, the length of the base is $\begin{align*}-4 + \sqrt{21}\end{align*}$ .
::然而,由于 x 是平行图基数的长度, 它必须是正值。只有 - 4+21 得出正值。因此, 基数的长度是 - 4+21 。

Example 2
::例2

Find the value of $\begin{align*}c\end{align*}$ that would make $\begin{align*}x^2-2x+c\end{align*}$ a perfect square trinomial. Then, factor the trinomial .
::查找 c 的值, 使 x2 - 2x+c 成为完全平方的三角。然后, 乘以三角。

$\begin{align*}c=\left(\frac{b}{2}\right)^2=\left(\frac{2}{2}\right)^2=1^2=1\end{align*}$ . The factors of $\begin{align*}x^2-2x+1\end{align*}$ are $\begin{align*}(x - 1)(x - 1)\end{align*}$ or $\begin{align*}(x - 1)^2\end{align*}$ .
::c=(b2)2=(222)2=12=1. x2-2x+1的系数为(x-1)(x-1)-1或(x-1)2。

Example 3
::例3

Solve the following quadratic equation by completing the square: $\begin{align*}x^2+10x+21=0\end{align*}$ .
::通过完成正方形(x2+10x+21=0)来解决以下二次方程。

Use the steps from the examples above.
::使用上述例子中的步骤。

$\begin{aligned} x^{2} + 10 x + 21 & = 0 \\ x^{2} + 10 x & = - 21 \\ x^{2} + 10 x + {(\frac{10}{2})}^{2} & = - 21 + {(\frac{10}{2})}^{2} \\ x^{2} + 10 x + 25 & = - 21 + 25 \\ (x + 5)^{2} & = 4 \\ x + 5 & = \pm 2 \\ x & = - 5 \pm 2 \\ x & = - 7, - 3 \end{aligned}$
$\begin{align*}x^2+10x+21 &=0\\ x^2+10x &=-21\\ x^2+10x+\left(\frac{10}{2}\right)^2 &=-21+\left(\frac{10}{2}\right)^2\\ x^2+10x+25 &=-21+25\\ (x+5)^2 &=4\\ x+5 &=\pm 2\\ x &=-5\pm 2\\ x &=-7,-3\end{align*}$
::x2+10xx+21=0x2+10x21x2+10x+(102)221+(102)2x2+10x+25}#21+25(x+5)2=4x+5}2=5x2x2x2x_2x7 -3

Example 4
::例4

Solve the following quadratic equation by completing the square: $\begin{align*}x-5x=12\end{align*}$ .
::通过完成正方形(x-5x=12)来解析以下二次方程。

Use the steps from the examples above.
::使用上述例子中的步骤。

$\begin{aligned} x^{2} - 5 x & = 12 \\ x^{2} - 5 x + {(\frac{5}{2})}^{2} & = 12 + {(\frac{5}{2})}^{2} \\ x^{2} - 5 x + \frac{25}{4} & = \frac{48}{4} + \frac{25}{4} \\ {(x - \frac{5}{2})}^{2} & = \frac{73}{4} \\ x - \frac{5}{2} & = \pm \frac{\sqrt{73}}{2} \\ x & = \frac{5}{2} \pm \frac{\sqrt{73}}{2} \end{aligned}$
$\begin{align*}x^2-5x &=12\\ x^2-5x+\left(\frac{5}{2}\right)^2 &=12+\left(\frac{5}{2}\right)^2\\ x^2-5x+\frac{25}{4} &=\frac{48}{4}+\frac{25}{4}\\ \left(x-\frac{5}{2}\right)^2 &=\frac{73}{4}\\ x-\frac{5}{2} &=\pm \frac{\sqrt{73}}{2}\\ x &=\frac{5}{2}\pm \frac{\sqrt{73}}{2}\end{align*}$
::x2-5x=12x2-5x+(522)2=12+(522)2x2-5x+254=484+254(x-52)2=734x-52=732x=52_732x

Review
::回顾

Determine the value of $\begin{align*}c\end{align*}$ that would complete the perfect square trinomial.
::确定C值,以完成完美的平方三角。

$\begin{align*}x^2+4x+c\end{align*}$
::x2+4x+c x2+4x+c

$\begin{align*}x^2-2x+c\end{align*}$
::x2-2x+c x2-2x+c

$\begin{align*}x^2+16x+c\end{align*}$
::x2+16x+c x2+16x+c

Rewrite the perfect square trinomial as a square of a binomial.
::将完美的正方形三角形重写为二进制的正方形。

$\begin{align*}x^2+6x+9\end{align*}$
::x2+6x+9

$\begin{align*}x^2-7x+\frac{49}{4}\end{align*}$
::x2-7x+494

$\begin{align*}x^2-\frac{1}{2}x+\frac{1}{16}\end{align*}$
::x2 - 12x+116

Solve the following quadratic equations by completing the square.
::通过完成广场来解决以下四方方程式。

$\begin{align*}x^2+6x-15=0\end{align*}$
::x2+6x- 15=0

$\begin{align*}x^2+10x+29=0\end{align*}$
::x2+10x+29=0

$\begin{align*}x^2-14x+9=-60\end{align*}$
::x2 - 14x+9*% 60

$\begin{align*}x^2+3x+18=-2\end{align*}$
::x2+3x+18 @%2

$\begin{align*}x^2-9x-5=23\end{align*}$
::x2-9x-5=23

$\begin{align*}x^2-20x=60\end{align*}$
::x2-20x=60

Solve the following quadratic equations by factoring, square roots, or completing the square.
::通过保理、平方根或完成方形,解决以下四方方程式。

$\begin{align*}x^2+x-30=0\end{align*}$
::x2+x-30=0

$\begin{align*}x^2-18x+90=0\end{align*}$
::x2 - 18x+90=0

$\begin{align*}x^2+15x+56=0\end{align*}$
::x2+15x+56=0

$\begin{align*}x^2+3x-24=12\end{align*}$
::x2+3x-24=12

$\begin{align*}(x-2)^2-20=-45\end{align*}$
:x-2)2 - 2045

$\begin{align*}x^2+24x+44=-19\end{align*}$
::x2+24x+4419

Solve $\begin{align*}x^2+7x-44=0\end{align*}$ by factoring and completing the square. Which method do you prefer?
::通过乘数和完成正方形来解决 x2+7x-44=0。您喜欢哪种方法 ?

Challenge Solve $\begin{align*}x^2+\frac{17}{8}x-2=-9\end{align*}$ .
::挑战解决 x2+178x-29.

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

常规

当领导系数等于1时完成广场

Completing the Square ::完成广场

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)

Completing the Square
::完成广场

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)