Course: 5.13 提取和使用四压公式

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输出和使用二次曲线公式
The profit on your school fundraiser is represented by the quadratic expression $\begin{align*}-3p^2 + 200p - 3000\end{align*}$ , where p is your price point. What is your break-even point (i.e., the price point at which you will begin to make a profit)? Hint: Set the equation equal to zero.
::学校募款人的利润由四边表达式 - 3p2+200p - 3000 表示, p是您的价格点。您的折中点是多少( 即您开始盈利的价格点) ? 提示 : 设置等式为零。

Quadratic Formula
::二次曲线公式

The last way to solve a quadratic equation is the Quadratic Formula . This formula is derived from for the equation $\begin{align*}ax^2+bx+c=0\end{align*}$ . We will derive the formula here.
::解析二次方程的最后方法就是二次方程。这个公式来自方程 x2+bx+c=0。我们将在此得出公式。

Deriving the Quadratic Formula
::生成二次曲线公式

Let's walk through each step of completing the square of $\begin{align*}ax^2+bx+c=0\end{align*}$ .
::让我们走过每一步完成x2+bx+c=0的正方形。

Step 1: Move the constant to the right side of the equation. $\begin{align*}ax^2+bx=-c\end{align*}$
::第1步:将常数移动到方程的右侧。 ax2+bxc

Step 2: “Take out” $\begin{align*}a\end{align*}$ from everything on the left side of the equation. $\begin{align*}a\left(x^2+\frac{b}{a}x\right)=-c\end{align*}$
::步骤2:从方程左侧的一切中“取出”一个。 a(x2+bax) {c

Step 3: Complete the square using $\begin{align*}\frac{b}{a}\end{align*}$ . $\begin{align*}\left(\frac{b}{2}\right)^2=\left(\frac{b}{2a}\right)^2=\frac{b^2}{4a^2}\end{align*}$
::第3步:使用ba.(b2)2=(b2a)2=b24a2完成广场

Step 4: Add this number to both sides. Don’t forget on the right side, you need to multiply it by $\begin{align*}a\end{align*}$ (to account for the $\begin{align*}a\end{align*}$ outside the parenthesis). $\begin{align*}a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\right)=-c+\frac{b^2}{4a}\end{align*}$
::步骤 4 : 将这个数字添加到两边。不要忘记右边的右边, 您需要将其乘以一个( 以说明外括号) a (x2+bax+b24a2) \\ c+b24a) \ c+b24a 。

Step 5: Factor the quadratic equation inside the parenthesis and give the right hand side a common denominator. $\begin{align*}a\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a}\end{align*}$
::第5步:将括号内的二次方程乘以,给右手侧一个共同分母。 a(x+b2a)2=b2-4ac4a

Step 6: Divide both sides by $\begin{align*}a\end{align*}$ . $\begin{align*}\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\end{align*}$
::第6步:将双方除以a.(x+b2a)2=b2-4ac4a2

Step 7: Take the square root of both sides. $\begin{align*}x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}\end{align*}$
::步骤7:采取双方的平方根。 x+b2a_b2-4ac2a

Step 8: Subtract $\begin{align*}\frac{b}{2a}\end{align*}$ from both sides to get $\begin{align*}x\end{align*}$ by itself. $\begin{align*}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\end{align*}$
::步骤8:从两边减去b2a,以获得x本身。 xbb2-4ac2a

This formula will enable you to solve any quadratic equation as long as you know $\begin{align*}a, b\end{align*}$ , and $\begin{align*}c\end{align*}$ (from $\begin{align*}ax^2+bx+c=0\end{align*}$ ).
::此公式将允许您解析任何二次方程, 只要您知道 a、 b 和 c (来自 x2+bx+c=0) 。

Solve the quadratic equation using the Quadratic Formula.
::使用二次方程公式解决二次方程。

$\begin{aligned} 9 x^{2} - 30 x + 26 = 0 \end{aligned}$
$\begin{align*}9x^2-30x+26=0\end{align*}$
::9x2-30x+26=0

First, make sure one side of the equation is zero. Then, find $\begin{align*}a, b,\end{align*}$ and $\begin{align*}c\end{align*}$ . $\begin{align*}a = 9, b = -30, c = 26\end{align*}$ . Now, plug in the values into the formula and solve for $\begin{align*}x\end{align*}$ .
::首先, 确保方程的一面为零。然后, 找到 a, b, 和 c. a= 9, b30, c=26。现在, 在公式中插入值, 并解决 x 。

$\begin{aligned} x & = \frac{- (- 30) \pm \sqrt{(- 30)^{2} - 4 (9) (26)}}{2 (9)} \\ = \frac{30 \pm \sqrt{900 - 936}}{18} \\ = \frac{30 \pm \sqrt{- 36}}{18} \\ = \frac{30 \pm 6 i}{18} \\ = \frac{5}{3} \pm \frac{1}{3} i \end{aligned}$
$\begin{align*}x &=\frac{-(-30)\pm \sqrt{(-30)^2-4(9)(26)}}{2(9)}\\ &=\frac{30\pm \sqrt{900-936}}{18}\\ &=\frac{30\pm \sqrt{-36}}{18}\\ &=\frac{30\pm 6i}{18}\\ &=\frac{5}{3} \pm \frac{1}{3}i\end{align*}$
::X(-)-30-(-)-(-)2-4(9)-(262)-(9)-(9)-(30)-(900)-(93618)-(30)-3618=(30)-3618=(30)-6i18=(53)/13i

Solve the quadratic equation using the Quadratic Formula.
::使用二次方程公式解决二次方程。

$\begin{aligned} 2 x^{2} + 5 x - 15 = - x^{2} + 7 x + 2 \end{aligned}$
$\begin{align*}2x^2+5x-15=-x^2+7x+2\end{align*}$
::2x2+5x- 15_ x2+7x+2

Let’s get everything onto the left side of the equation.
::让我们把一切都放在等式的左边。

$\begin{aligned} 2 x^{2} + 5 x - 15 & = - x^{2} + 7 x + 2 \\ 3 x^{2} - 2 x - 17 & = 0 \end{aligned}$
$\begin{align*}2x^2+5x-15 &=-x^2+7x+2\\ 3x^2-2x-17 &=0\end{align*}$
::2x2+5x- 152+7x+23x2- 2x- 17=0

Now, use $\begin{align*}a = 3, b = -2,\end{align*}$ and $\begin{align*}c = -17\end{align*}$ and plug them into the Quadratic Formula.
::现在,使用a=3,b2,和c17 并把它们插入二次曲线公式。

$\begin{aligned} x & = \frac{- (- 2) \pm \sqrt{(- 2)^{2} - 4 (3) (- 17)}}{2 (3)} \\ = \frac{2 \pm \sqrt{4 + 204}}{6} \\ = \frac{2 \pm \sqrt{208}}{6} \\ = \frac{2 \pm 4 \sqrt{13}}{6} \\ = \frac{1 \pm 2 \sqrt{13}}{3} \end{aligned}$
$\begin{align*}x &=\frac{-(-2) \pm \sqrt{(-2)^2-4(3)(-17)}}{2(3)}\\ &=\frac{2 \pm \sqrt{4+204}}{6}\\ &=\frac{2 \pm \sqrt{208}}{6}\\ &=\frac{2 \pm 4\sqrt{13}}{6}\\ &=\frac{1 \pm 2\sqrt{13}}{3}\end{align*}$
::X(-)2-(-2)-(-2)-2-4(3)-(17)-2(3)=24+2046=22086=24136=12133

Let's solve $\begin{align*}x^2+20x+51=0\end{align*}$ by factoring, completing the square, and using the Quadratic Formula.
::让我们通过乘数、完成方形和使用二次曲线公式来解析 x2+20x+51=0。

While it might not look like it, 51 is not a prime number. Its factors are 17 and 3, which add up to 20.
::虽然看起来可能不象,但51不是质数,其因数是17和3,加起来达20。

$\begin{aligned} x^{2} + 20 x + 51 & = 0 \\ (x + 17) (x + 13) & = 0 \\ x & = - 17, - 3 \end{aligned}$
$\begin{align*}x^2+20x+51 &=0\\ (x+17)(x+13) &=0\\ x &=-17, -3\end{align*}$
::x2+20x+51=0(x+17)(x+13)=0x%17,-3

Now, solve by completing the square.
::现在,通过完成广场解决。

$\begin{aligned} x^{2} + 20 x + 51 & = 0 \\ x^{2} + 20 x & = - 51 \\ x^{2} + 20 x + 100 & = - 51 + 100 \\ (x + 10)^{2} & = 49 \\ x + 10 & = \pm 7 \\ x & = - 10 \pm 7 \to - 17, - 3 \end{aligned}$
$\begin{align*}x^2+20x+51 &=0\\ x^2+20x &=-51\\ x^2+20x+100 &=-51+100\\ (x+10)^2 &=49\\ x+10 &=\pm 7\\ x &=-10 \pm 7 \rightarrow -17, -3\end{align*}$
::x2+20x+51=0x2+20x51x2+20x+100*51+100(x+1010)2=49x+10=7x_10_7*17,-3

Lastly, let’s use the Quadratic Formula. $\begin{align*}a = 1, b = 20, c = 51\end{align*}$ .
::最后,让我们使用二次曲线公式。 a=1,b=20,c=51。

$\begin{aligned} x & = \frac{- 20 \pm \sqrt{20^{2} - 4 (1) (51)}}{2 (1)} \\ = \frac{- 20 \pm \sqrt{400 - 204}}{2} \\ = \frac{- 20 \pm \sqrt{196}}{2} \\ = \frac{- 20 \pm 14}{2} \\ = - 17, - 3 \end{aligned}$
$\begin{align*}x &=\frac{-20 \pm \sqrt{20^2-4 (1)(51)}}{2(1)}\\ &=\frac{-20 \pm \sqrt{400-204}}{2}\\ &=\frac{-20 \pm \sqrt{196}}{2}\\ &=\frac{-20 \pm 14}{2}\\ &=-17, -3\end{align*}$
::

Notice that no matter how you solve this, or any, quadratic equation, the answer will always be the same.
::请注意,无论你如何解决这个或者任何二次方程, 答案总是相同的。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the break-even point.
::早些时候,有人要求你找到平衡点。

The break-even point is the point at which the equation equals zero. So use the Quadratic Formula to solve $\begin{align*}-3p^2 + 200p - 3000\end{align*}$ for p .
::平衡点是方程式等于零的点。因此, p 使用二次曲线公式解析 - 3p2+200p-3000 。

$\begin{aligned} - 3 p^{2} + 200 p - 3000 = 0 \end{aligned}$
$\begin{align*}-3p^2 + 200p - 3000 = 0\end{align*}$
::-3p2+200p-3000=0

Now, use $\begin{align*}a = -3, b = 200,\end{align*}$ and $\begin{align*}c = -3000\end{align*}$ and plug them into the Quadratic Formula.
::现在,使用a++3,b=200,c+++3000 并插入二次曲线公式。

$\begin{aligned} p & = \frac{- (200) \pm \sqrt{(200)^{2} - 4 (- 3) (- 3000)}}{2 (- 3)} \\ = \frac{- 200 \pm \sqrt{40000 - 36000}}{- 6} \\ = \frac{- 200 \pm \sqrt{4000}}{- 6} \\ = \frac{- 200 \pm 20 \sqrt{10}}{- 6} \\ = \frac{100}{3} \pm \frac{10 \sqrt{10}}{3} \end{aligned}$
$\begin{align*}p &=\frac{-(200) \pm \sqrt{(200)^2-4(-3)(-3000)}}{2(-3)}\\ &=\frac{-200\pm \sqrt{40000-36000}}{-6}\\ &=\frac{-200\pm \sqrt{4000}}{-6}\\ &=\frac{-200\pm 20\sqrt{10}}{-6}\\ &=\frac{100}{3} \pm \frac{10\sqrt{10}}{3}\end{align*}$
:200)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Therefore , there are two break-even points: $\begin{align*}\frac{100}{3} \pm \frac{10\sqrt{10}}{3}\end{align*}$ .
::因此,有两个平衡点:100310103。

Example 2
::例2

Solve $\begin{align*}-6x^2+15x-22=0\end{align*}$ using the Quadratic Formula.
::使用二次曲线公式解决 - 6x2+15x- 22=0。

$\begin{align*}a = -6, b = 15,\end{align*}$ and $\begin{align*}c = -22\end{align*}$
::a6,b=15,c=22

$\begin{aligned} x & = \frac{- 15 \pm \sqrt{15^{2} - 4 (- 6) (- 22)}}{2 (- 6)} \\ = \frac{- 15 \pm \sqrt{225 - 528}}{- 12} \\ = \frac{- 15 \pm i \sqrt{303}}{- 12} \\ = \frac{5}{4} \pm \frac{\sqrt{303}}{12} i \end{aligned}$
$\begin{align*}x &=\frac{-15 \pm \sqrt{15^2-4(-6)(-22)}}{2(-6)}\\ &=\frac{-15 \pm \sqrt{225-528}}{-12}\\ &=\frac{-15 \pm i \sqrt{303}}{-12}\\ &=\frac{5}{4} \pm \frac{\sqrt{303}}{12}i\end{align*}$
::-15 -152 -4(-6)(-6)(-22)(-6)(-6)(-6)_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Example 3
::例3

Solve $\begin{align*}2x^2-x-15=0\end{align*}$ using all three methods.
::使用所有三种方法解决 2x2 - x - 15=0 。

Factoring : $\begin{align*}ac = -30\end{align*}$ . The factors of -30 that add up to -1 are -6 and 5. Expand the $\begin{align*}x-\end{align*}$ term .
::乘数 : ac30 。 -30 乘以 -1的乘数为 -6 和 5. 扩大 x- term 。

$\begin{aligned} 2 x^{2} - 6 x + 5 x - 15 & = 0 \\ 2 x (x - 3) + 5 (x - 3) & = 0 \\ (x - 3) (2 x + 5) & = 0 \\ x & = 3, - \frac{5}{2} \end{aligned}$
$\begin{align*}2x^2-6x+5x-15 &=0\\ 2x(x-3)+5(x-3) &=0\\ (x-3)(2x+5) &=0\\ x &=3, -\frac{5}{2}\end{align*}$
::2x2-6x+5x-15=02x(x-3)+5(x-3)=0(x-3)(2x+5)=0x=3,-52

Complete the square
::完成正方形

$\begin{aligned} 2 x^{2} - x - 15 & = 0 \\ 2 x^{2} - x & = 15 \\ 2 (x^{2} - \frac{1}{2} x) & = 15 \\ 2 (x^{2} - \frac{1}{2} x + \frac{1}{16}) & = 15 + \frac{1}{8} \\ 2 {(x - \frac{1}{4})}^{2} & = \frac{121}{8} \\ {(x - \frac{1}{4})}^{2} & = \frac{121}{16} \\ x - \frac{1}{4} & = \pm \frac{11}{4} \\ x & = \frac{1}{4} \pm \frac{11}{4} \to 3, - \frac{5}{2} \end{aligned}$
$\begin{align*}2x^2-x-15 &=0\\ 2x^2-x &=15\\ 2\left(x^2-\frac{1}{2}x\right) &=15\\ 2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right) &=15+\frac{1}{8}\\ 2\left(x-\frac{1}{4}\right)^2 &=\frac{121}{8}\\ \left(x-\frac{1}{4}\right)^2 &=\frac{121}{16}\\ x-\frac{1}{4} &= \pm \frac{11}{4}\\ x &=\frac{1}{4} \pm \frac{11}{4} \rightarrow 3, -\frac{5}{2}\end{align*}$
::2x2 - x - x - 15=02x2 - 15=152(x2 - 12x)=152(x2 - 12x+116)=15+182(x- 14)2=1218(x- 14)2=1216x - 14* 14* 114x=14_ 114_ 114_ 3,-52

Quadratic Formula
::二次曲线公式

$\begin{aligned} x & = \frac{1 \pm \sqrt{1^{2} - 4 (2) (- 15)}}{2 (2)} \\ = \frac{1 \pm \sqrt{1 + 120}}{4} \\ = \frac{1 \pm \sqrt{121}}{4} \\ = \frac{1 \pm 11}{4} \\ = \frac{12}{4}, - \frac{10}{4} \to 3, - \frac{5}{2} \end{aligned}$
$\begin{align*}x &=\frac{1 \pm \sqrt{1^2-4(2)(-15)}}{2(2)}\\ &=\frac{1 \pm \sqrt{1+120}}{4}\\ &=\frac{1 \pm \sqrt{121}}{4}\\ &=\frac{1 \pm 11}{4}\\ &=\frac{12}{4}, -\frac{10}{4} \rightarrow3, -\frac{5}{2}\end{align*}$
::x=112-4(2)-2-152(2)=11+1204=11214=1114=124,-1043,-52

Review
::回顾

Solve the following equations using the Quadratic Formula.
::使用“二次曲线公式”解决以下方程式。

$\begin{align*}x^2+8x+9=0\end{align*}$
::x2+8x+9=0

$\begin{align*}4x^2-13x-12=0\end{align*}$
::4x2-13x-12=0

$\begin{align*}-2x^2+x+5=0\end{align*}$
::− 2x2+x+5=0

$\begin{align*}7x^2-11x+12=0\end{align*}$
::7x2-11x+12=0

$\begin{align*}3x^2+4x+5=0\end{align*}$
::3x2+4x+5=0

$\begin{align*}x^2-14x+49=0\end{align*}$
::x2 - 14x+49=0

Choose any method to solve the equations below.
::选择任何解析以下方程式的方法。

$\begin{align*}x^2+5x-150=0\end{align*}$
::x2+5x-150=0

$\begin{align*}8x^2-2x-3=0\end{align*}$
::8x2-2-2x-3=0

$\begin{align*}-5x^2+18x-24=0\end{align*}$
::-5x2+18x-24=0

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

$\begin{align*}x^2-16x+4=0\end{align*}$
::x2 - 16x+4=0

$\begin{align*}9x^2-196=0\end{align*}$
::9x2- 196=0

Solve the following equations using all three methods.
::使用所有三种方法解决以下方程式。

$\begin{align*}4x^2+20x+25=0\end{align*}$
::4x2+20x+25=0

$\begin{align*}x^2-18x-63=0\end{align*}$
::x2-18x-63=0

Writing Explain when you would use the different methods to solve different types of equations. Would the type of answer (real or imaginary) help you decide which method to use? Which method do you think is the easiest?
::写入解释您何时会使用不同方法解析不同的方程式类型。答案类型( 真实的或想象的) 会帮助您决定使用哪种方法吗 ? 您认为哪种方法最简单 ?

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

输出和使用二次曲线公式

Quadratic Formula ::二次曲线公式

Deriving the Quadratic Formula ::生成二次曲线公式

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 Formula
::二次曲线公式

Deriving the Quadratic Formula
::生成二次曲线公式

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)