Section: 使用二次曲线公式解决二次赤道等量 | 8.5 使用二次曲线公式解决二次赤道等量

Section outline

A RICE chart allows chemists to keep track of the concentrations of different substances in a chemical reaction ¹ . Cave formation is an example of a chemical reaction in nature. To form a cave, groundwater—which is a weak acid containing carbon dioxide dissolved in water—seeps into the ground and reacts with limestone. Once this mixture reaches the open air of the cave, the reaction reverses and a solid deposit of calcium carbonate is left behind, forming stalagmites and stalactites ² .
::RICE 图表使化学家能够追踪化学反应中不同物质的浓度。洞穴形成是化学反应的一个例子。要形成洞穴,地下水(水中溶解的含二氧化碳的微弱酸)渗入地面,与石灰石发生反应。一旦这种混合物到达洞穴的露天空气,反应反射和碳酸钙的固体沉积就落在后面,形成炉渣和硫化物2。

The result of a RICE chart for a chemical reaction with a weak acid is a quadratic equation , say, for instance, $\begin{aligned} x^{2} + 10^{- 5} x - 10^{- 6} = 0 \end{aligned}$ . It would be difficult to apply our earlier techniques to this equation . However, we have one more technique for solving quadratic equations in this section.
::RICE 微酸化学反应图表的结果是一个二次方程,例如, x2+10-5-5x-10-6=0。在这个方程上很难应用我们先前的技术。然而,我们在本节中还有一种解决二次方程的技术。

Solving Quadratic Equations by the Quadratic Formula
::以二次曲线公式解决二次赤道等量

The last way to solve a quadratic equation is by using the Quadratic Formula . This formula is derived by for the equation $\begin{aligned} a x^{2} + b x + c = 0 \end{aligned}$ . We will derive the formula here.
::解析二次方程的最后方法就是使用二次方程式。这个公式是用方程式 x2+bx+c=0 来推导的。我们将在此得出公式。

Investigation: Deriving the Quadratic Formula
::调查:提出“四压公式”

1. Move the constant to the right side of the equation: $\begin{aligned} a x^{2} + b x = - c \end{aligned}$ .
::1. 将常数移动到方程式的右侧:x2+bx=-c。

2. Factor out $\begin{aligned} a \end{aligned}$ from everything on the left side of the equation: $\begin{aligned} a (x^{2} + \frac{b}{a} x) = - c \end{aligned}$ .
::2. 从方程左侧的所有东西(a(x2+bax)=-c)中计算出一个值。

3. Complete the square using $\begin{aligned} \frac{b}{a} \end{aligned}$ : $\begin{aligned} {(\frac{b}{2})}^{2} = {(\frac{\frac{b}{a}}{2})}^{2} = {(\frac{b}{2 a})}^{2} = \frac{b^{2}}{4 a^{2}} \end{aligned}$ .
::3. 使用bab2)2=(ba2)2=(b2a)2=(b2a)2=b24a2,完成广场。

4. Add this expression to on the left side. On the right side, you need to multiply it by $\begin{aligned} a \end{aligned}$ (to account for the $\begin{aligned} a \end{aligned}$ outside the parentheses): $\begin{aligned} a (x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}}) = - c + \frac{b^{2}}{4 a} . \end{aligned}$
::4. 向左侧添加此表达式。在右侧, 您需要将其乘以一个( 用于括号外的计算): a( x2+bax+b24a2) =- c+b24a 。

5. Factor the quadratic expression inside the " data-term="Parentheses" role="term" tabindex="0"> parentheses and find the common denominator on the right-hand side: $\begin{aligned} a {(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a} . \end{aligned}$
::5. 将括号内的二次表达式乘以,在右侧找到共同分母x+b2a)2=b2-4ac4a。

6. Divide both sides by $\begin{aligned} a \end{aligned}$ : $\begin{aligned} {(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}} \end{aligned}$ .
::6. 将双方除以ax+b2a)2=b2-4ac4a2。

7. Take the square root of both sides: $\begin{aligned} x + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ .
::7. 以两侧的平方根为平方根: x+b2ab2-4ac2a。

8. Subtract $\begin{aligned} \frac{b}{2 a} \end{aligned}$ from both sides to get $\begin{aligned} x \end{aligned}$ by itself: $\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ .
::8. 从两边减去b2a以获得x本身:x=-bb2-4ac2a。

This formula can be used to solve any quadratic equation.
::此公式可用于解析任何二次方程。

How To Using the Quadratic Formula
::如何使用二次曲线公式

1. Identify $\begin{aligned} a, b \end{aligned}$ , and $\begin{aligned} c \end{aligned}$ from $\begin{aligned} a x^{2} + b x + c = 0 \end{aligned}$ .
::1. 从x2+bx+c=0中识别a、b和c。

2. Substitute the values into the formula and simplify the expression.
::2. 将数值替换为公式并简化表达式。

$\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$

::x=-bb2-4ac2a

by Mathispower4u demonstrates how to derive the quadratic formula.
::由 Mathispower4u 演示如何得出二次方程式。

Example 1
::例1

Solve $\begin{aligned} 2 x^{2} - x - 15 = 0 \end{aligned}$ .
::溶解 2x2 -x -15=0 。

Solution: Identify a,b, and c, and then substitute into the formula.
::解决办法:确定a、b和c,然后替换成公式。

$\begin{aligned} x & = \frac{1 \pm \sqrt{1^{2} - 4 (2) (- 15)}}{2 (2)} \\ = \frac{1 \pm \sqrt{1 + 120}}{4} \\ = \frac{1 \pm \sqrt{121}}{4} \\ = \frac{1 \pm 11}{4} \\ = \frac{12}{4}, - \frac{10}{4} \to 3, - \frac{5}{2} \end{aligned}$

::x=112-4(2)-152(2)(2)=11+1204=11214=1114=124,-1043,-52

This video by CK-12 demonstrates how to use the quadratic formula to solve quadratic equations.
::CK-12的这段影片展示了如何使用二次方程式解决二次方程式。

Example 2
::例2

Solve $\begin{aligned} 2 x^{2} + 5 x - 15 = - x^{2} + 7 x + 2 \end{aligned}$ using the Quadratic Formula.
::使用二次曲线公式解决 2x2+5x-15=-x2+7x+2。

Solution: Get everything onto the left side of the equation.
::解决方案:把所有东西都放在方程的左侧。

$\begin{aligned} 2 x^{2} + 5 x - 15 & = - x^{2} + 7 x + 2 \\ 3 x^{2} - 2 x - 17 & = 0 \end{aligned}$

::2x2+5x-15=-x2+7x+23x2-2x-17=0

Now, use $\begin{aligned} a = 3, b = - 2, \end{aligned}$ and $\begin{aligned} c = - 17 \end{aligned}$ and plug them into the Quadratic Formula.
::现在,使用a=3,b=2,和c=17,并插入二次曲线公式。

$\begin{aligned} x & = \frac{- (- 2) \pm \sqrt{(- 2)^{2} - 4 (3) (- 17)}}{2 (3)} \\ = \frac{2 \pm \sqrt{4 + 204}}{6} \\ = \frac{2 \pm \sqrt{208}}{6} \\ = \frac{2 \pm 4 \sqrt{13}}{6} \\ = \frac{1 \pm 2 \sqrt{13}}{3} \end{aligned}$

::x=(-2)-(-2)-(-2)-(-2)-2-4(3)(-17)-(3)-(3)-(7)-2(3)=(2)-4+2046=(2)-2086=(2)-2086=(2)-4136=(1)/2133)

by Tyler Wallace demonstrates how to use the quadratic formula to solve quadratic equations.
::泰勒·华莱士展示了如何使用二次方程式解决二次方程式。

Example 3
::例3

Solve $\begin{aligned} 9 x^{2} - 30 x - 24 = 0 \end{aligned}$ using the Quadratic Formula.
::使用二次曲线公式解决 9x2- 30x- 24=0 。

Solution: First make sure one side of the equation is zero. Then find $\begin{aligned} a, b, \end{aligned}$ and $\begin{aligned} c \end{aligned}$ . $\begin{aligned} a = 9, b = - 30, c = - 24 \end{aligned}$ . Now, plug the values into the formula and solve for $\begin{aligned} x \end{aligned}$ .
::解决方案 : 首先确保方程式的一面为零。然后找到 a, b, 和 c. a= 9, b= 30, c= 24。现在, 将值插入公式中, 并解决 x 。

$\begin{aligned} x & = \frac{- (- 30) \pm \sqrt{(- 30)^{2} - 4 (9) (- 24)}}{2 (9)} \\ = \frac{30 \pm \sqrt{900 + 864}}{18} \\ = \frac{30 \pm \sqrt{1, 764}}{18} \\ = \frac{30 \pm 42}{18} \\ = 4, - \frac{2}{3} \end{aligned}$

::x( 30) ( 30) ( 30) 2- 4( 9) (9)( 242( 9) = 30900+86418= 301, 76418= 30 4218=4, 23

Example 4
::例4

The result of a RICE chart for a chemical reaction with a weak acid is a quadratic equation, say for instance, $\begin{aligned} x^{2} + 10^{- 5} x - 10^{- 6} = 0 \end{aligned}$ . Determine the amount of the substance, x.
::RICE 与微酸发生化学反应的 RICE 图表的结果是一个二次方程,例如 x2+10-5x-10-6=0。确定物质的数量, x。

Solution: Note $\begin{aligned} a = 1, b = 10^{- 5} \end{aligned}$ and $\begin{aligned} c = - 10^{- 6} \end{aligned}$ . Substituting these values into the quadratic formula yields
::溶液:注a=1,b=10-5和c10-6。

$\begin{aligned} x & = \frac{- 10^{- 5} \pm \sqrt{(10^{- 5})^{2} - 4 (1) (- 10^{- 6})}}{2 (1)} \\ = \frac{- 10^{- 5} \pm \sqrt{10^{- 10} + 4 \times 10^{- 6}}}{2} \\ = \frac{- 10^{- 5} \pm \sqrt{4.0001 \times 10^{- 6}}}{2} \\ = \frac{- 10^{- 5} \pm 2.000025 \times 10^{- 3}}{2} \\ = \frac{- 10^{- 5} + 2.000025 \times 10^{- 3}}{2} reject negative solution, amount is positive \\ = \frac{1.990025 \times 10^{- 3}}{2} \\ = 9.950125 \times 10^{- 4} \end{aligned}$

::x=-10-5-(10-5)2-4-4(1)(-10-6)2(1)=----5-10-10-10+4x10-10-62=-10-5-00.0001x10-62=-5-0.0010-62=--5-2.0025x10-32=----5-5+2.000025x10-32-反转负溶液,正数=1.990025x10-32-32=9.950125x10-4

WARNINGS
::警告

If b is negative, remember to change the sign in the 1st term in the formula: $\begin{aligned} b = - 2 ⟶ - b = - (- 2) = 2 \end{aligned}$ .
::如果 b 是负, 则记住更改公式中第一学期的符号 : b=-2 - b=- (-2)= 2。

Divide both terms by 2 a .
::将两个条件除以 2a。

a, b, and c represent the coefficients. Do not include the variable with them.
::a、b和c 表示系数,不包括变量。

Simplify the radical 1st, then reduce the fraction .
::简化1号基体,然后降低分数。

Summary
::摘要

To solve quadratic equations using the quadratic formula, identify the coefficients in the equation and then substitute those values into the formula: $\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ .
::使用二次方程解决二次方程,确定方程中的系数,然后将这些数值替换成公式:x=-bb2-4ac2a。

Review
::回顾

Solve the following equations using the Quadratic Formula:
::使用二次曲线公式解决下列方程式:

1. $\begin{aligned} x^{2} + 8 x + 9 = 0 \end{aligned}$
::1. x2+8x+9=0

2. $\begin{aligned} - 2 x^{2} + x + 5 = 0 \end{aligned}$
::2. -2x2+x+5=0

3. $\begin{aligned} 3 x^{2} = - 4 x + 5 \end{aligned}$
::3. 3x2=-4x+5

4. $\begin{aligned} x^{2} + 5 x - 150 = 0 \end{aligned}$
::4. x2+5x-150=0

5. $\begin{aligned} 8 x^{2} = 2 x + 3 \end{aligned}$
::5. 8x2=2x+3

6. $\begin{aligned} - 5 x^{2} + 18 x - 24 = 0 \end{aligned}$
::6.5x2+18x-24=0

7. $\begin{aligned} 10 x^{2} + x - 2 = 0 \end{aligned}$
::7. 10x2+x-2=0

8. $\begin{aligned} x^{2} + 4 = 16 x \end{aligned}$
::8. x2+4=16x

9. $\begin{aligned} 4 x^{2} + 20 x + 25 = 0 \end{aligned}$
::9. 4x2+20x+25=0

10. $\begin{aligned} x^{2} - 18 x - 63 = 0 \end{aligned}$
::10. x2-18x-63=0

Explore More
::探索更多

1. The profit on your fundraiser is represented by the quadratic expression $\begin{aligned} - 3 p^{2} + 200 p - 3, 000 \end{aligned}$ , 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.
::1. 您筹款人的利润由四边表达式 - 3p2+200p- 3,000表示, p是您的价格点。您的收支平衡点( 即您开始盈利的价格点) 是什么 ? 提示 : 将方程式设为零。

2. A plane travels at a speed of 160 mph in calm air. Flying with a tailwind, the plane is clocked over a distance of 550 miles. Flying against a headwind, it takes 2 hours longer to complete the return trip. What was the wind velocity? (Round your answer to the nearest tenth.)
::2. 飞机飞行速度为160英里,在平静的空气中飞行速度为160英里,飞行时有尾风,飞行时长为550英里,飞行时长为2小时,完成返回旅行需要2小时。风速是多少? (你对最近的10英里的答复是多少? )

3. You want to build a fence around your rectangular yard. The area of your yard is 3,860 ft ² . The width of your neighbor's yard is 7 feet longer than the length of your yard. What are the dimensions of your yard? Round your answer to the nearest hundredth.
::3. 您想在您的长方形院子周围修建一道栅栏,您的院子面积为3 860平方英尺。您的邻居院子宽度比您的院子长7英尺。您的院子的尺寸是多少? 您的院子的尺寸是多少? 将您的答案绕到最近的100英尺处。

4. Robert stands on the topmost tier of bleachers in a gymnasium, and throws a basketball out onto the basketball court with a vertical upward velocity of 60 ft/s. The ball is 50 feet above the ground at the moment he releases the ball. When does the ball land?
::4. 罗伯特站在健身房最顶端的露天机上,向篮球场投掷篮球,垂直向上速度为60英尺/秒,球在球发球时在地上方50英尺处,球何时落地?

5. Find the sum and the product of the two solutions $\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ . Note how this relates to how we solve a quadratic equation by factoring.
::5. 找出两种溶液x=-bb2-4ac2a的总和和产品,注意这与我们如何通过保理处理四方方程式有何关系。

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

Please see the Appendix.
::请参看附录。

References
::参考参考资料

1. "RICE Chart," last edited March 1, 2017,
::1. 2017年3月1日编辑的“RICE图表”,

2. " How Caves Form | Caves and Karst | Foundations of the Mendips," by British Geological Survey Natural Environment Research Council, 2017,
::2. “洞穴如何形成洞穴和门底ps的卡斯特基金会”,英国地质调查局自然环境研究理事会,2017年,

Section outline

Solving Quadratic Equations by the Quadratic Formula ::以二次曲线公式解决二次赤道等量

How To Using the Quadratic Formula ::如何使用二次曲线公式

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

WARNINGS ::警告

Summary ::摘要

Review ::回顾

Explore More ::探索更多

Answers for Review and Explore More Problems ::回顾和探讨更多问题的答复

References ::参考参考资料

Solving Quadratic Equations by the Quadratic Formula
::以二次曲线公式解决二次赤道等量

How To Using the Quadratic Formula
::如何使用二次曲线公式

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

WARNINGS
::警告

Summary
::摘要

Review
::回顾

Explore More
::探索更多

Answers for Review and Explore More Problems
::回顾和探讨更多问题的答复

References
::参考参考资料