Course: 8.2 通过保理办法解决赤道等量 | The Way To Learn

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通过保理解决夸量等量
A rocket firework is launched from a platform and travels at a height of $\begin{aligned} - 5 t^{2} + 30 t \end{aligned}$ , where t is in seconds . To keep safe, the person who fired the firework needs to get out of the zone where the firework falls. How much time does the person have to evacuate the fallout zone ¹ ? In this section, we discuss how to using the factoring techniques we considered in the previous chapter.
::从平台发射火箭,在 - 5t2+30t的高度飞行, 高度为 - 5t2+30t, 时间在几秒钟之内。为了安全起见, 发射烟火的人需要离开烟火坠落区。需要多少时间才能撤离坠落区1 ? 本节讨论如何使用我们在前一章中考虑的计数技术。

Quadratic Equations
::二次赤道

In this chapter, we will cover polynomial functions and concentrate on quadratic functions. It is often helpful to determine when the function is equal to zero. We call this a polynomial equation , and in the case of quadratics, a quadratic equation .
::在本章中,我们将涵盖多元函数,并集中关注二次函数。确定该函数何时等于零通常是有益的。我们称此为多元等式,对于二次函数,则称其为二次等式。

Quadratic Equations
::二次赤道

A quadratic equation is an equation of the form
::二次方程是窗体的方程

$\begin{aligned} a x^{2} + b x + c = 0, \end{aligned}$
where $\begin{aligned} a \end{aligned}$ , $\begin{aligned} b \end{aligned}$ , and $\begin{aligned} c \end{aligned}$ are real numbers, and $\begin{aligned} a \neq 0 \end{aligned}$ .
::ax2+bx+c=0,a、b和c为实际数字,a+bx+c=0。

The solutions of a quadratic (or polynomial) equation are called the roots or zeros of the function .
::二次方程(或多面方程)的解决方案称为函数的根或零。

Solving Quadratic Equations by Factoring
::通过保理解决夸量等量

Chapter 7 covered several techniques to factor quadratics and certain that can be expressed in a quadratic form . In the Explore More problems in that section, we used the zero product property . We formalize how to use it in this section.
::在第7章中,我们使用了零产品属性,并在这一节中正式确定了如何使用。

Zero-Product Property
::零建财产

If $\begin{aligned} a b = 0 \end{aligned}$ , then either $\begin{aligned} a = 0 \end{aligned}$    or $\begin{aligned} b = 0 \end{aligned}$ .

To solve quadratic equations by factoring, we factor as in the previous chapter and set each factor equal to 0.
::为了通过保理解决二次方程,我们按照前一章中的做法,将每个因数设定为0。

How To Solve Quadratic Equations By Factoring
::如何通过乘法解决赤道等量

1. Arrange the terms in the equation so that one side of the equation is equal to 0.
::1. 在方程中排列条件,使方程的一面等于0。

2. Factor the algebraic expression .
::2. 乘以代数表达式。

3. Using the zero product property, set each factor equal to 0 and solve each equation.
::3. 使用零产品属性,设定每个系数等于0,并解决每个方程式。

Greatest Common Factor
::最大共同因素

The first factoring technique we learned was factoring out the greatest common factor .
::我们所学的第一种保理技术是把最大的共同因素考虑在内。

Example 1
::例1

Solve $\begin{aligned} 10 x^{2} - 25 x = 0 \end{aligned}$ .
::解决 10x2 - 25x=0 。

Solution: The first part of factoring is to look for a GCF to factor out. Here, we can factor out the GCF, $\begin{aligned} 5 x \end{aligned}$ .
::解决办法:考虑因素的第一部分是寻找一个全球合作框架来考虑。在这里,我们可以将全球合作框架的5x因素考虑在内。

$\begin{aligned} 10 x^{2} - 25 x & = 0 \\ 5 x (2 x - 5) & = 0 \end{aligned}$

::10x2-25x=05x(2x-5)=0

Set the two factors equal to 0 and solve.
::设定等于 0 并解决的两个系数。

$\begin{aligned} 5 x & = 0 2 x - 5 = 0 \\ x & = 0 or 2 x = 5 \\ x = \frac{5}{2} \end{aligned}$

::5x=02x-5=0x=0x=0or 2x=5x=52

To check, we substitute each value back into the original equation.
::为了检查,我们把每个值都换回原来的方程

$\begin{aligned} 10 (0)^{2} - 25 (0) = 0 10 {(\frac{5}{2})}^{2} - 25 (\frac{5}{2}) = 0 \\ 0 = 0 or 10 \cdot \frac{25}{4} - \frac{125}{2} = 0 \\ \frac{125}{2} - \frac{125}{2} = 0 \end{aligned}$

::10(0)2-25(0)=010(52)2-25(52)=00=0=0or10_254-1252=01252-1252=0

WARNING
::警告

Do not divide by x. It can be equal to 0, and dividing by 0 is an undefined operation . The following is incorrect.
::不除以 x。它可以等于 0, 除以 0 是一个未定义的操作。以下不正确。

$\begin{aligned} 10 x^{2} - 25 x & = 0 \\ \frac{10 x^{2} - 25 x}{x} & = \frac{0}{x} This operation is illegal. \\ 10 x - 25 & = 0 \\ x & = \frac{5}{2} Notice the solution where x=0 is missing. \end{aligned}$

::10x2- 25x=010x2- 25xxx=0xx 此操作是非法的 10x- 25=0x=52x 注意到 X=0缺失的溶液。

Example 2
::例2

A rocket firework is launched from a platform and travels at a height of $\begin{aligned} - 5 t^{2} + 30 t \end{aligned}$ , where t is in seconds. To keep safe, the person who fired the firework needs to get out of the zone where the firework falls. How much time does the person have to evacuate the fallout zone?
::火箭从平台发射,在5-2+30特高度飞行,时间在几秒钟之内。为了安全起见,发射烟火的人需要离开烟火坠落区。要撤离沉降区需要多少时间?

Solution: The height of the rocket is 0 when it is on the ground, so we can set the height formula equal to 0 to solve.
::解答:当火箭在地面上时,高度是0, 所以我们可以设定高度公式等于 0 来解答。

$\begin{aligned} - 5 t^{2} + 30 t & = 0 \\ - 5 t (t - 6) & = 0 \\ - 5 t = 0 & t - 6 = 0 \\ t = \frac{0}{- 5} = 0 & t = 6 \end{aligned}$

::-5t2+30t=0-5t(t-6)=0-5t=0-6=0t-6=0t=0t=0T=0-5=0t=6

To check, we substitute each value back into the original equation.
::为了检查,我们把每个值都换回原来的方程

$\begin{aligned} - 5 (0)^{2} + 30 (0) = 0 - 5 (6)^{2} + 30 (6) = 0 \\ 0 = 0 - 180 + 180 = 0 \\ 0 = 0 \end{aligned}$

The 1st solution represents the initial launch of the firework. The 2nd solution represents the time when the firework comes back down to the ground, so the person has 6 seconds to evacuate the fallout zone.
::第一个解决方案代表着最初的烟火发射。第二个解决方案代表着烟火返回地面的时间,所以这个人有6秒钟的时间撤离沉降区。

This video by CK-12 demonstrates how to solve an equation by factoring out the greatest common factor.
::CK-12的这段影片展示了如何通过考虑到最大的共同因素来解决方程式问题。

Perfect Square Trinomials and Difference of Two Squares
::完美的三维广场和两个广场的差异

We considered some special cases of factoring quadratics in the previous chapter. They were:
::在前一章中,我们考虑了某些特殊因素,即保理等离子体。

Perfect Square Trinomials and Difference of Two Squares
::完美的三维广场和两个广场的差异

$\begin{aligned} a^{2} \pm 2 a b + b^{2} = (a \pm b)^{2} Perfect Square Trinomial \\ a^{2} - b^{2} = (a + b) (a - b) Difference of Two Squares \end{aligned}$

Example 3
::例3

Solve $\begin{aligned} x^{2} - 16 x + 64 = 0 \end{aligned}$ .
::解决 x2 - 16x+64=0 。

Solution: The expression on the left-hand side of the equation is a perfect square trinomial . To determine the two repeated factors, find the square roots of 64, which are 8 and -8. Since 2 times -8 is -16, the two repeated factors are
::解决方案: 方程左侧的表达式是完全平方方方三角。要确定两个重复的因素, 找到64的正方根, 即8和8。由于2乘8是16, 两个重复的因素是:

$\begin{aligned} x^{2} - 16 x + 64 = 0 \\ (x - 8)^{2} = 0 \\ x - 8 = 0 \\ x = 8. \end{aligned}$

::x2-16x+64=0(x-8)2=0x-8=8=0x=8。

We do not need to set up an equation for each of the factors here, because it would be two equations that are the same. Likewise, we only need to do one check.
::我们不需要为这里的每一个因素设置一个等式,因为两个等式是相同的,同样,我们只需要做一个检查。

$\begin{aligned} (8)^{2} - 16 (8) + 64 = 0 \\ 64 - 128 + 64 = 0 \\ 0 = 0 \end{aligned}$

Example 4
::例4

Solve $\begin{aligned} x^{2} - 121 = 0 \end{aligned}$ .
::解决 x2 - 1221=0 。

Solution: Using the difference of two squares formula, we have
::解决办法:使用两个方形公式的差别,我们

$\begin{aligned} x^{2} - 121 & = 0 \\ (x + 11) (x - 11) & = 0 \end{aligned}$

::x2 - 121=0(x+11)(x- 11)=0

Checking the solution, we have
::检查解决方案,我们有

$\begin{aligned} x + 11 = 0 & x - 11 = 0 \\ x = - 11 & x = 11 \end{aligned}$

::x+11=0x-11=0x=11x=11x=11x

Factoring When the Lead Coefficient Is 1
::含铅系数为 1

Example 5
::例5

Solve $\begin{aligned} x^{2} - 9 x + 18 = 0 \end{aligned}$ .
::解决 x2- 9x+18=0 。

Solution: Because $\begin{aligned} a = 1 \end{aligned}$ , determine the two factors of 18 that add up to -9.
::解答: 因为 a=1, 确定 18 中两个系数加到 -9 。

$\begin{aligned} x^{2} - 9 x + 18 & = 0 \\ (x - 6) (x - 3) & = 0 \end{aligned}$

::x2- 9x+18=0(x-6)(x-3)=0

So, $\begin{aligned} x - 6 = 0 \end{aligned}$ OR $\begin{aligned} x - 3 = 0 \end{aligned}$ . Therefore , $\begin{aligned} x = 6 \end{aligned}$ or $\begin{aligned} x = 3 \end{aligned}$ .
::所以, x-6=0 OR x-3=0。因此, x=6 或 x=3。

Check ing the solutions , we have
::在检查解决办法时,我们有:

$\begin{aligned} 6^{2} - 9 (6) + 18 & = 0 or 3^{2} - 9 (3) + 18 = 0 \\ 36 - 54 + 18 & = 0 9 - 27 + 18 = 0 \end{aligned}$

::62-9(6)+18=0or32-9(3)+18=036-54+18=09-27+18=0

This video by CK-12 demonstrates how to solve equations by simplifying, rearranging, and factoring.
::CK-12的这段影片展示了如何通过简化、重新安排和保理来解析方程式。

Factoring When the Lead Coefficient Is Not Equal to 1
::含铅系数不等于1时的保理

Example 6
::例6

Solve $\begin{aligned} 6 x^{2} + x - 4 = 11 \end{aligned}$ .
::解决 6x2+x-4=11。

Solution: Before we factor, we must combine like terms on one side of the equation. The zero product property states that one side of the equation must be 0.
::解答: 在我们考虑之前, 我们必须在方程式的一面结合类似条件。零产品属性表示方程式的一面必须是 0 。

$\begin{aligned} 6 x^{2} + x - 4 = 11 \\ \underline{- 11 = - 11} \\ 6 x^{2} + x - 15 = 0 \end{aligned}$

::6x2+x-4=11-11=11_6x2+x-15=0

Here, $\begin{aligned} a \neq 1 \end{aligned}$ . The product of $\begin{aligned} a c \end{aligned}$ is -90. What are the two factors of -90 that add up to 1? 10 and -9. Expand the x- term and factor.
::a++1. ac 的产物是 -90。 -90 的两种因素是多少? 10 和 -9. 扩大 x 期和系数。

$\begin{aligned} 6 x^{2} + x - 15 & = 0 \\ 6 x^{2} - 9 x + 10 x - 15 & = 0 \\ 3 x (2 x - 3) + 5 (2 x - 3) & = 0 \\ (2 x - 3) (3 x + 5) & = 0 \end{aligned}$

::6x2+x-15=06x2-9x+10x-15=03x(2x-3)+5(2x-3)=0(2x-3)(3x+5)=0

Lastly, set each factor equal to 0 and solve.
::最后,设定每个系数等于 0 并解决。

$\begin{aligned} 2 x - 3 & = 0 3 x + 5 = 0 \\ 2 x & = 3 or 3 x = - 5 \\ x & = \frac{3}{2} x = - \frac{5}{3} \end{aligned}$

::2x-3=0 3x+5=02x=3or3x=5x=32x=-53

Checking the solutions, we have:
::在检查解决方案时,我们有:

$\begin{aligned} 6 {(\frac{3}{2})}^{2} + \frac{3}{2} - 4 & = 11 6 {(- \frac{5}{3})}^{2} - \frac{5}{3} - 4 = 11 \\ 6 \cdot \frac{9}{4} + \frac{3}{2} - 4 & = 11 or 6 \cdot \frac{25}{9} - \frac{5}{3} - 4 = 11 \\ \frac{27}{2} + \frac{3}{2} - 4 & = 11 \frac{50}{3} - \frac{5}{3} - 4 = 11 \\ 15 - 4 & = 11 15 - 4 = 11 \end{aligned}$

::6(32)2+32-4=11 6(-532)+32-4=111 6(-53)2-53-4=4=11(1)(118)94+32-4=411or 62559-53-44=4=1172+32-4=11503-534=1115-4=1115-4=11

WARNING
::警告

$\begin{aligned} a b = 11 \end{aligned}$ does not imply $\begin{aligned} a = 11 \end{aligned}$ or $\begin{aligned} b = 11 \end{aligned}$ .

Example 7
::例7

Solve $\begin{aligned} 12 x^{2} + 13 x + 7 = 12 - 4 x \end{aligned}$ .
::解决 12x2+13x+7=12-4x。

Solution: Combine all the like terms onto the same side of the equals sign, so that one side is 0.
::解答: 将所有类似术语合并到相等符号的同一侧, 这样一面是 0 。

$\begin{aligned} 12 x^{2} + 13 x + 7 & = 12 - 4 x \\ 12 x^{2} + 17 x - 5 & = 0 \end{aligned}$

::12x2+13x+7=12-4x12x2+17x-5=0

$\begin{aligned} a c = - 60 \end{aligned}$ . The factors of -60 that add up to 17 are 20 and -3. Expand the x- term and factor.
::ac=-60。 -60系数加17是20和-3。扩大x期和系数。

$\begin{aligned} 12 x^{2} + 17 x - 5 & = 0 \\ 12 x^{2} + 20 x - 3 x - 5 & = 0 \\ 4 x (3 x + 5) - 1 (3 x + 5) & = 0 \\ (3 x + 5) (4 x - 1) & = 0 \end{aligned}$

::12x2+17x-5=012x2+20x-3x-5=04x(3x+5)-1(3x+5)=0(3x+5)-4x-1=0

Solve the equation with each factor for $\begin{aligned} x \end{aligned}$ .
::以 x 的每个因数解决方程式。

$\begin{aligned} 3 x + 5 & = 0 4 x - 1 = 0 \\ 3 x & = - 5 or 4 x = 1 \\ x & = - \frac{5}{3} x = \frac{1}{4} \end{aligned}$

::3x+5=04x-1=03x=-5or4x=1x=-53x=14

Checking the solutions, we have
::在检查解决办法时,我们有:

$\begin{aligned} 12 (- \frac{5}{3})^{2} + 13 (- \frac{5}{3}) + 7 & = 12 - 4 (- \frac{5}{3}) \\ 12 (\frac{25}{9}) + 13 (- \frac{5}{3}) + 7 & = 12 - 4 (- \frac{5}{3}) \\ \frac{100}{3} - \frac{65}{3} + 7 & = 12 + \frac{20}{3} \\ \frac{100}{3} - \frac{65}{3} + \frac{21}{3} & = \frac{36}{3} + \frac{20}{3} \\ \frac{56}{3} & = \frac{56}{3} \\ 12 (\frac{1}{4})^{2} + 13 (\frac{1}{4}) + 7 & = 12 - 4 (\frac{1}{4}) \\ 12 (\frac{1}{16}) + 13 (\frac{1}{4}) + 7 & = 12 - 4 (\frac{1}{4}) \\ \frac{3}{4} + \frac{13}{4} + 7 & = 12 - 1 \\ \frac{3}{4} + \frac{13}{4} + \frac{28}{4} & = 11 \\ \frac{44}{4} & = 11 \\ 11 & = 11 \end{aligned}$

by Mathispower4u demonstrates how to solve quadratic equations by factoring.
::Mathispower4u 演示如何通过乘法解析二次方程。

Factoring an Expression that is Quadratic in Form
::计算表列中的二次曲线表达式

Example 8
::例8

Solve $\begin{aligned} x^{4} - 2 x^{2} - 8 = 0 \end{aligned}$ .
::解决 x4 - 2x2 - 8=0 。

Solution: This is not a quadratic equation, but it is an equation that is quadratic in form: $\begin{aligned} x^{4} = (x^{2})^{2} . \end{aligned}$ Using a dummy variable , u, to represent the unknown value of $\begin{aligned} x^{2}, \end{aligned}$ we have:
::解析度: 这不是一个二次方程, 而是一个四方程, 形式为 x4=( x2) 。使用一个假变量u, 来表示 x2 的未知值 :

$\begin{aligned} x^{4} - 2 x^{2} - 8 = 0 \\ u = x^{2} & Define dummy variable \\ u^{2} - 2 u - 8 = 0 & Substitute x^{2} for u . \\ (u - 4) (u + 2) = 0 \\ u - 4 = 0 u + 2 = 0 \\ x^{2} - 4 = 0 x^{2} + 2 = 0 \\ (x + 2) (x - 2) = 0 no solutions \\ x + 2 = 0 x - 2 = 0 \\ x = - 2, x = 2 \end{aligned}$

::x4-2x2 - 8x2 - 8= 0u=x2 定义模拟变量2 - 2u - 8= 0 u. (u-4)(u+2)= 0u-4= 0u-2= 0x2= 0x2 - 4= 0x2+2=0(x+2)(x-2)=0no溶液x+2=0x-2=0x=2x=2x=2

Checking our two solutions, we have
::检查我们的两个解决方案,我们有

$\begin{aligned} 2^{4} - 2 (2^{2}) - 8 = 0 & (- 2)^{4} - 2 ((- 2)^{2}) - 8 = 0 \\ 16 - 8 - 8 = 0 & 16 - 8 - 8 = 0 \\ 0 = 0 & 0 = 0 \end{aligned}$

by Mathispower4u demonstrates how to solve an equation by writing it in quadratic form and solving by factoring.
::由 Mathispower4u 来演示如何以二次形式写出方程式,

How To Solve Quadratic Equations With Desmos
::如何用脱色来解析赤道等量

To solve a quadratic equation like $\begin{aligned} x^{2} - 3 x + 4 = 0 \end{aligned}$ by graphing, input the function into Desmos and find the points where the graph intersects the x -axis.
::要通过图形化解决 x2 - 3x+4=0 这样的二次方程,请将函数输入 Desmos 并找到图形交叉 x 轴的点。

The solutions to this equation are $\begin{aligned} x = - 4 \end{aligned}$ and $\begin{aligned} x = 1 \end{aligned}$ .
::此方程式的解决方案是 x= 4 和 x= 1 。



How To Solve Quadratic Equations With a TI-83/84
::如何用TI-83/84解决赤道等同

1. Press Y= and input the function. Use X for the variable and the caret button, ^ , or the square button, x 2 , for the exponent .
::1. 按 Y = 键并输入函数。对变量使用 X,对引号使用 X 和照顾按钮, , 或平方按钮, x2 。

2. Press GRAPH .
::2. 按GRAPH。

3. Press 2nd , TRACE , and then under the CALCULATE menu choose 2: zeros.
::3. 按2,TRACE,然后在CALCULATE菜单下选择2:零。

4. Trace to the left of the point and press ENTER . Trace to the right of the point and press ENTER . Press ENTER a 3rd time for the guess. The output should be one solution.
::4. 追踪点的左侧并按 ENTER. 跟踪点右侧并按 ENTER. 按 ENTER. 第三次按 ENTER 进行猜测, 输出应该是一个解决方案。

by JenniStout demonstrates how to find the solutions to a quadratic equation using a TI-84.
::JenniStout用TI-84演示如何找到四边方程式的解决方案。

Summary
::摘要

To solve a quadratic equations by factoring, arrange all the terms on one side of the equation so the other side equals 0, factor the expression, set each factor equal to 0, and solve each equation.
::要通过乘法解决二次方程, 将方程一边的所有条件排列为一, 使另一方等于 0, 乘以表达式, 设定每个系数等于 0, 并解析每个方程。

Review
::回顾

Solve the following quadratic equations by factoring:
::通过乘数解决下列二次方程:

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

2. $\begin{aligned} x^{2} + 6 x = 0 \end{aligned}$
::2. x2+6x=0

3. $\begin{aligned} 2 x^{2} - 5 x = 12 \end{aligned}$
::3. 2x2-5x=12

4. $\begin{aligned} 12 x^{2} + 7 x - 10 = 0 \end{aligned}$
::4. 12x2+7x-10=0

5. $\begin{aligned} x^{2} + x = 56 \end{aligned}$
::5. x2+x=56

6. $\begin{aligned} 16 x = 32 x^{2} \end{aligned}$
::6. 16x=32x2

7. $\begin{aligned} 3 x^{2} + 13 x = - 12 \end{aligned}$
::7. 3x2+13x12

8. $\begin{aligned} 36 x^{2} - 48 = 1 \end{aligned}$
::8. 36x2-48=1

9. $\begin{aligned} 5 x^{2} + 12 x + 4 = 0 \end{aligned}$
::9. 5x2+12x+4=0

10. $\begin{aligned} 8 x^{2} + - 30 x - 6 = 41 x + 3 \end{aligned}$
::10. 8x230x-6=41x+3

Explore More
::探索更多

1. The height of a ball that is thrown straight up in the air from a height of 2 meters above the ground, with a velocity of 9 meters per second, is given by the quadratic equation $\begin{aligned} h = - 5 t^{2} + 9 t + 2 \end{aligned}$ , where t is the time in seconds. How long does it take the ball to hit the ground?
::1. 一个球从地面上2米高处直直投到空中,速度为每秒9米的球的高度由四方形方程式 h=-5t2+9t+2给予,其时间为秒。球撞击地面需要多长时间?

2. George is building a fence for the backyard. The total area of the backyard is 1,600 square feet. The width of the house is half the length of the yard, plus 7 feet. How much fencing does George need to buy?
::2. George正在为后院修建围栏,后院总面积为1 600平方英尺,房屋宽度为院子长度的一半,加上7英尺。

3. Suppose you own a small business that currently charges $12 per item you sell, and you average 36 sales per day. Also, suppose you read a newspaper article in which the author states that for every 50-cent decrease in the cost of a product being sold, the average business can expect to gain two sales per day. Use this information to attempt to maximize your income. What should you charge?
::3. 假设你拥有一个小企业,目前每售一件物品要收费12美元,平均每天销售36美元;还假设你读了一篇报纸文章,其中撰文者说,每售出一项产品每减少50%,平均商业就可望每天获得两笔销售,利用这一信息尽量扩大收入。你应收取什么费用?

4. The members of a ski club were going to rent a bus to drive them to a ski resort at a total cost of $2,420, which was to be divided equally among the members. At the last minute, two members decided to drive their own cars. The cost to the remaining members increased $11 each. How many members rode the bus?
::4. 一家滑雪俱乐部的成员打算租一辆公共汽车,将他们开到滑雪胜地,总费用为2 420美元,由成员平分,最后一刻,两名成员决定自己开车,其余成员每人增加11美元,有多少成员坐车?

5. The members of an adventure club were planning a snorkeling trip at a total cost of $7,020, which was to be divided equally among the members. At the last minute, two members informed the club that they would not be able to come on the trip. The cost to the remaining members increased $5 each. How many members went on the snorkeling trip?
::5. 探险俱乐部成员正计划以总费用7 020美元进行潜水旅行,这笔费用将在成员之间平均分配,最后一刻,两名成员通知俱乐部他们无法前来旅行,其余成员每人的费用增加5美元,有多少成员参加了潜水旅行?

6. Find two positive numbers whose difference is 16 and whose product is 420.
::6. 发现两个正数,其差异为16,其产品为420。

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。

PLIX
::PLIX

Try this interactive that reinforces the concepts explored in this section:
::尝试这一互动,强化本节所探讨的概念:

References
::参考参考资料

1. "Safety Tips: The National Council on Fireworks Safety," last accessed May 26, 2017,
::1. “安全提示:全国烟火安全委员会”, 2017年5月26日最后一次访问,

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