Section: 比较解决四方形的比较方法 | 10.9 解决四方办法的比较

Section outline

What if you and a friend started running from the same spot. You ran west and your friend ran south. After two hours, you had run 10 miles. The distance at that point between you and your friend was three times the distance your friend ran plus 2 miles? How could you determine how many miles your friend ran?
::如果你和一个朋友从同一个地点跑出来,你会怎么样?你跑到西边,你朋友往南跑。过了两个小时,你跑了10英里。当时你和你朋友之间的距离是你的朋友跑了3倍加2英里的距离?你怎么能确定你朋友跑了多少英里?

Comparing Methods for Solving Quadratics
::比较解决四方形的比较方法

In mathematics, you’ll need to that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation :
::在数学方面,你需要描述应用问题或更复杂问题的一部分。你已经学会了解决四方方程式的四种方法:

Factoring
::保理

Taking the square root
::取平方根

Quadratic formula
::二次曲线公式

Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in different situations.
::通常你必须自己决定使用哪种方法。但是,这里有一些指导准则,说明不同情况下哪种方法更好。

Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.
::如果二次表达式易于因数,则乘数总是最好的。总是值得检查您能否因数,因为这是最快的方法。许多表达式是不可因数的,因此这种方法在实践中不会经常使用。

Taking the square root is best used when there is no $\begin{aligned} x - \end{aligned}$ term in the equation .
::当方程式中没有 x- term 时, 取平方根是最佳使用法。

Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.
::二次曲线公式是用来解决二次方程的最常见的方法。如果直接通过平方根和保理法解决的方法行不通,这是大多数人喜欢使用的方法。

Completing the square can be used to solve any quadratic equation. This is usually not any better than using (in terms of difficult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in algebra II, trigonometry, physics, calculus, and beyond).
::完成方形可以用来解决任何二次方程。这通常不会比使用(在困难的计算中)更好,但如果您需要以顶点形式重写二次函数,则非常有用。它也用来以标准形式重写圆形、椭圆形和双光形的方程(在代数II、三角测量、物理、微积分等等方面你将会做的事情 ) 。

If you are using factoring or the quadratic formula, make sure that the equation is in standard form.
::如果您使用乘数公式或二次公式,请确保公式为标准格式。

Solving Quadratic Equations
::解析二次赤道

Solve each quadratic equation.
::解决每个二次方程式

a) $\begin{aligned} x^{2} - 4 x - 5 = 0 \end{aligned}$
::a) x2-4x-5=0

This expression if easily factorable so we can factor and apply the zero-product property :
::这个表达式如果容易因数,我们就可以对零产品属性进行系数和适用:

$\begin{aligned} Factor: & (x - 5) (x + 1) = 0 \\ Apply zero-product property: & x - 5 = 0 and x + 1 = 0 \\ Solve: & x = 5 and x = - 1 \end{aligned}$

::系数 : (x- 5)(x+1) = 0Apply 零产品属性: x-5= 0andx+1=0Solve: x=5 andx @%1

Answer: $\begin{aligned} x = 5 \end{aligned}$ and $\begin{aligned} x = - 1 \end{aligned}$
::答复:x=5和x1

b) $\begin{aligned} x^{2} = 8 \end{aligned}$
::b) x2=8

Since the expression is missing the $\begin{aligned} x \end{aligned}$ term we can take the square root:
::由于表达式缺失 x 术语, 我们可以选择平方根 :

Take the square root of both sides: $\begin{aligned} x = \sqrt{8} \end{aligned}$ and $\begin{aligned} x = - \sqrt{8} \end{aligned}$
::以两侧的平方根为平方根: x=8 和 x8

Answer: $\begin{aligned} x = 2.83 \end{aligned}$ and $\begin{aligned} x = - 2.83 \end{aligned}$
::答复:x=2.83和x2.83

c) $\begin{aligned} - 4 x^{2} + x = 2 \end{aligned}$
:c)-4x2+x=2

Re-write the equation in standard form: $\begin{aligned} - 4 x^{2} + x - 2 = 0 \end{aligned}$
::以标准格式重写公式 : - 4x2+x-2=0

It is not apparent right away if the expression is factorable so we will use the quadratic formula:
::如果该表达式是可因数的,那么我们就会使用四边式的公式。

$\begin{aligned} Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = - 4, b = 1, c = - 2 : & x & = \frac{- 1 \pm \sqrt{1^{2} - 4 (- 4) (- 2)}}{2 (- 4)} \\ Simplify: & x & = \frac{- 1 \pm \sqrt{1 - 32}}{- 8} = \frac{- 1 \pm \sqrt{- 31}}{- 8} \end{aligned}$

::二次曲线公式:xbb2-4ac2a+lug, 数值为a4, b=1, c2:x112-4(4)(-2)(-2)(-4)(-4)(-4)(-4) 简化:x11-1-32-8131-8。

Answer: no real solution
::答复:没有真正的解决办法

d) $\begin{aligned} 25 x^{2} - 9 = 0 \end{aligned}$
:d) 25x2-9=0

This problem can be solved easily either with factoring or taking the square root. Let’s take the square root in this case:
::这个问题可以通过保理或平方根来轻易解决。

$\begin{aligned} Add 9 to both sides of the equation: & 25 x^{2} & = 9 \\ Divide both sides by 25 : & x^{2} & = \frac{9}{25} \\ Take the square root of both sides: & x & = \sqrt{\frac{9}{25}} and x = - \sqrt{\frac{9}{25}} \\ Simplify: & x & = \frac{3}{5} and x = - \frac{3}{5} \end{aligned}$

::在方程的两侧添加 9: 25x2= 9 以 25: x2= 925 以 25: x2= 925 将两侧的平方根 : x= 925 和 x\\\\ 925 以简单化: x= 35 和 x35 以 25: x= 925 以 25: x2= 925 以两侧的平方根 : x= 925 和 x*925 以简单化: x= 35 和 x\ *35

Answer: $\begin{aligned} x = \frac{3}{5} \end{aligned}$ and $\begin{aligned} x = - \frac{3}{5} \end{aligned}$
::答复:x=35和x=35

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

$\begin{aligned} Re-write the equation in standard form: & 3 x^{2} - 8 x & = 0 \\ Factor out common x term: & x (3 x - 8) & = 0 \\ Set both terms to zero: & x & = 0 and 3 x = 8 \\ Solve: & x & = 0 and x = \frac{8}{3} = 2.67 \end{aligned}$

::以标准窗体: 3x2- 8x= 0 格式重写方程式 : 3x2- 8x= 0Fictor 排除常见的 x 术语: x( 3x- 8) = 0Set 两种条件都为 0: x= 0 和 3x= 8Solve: x=0 和 x=83= 2. 67

Answer: $\begin{aligned} x = 0 \end{aligned}$ and $\begin{aligned} x = 2.67 \end{aligned}$
::答复:x=0和x=2.67

Solving Real-World Problems by Completing the Square
::通过完成广场解决现实世界问题

In the last section you learned that an object that is dropped falls under the influence of gravity. The equation for its height with respect to time is given by $\begin{aligned} y = \frac{1}{2} g t^{2} + y_{0} \end{aligned}$ , where $\begin{aligned} y_{0} \end{aligned}$ represents the initial height of the object and $\begin{aligned} g \end{aligned}$ is the coefficient of gravity on earth, which equals $\begin{aligned} - 9.8 m / s^{2} \end{aligned}$ or $\begin{aligned} - 32 f t / s^{2} \end{aligned}$ .
::在最后一节中,您了解到一个被丢弃的物体处于重力影响之下。相对于时间而言,其高度的方程式由 y= 12gt2+y0 给出,其中 y0 表示对象的初始高度, g 表示地球重力系数,该系数等于- 9.8 m/s2 或- 32 ft/s2。

On the other hand, if an object is thrown straight up or straight down in the air, it has an initial vertical velocity. This term is usually represented by the notation $\begin{aligned} v_{0 y} \end{aligned}$ . Its value is positive if the object is thrown up in the air and is negative if the object is thrown down. The equation for the height of the object in this case is
::另一方面,如果一个物体被直向上或直向向空中投掷,则其最初的垂直速度为垂直速度。该词通常用注注表示。如果物体被投向空中,其值为正值,如果物体被抛向空中,则其值为负值。在这种情况下,物体的高度的方程式是正值。

$\begin{aligned} y = \frac{1}{2} g t^{2} + v_{0 y} t + y_{0} \end{aligned}$

::y= 12gt2+v0yt+y0

Plugging in the appropriate value for $\begin{aligned} g \end{aligned}$ turns this equation into
::g 将此方程式转换为

$\begin{aligned} y = - 4.9 t^{2} + v_{0 y} t + y_{0} \end{aligned}$ if you wish to have the height in meters
::y4.9t2+v0yt+y0 如果您想要以米计高度

$\begin{aligned} y = - 16 t^{2} + v_{0 y} t + y_{0} \end{aligned}$ if you wish to have the height in feet
::y16t2+v0yt+Y0 如果您想要高高的脚

Real-World Application: Velocity
::现实世界应用:速度

An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s.
::箭头从2米高处直射直射,速度50米/秒。

Since we are given the velocity in m/s, use: $\begin{aligned} y = - 4.9 t^{2} + v_{0 y} t + y_{0} \end{aligned}$
::既然我们被给定了 m/ s 的速度, 请使用: y4. 9t2+v0yt+y0

We know $\begin{aligned} v_{0 y} = 50 m / s \end{aligned}$ and $\begin{aligned} y_{0} = 2 \end{aligned}$ meters so: $\begin{aligned} y = - 4.9 t^{2} + 50 t + 2 \end{aligned}$
::我们知道V0y=50米/秒和y0=2米,所以:y4.9t2+50t+2

a) How high will the arrow be 4 seconds after being shot? After 8 seconds?
:a) 箭头射中后4秒多高?

To find how high the arrow will be 4 seconds after being shot we plug in $\begin{aligned} t = 4 \end{aligned}$ :
::要找到箭头在被射中后4秒的高度, 我们插入 t=4 :

$\begin{aligned} y & = - 4.9 (4)^{2} + 50 (4) + 2 \\ = - 4.9 (16) + 200 + 2 = \underline{\underline{123.6 m e t e r s}} \end{aligned}$

::y4.9(4)2+50(4)+24.9(16)+200+2=123.6米__

we plug in $\begin{aligned} t = 8 \end{aligned}$ :
::我们插入 t=8 :

$\begin{aligned} y & = - 4.9 (8)^{2} + 50 (8) + 2 \\ = - 4.9 (64) + 400 + 2 = \underline{\underline{88.4 m e t e r s}} \end{aligned}$

::y4.9(8)2+50(8)+24.9(64)+400+2=88.4米__

b) At what time will the arrow hit the ground again?
:b) 箭头何时会再次落地?

The height of the arrow on the ground is $\begin{aligned} y = 0 \end{aligned}$ , so: $\begin{aligned} 0 = - 4.9 t^{2} + 50 t + 2 \end{aligned}$
::箭头在地面的高度为y=0, 所以: 04. 9t2+50t+2

Solve for $\begin{aligned} t \end{aligned}$ by completing the square:
::完成广场, 解决 t :

$\begin{aligned} - 4.9 t^{2} + 50 t & = - 2 \\ - 4.9 (t^{2} - 10.2 t) & = - 2 \\ t^{2} - 10.2 t & = 0.41 \\ t^{2} - 2 (5.1) t + (5.1)^{2} & = 0.41 + (5.1)^{2} \\ (t - 5.1)^{2} & = 26.43 \\ t - 5.1 & = 5.14 and t - 5.1 = - 5.14 \\ t & = \underline{\underline{10.2}} s e c and t = - 0.04 s e c \end{aligned}$

::-4.9t2+50t2-4.9(t2-10.2t)2t2-10.2t=0.412-2-2(5.1)t+(5.1)2=0.41+(5.1)2(t-5.1)2=26.43t-5.1=5.14和t-5.15.14t=10.2-秒和t0.04秒

The arrow will hit the ground about 10.2 seconds after it is shot.
::箭头在射出10.2秒后就会击中地面

c) What is the maximum height that the arrow will reach and at what time will that happen?
:c) 箭头将达到的最高高度是多少,何时会达到?

If we graph the height of the arrow with respect to time we would get an upside down parabola $\begin{aligned} (a < 0) \end{aligned}$ . The maximum height and the time when this occurs is really the vertex of this parabola: $\begin{aligned} (t, h) . \end{aligned}$
::如果我们按时间绘制箭头的高度, 我们就会在抛物线上得到一个反向( a<0 ) 。最大高度和发生这种情况的时间实际上是这个抛物线的顶点 t, h ) 。

$\begin{aligned} We re-write the equation in vertex form: & y = - 4.9 t^{2} + 50 t + 2 \\ y - 2 = - 4.9 t^{2} + 50 t \\ y - 2 = - 4.9 (t^{2} - 10.2 t) \\ Complete the square: & y - 2 - 4.9 (5.1)^{2} = - 4.9 (t^{2} - 10.2 t + (5.1)^{2}) \\ y - 129.45 = - 4.9 (t - 5.1)^{2} \end{aligned}$

::我们以顶层形式重写方程式 :y4.9t2+50t+2y-24.9t2+50ty-24.9(t2-10.2t) 完成正方形 :y-2-4.9(5.1)2}4.9(t2-10.2t+(5.1.1)2)y-129.45*4.9(t-5.1)2

The vertex is at (5.1, 129.45). In other words, when $\begin{aligned} t = 5.1 s e c o n d s \end{aligned}$ , the height is $\begin{aligned} y = 129 m e t e r s \end{aligned}$ .
::顶点在(5.1, 129.45),换句话说,当t=5.1秒时,高度为y=129米。

Another type of application problem that can be solved using quadratic equations is one where two objects are moving away from each other in perpendicular directions. Here is an example of this type of problem.
::另一种应用问题可以通过二次方程解决,一种是两个对象在垂直方向上相互移动。这里是这类问题的一个实例。

Real-World Application: Distance Between Two Objects
::Real- World 应用程序: 两个对象之间的距离

Two cars leave an intersection . One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.
::两辆车离开一个十字路口,一辆车向北行驶,另一辆车向东行驶,当往北行驶的汽车行驶30英里时,车距比往东行驶的汽车距离高出10英里,是车行行驶距离的两倍多。

Let $\begin{aligned} x = \end{aligned}$ the distance traveled by the car heading east
::让x=车向东行驶的距离

Then $\begin{aligned} 2 x + 10 = \end{aligned}$ the distance between the two cars
::然后 2x+10 = 两辆汽车之间的距离

Let’s make a sketch:
::让我们做个草图:

We can use the Pythagorean Theorem to find an equation for $\begin{aligned} x \end{aligned}$ :
::我们可以使用毕达哥里安理论来寻找 x 的方程式 :

$\begin{aligned} x^{2} + 30^{2} = (2 x + 10)^{2} \end{aligned}$

::x2+302=(2x+10)2

Expand " data-term="Parentheses" role="term" tabindex="0"> parentheses and simplify:
::扩大括号并简化:

$\begin{aligned} x^{2} + 900 & = 4 x^{2} + 40 x + 100 \\ 800 & = 3 x^{2} + 40 x \end{aligned}$

::x2+900=4x2+40x+100800=3x2+40x

Solve by completing the square:
::完成方块的解决方式 :

$\begin{aligned} \frac{800}{3} & = x^{2} + \frac{40}{3} x \\ \frac{800}{3} + {(\frac{20}{3})}^{2} & = x^{2} + 2 (\frac{20}{3}) x + {(\frac{20}{3})}^{2} \\ \frac{2800}{9} & = {(x + \frac{20}{3})}^{2} \\ x + \frac{20}{3} & = 17.6 and x + \frac{20}{3} = - 17.6 \\ x & = 11 and x = - 24.3 \end{aligned}$

::8003=x2+403x8003+(20322)2=x2+2(203)x+(203)22809=(x+203)2x203+203=17.6和x+203=176*176x=11和x24.3

Since only positive distances make sense here, the distance between the two cars is: $\begin{aligned} 2 (11) + 10 = 32 m i l e s \end{aligned}$
::由于这里只有正距离才有意义,两辆汽车之间的距离是:2(11)+10=32英里。

Solve Applications Using Quadratic Functions by any Method
::使用任何方法的二次函数解决应用程序

Here is an application problem that arises from number relationships and geometry applications.
::这是一个应用问题,来自数字关系和几何应用。

The product of two positive consecutive integers is 156. Find the integers.
::两个正连续整数的产物为156。查找整数。

Define: Let $\begin{aligned} x = \end{aligned}$ the smaller integer
::定义:让 x = 较小整数

Then $\begin{aligned} x + 1 = \end{aligned}$ the next integer
::然后 x+1 = 下一个整数

Translate: The product of the two numbers is 156. We can write the equation:
::翻译:这两个数字的产物是156。我们可以写出方程式:

$\begin{aligned} x (x + 1) = 156 \end{aligned}$

::x( x+1) =156

Solve:
::解决 :

$\begin{aligned} x^{2} + x & = 156 \\ x^{2} + x - 156 & = 0 \end{aligned}$

::x2+x=156x2+x-156=0

Apply the quadratic formula with: $\begin{aligned} a = 1, b = 1, c = - 156 \end{aligned}$
::应用四方形公式,其值为: a=1, b=1, c=156

$\begin{aligned} x & = \frac{- 1 \pm \sqrt{1^{2} - 4 (1) (- 156)}}{2 (1)} \\ x & = \frac{- 1 \pm \sqrt{625}}{2} = \frac{- 1 \pm 25}{2} \\ x & = \frac{- 1 + 25}{2} and x = \frac{- 1 - 25}{2} \\ x & = \frac{24}{2} = 12 and x = \frac{- 26}{2} = - 13 \end{aligned}$

::112-4(1)(-156)2(1)x162521252x1+252和1_252x=242=12andx26213

Since we are looking for positive integers, we want $\begin{aligned} x = 12 \end{aligned}$ . So the numbers are 12 and 13.
::由于我们正在寻找正整数, 我们需要 x=12。所以数字是 12 和 13 。

Check: $\begin{aligned} 12 \times 13 = 156 \end{aligned}$ . The answer checks out.
::检查: 12x13=156。答案检查出来。

Real-World Application: Fencing
::真实世界应用程序:

Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is $\begin{aligned} 200 f t^{2} \end{aligned}$ . How much fencing does Suzie need?
::Suzie想建造一座花园,有3个不同的长方形区段。她想在整个花园周围和每个区段之间围成围栏。地块宽度是2倍,总面积是200英尺。 Suzie需要多少围栏?

Define: Let $\begin{aligned} x = \end{aligned}$ the width of the plot
::定义: 让 x = 绘图宽度

Then $\begin{aligned} 2 x = \end{aligned}$ the length of the plot
::然后 2x=绘图的长度

Translate: area of a rectangle is $\begin{aligned} A = length \times width \end{aligned}$ , so
::翻译:矩形区域为 A=长xxwidth, 所以

$\begin{aligned} x (2 x) = 200 \end{aligned}$

:x2x) = 200

Solve: $\begin{aligned} 2 x^{2} = 200 \end{aligned}$
::解决: 2x2=200

Solve by taking the square root:
::以平方根解决:

$\begin{aligned} x^{2} & = 100 \\ x & = \sqrt{100} and x = - \sqrt{100} \\ x & = 10 and x = - 10 \end{aligned}$

::x2 = 100x= 100 和 x @ @% 100x= 10 和 x% 10

We take $\begin{aligned} x = 10 \end{aligned}$ since only positive dimensions make sense.
::我们用x=10,因为只有正维值才有意义。

The plot of land is $\begin{aligned} 10 f e e t \times 20 f e e t \end{aligned}$ .
::土地面积为10英尺x20英尺。

To fence the garden the way Suzie wants, we need 2 lengths and 4 widths $\begin{aligned} = 2 (20) + 4 (10) = 80 \end{aligned}$ feet of fence.
::为了按照Suzie的愿望围住花园,我们需要2长4宽=2(20)+4(10)=80英尺的围栏。

Check: $\begin{aligned} 10 \times 20 = 200 f t^{2} \end{aligned}$ and $\begin{aligned} 2 (20) + 4 (10) = 80 f e e t \end{aligned}$ . The answer checks out.
::检查: 10x20=200 平方英尺和 2( 20)+4( 10) = 80 英尺。答案检查出来。

Examples
::实例

Example 1
::例1

An isosceles triangle is enclosed in a square so that its base coincides with one of the sides of the square and the tip of the triangle touches the opposite side of the square. If the area of the triangle is $\begin{aligned} 20 i n^{2} \end{aligned}$ what is the length of one side of the square?
::方形三角形被封闭在一个方形内,使三角形的底部与正方形的两面之一相吻合,三角形的尖部与正方形的对面相接触。如果三角形的区域是20英寸2,则方形一面的长度是多少?

Draw a sketch:
::绘制草图 :

Define: Let $\begin{aligned} x = \end{aligned}$ base of the triangle
::定义:让 x=三角形的基底

Then $\begin{aligned} x = \end{aligned}$ height of the triangle
::然后 x= 三角形的高度

Translate: Area of a triangle is $\begin{aligned} \frac{1}{2} \times b a s e \times h e i g h t \end{aligned}$ , so $\begin{aligned} \frac{1}{2} \cdot x \cdot x = 20 \end{aligned}$
::翻译: 三角形的面积为 12xbasexh8, 所以 12xxxx=20

Solve: $\begin{aligned} \frac{1}{2} x^{2} = 20 \end{aligned}$
::溶解: 12x2=20

Solve by taking the square root:
::以平方根解决:

$\begin{aligned} x^{2} & = 40 \\ x & = \sqrt{40} and x = - \sqrt{40} \\ x & = 6.32 and x = - 6.32 \end{aligned}$

::x2=40x=40和 x40x=6.32和 x6.32

The side of the square is 6.32 inches. That means the area of the square is $\begin{aligned} (6.32)^{2} = 40 i n^{2} \end{aligned}$ , twice as big as the area of the triangle.
::方形的侧面是6.32英寸。这意味着方形的面积是(6.322)2=40英寸2,是三角形面积的两倍。

Check: It makes sense that the area of the square will be twice that of the triangle. If you look at the figure you can see that you could fit two triangles inside the square.
::选中 : 方形的区域将是三角形的两倍, 这是合理的。如果您查看数字, 您可以看到您可以在方形内安装两个三角形。

Example 2
::例2

The length of a rectangular pool is 10 meters more than its width. The area of the pool is 875 square meters. Find the dimensions of the pool.
::矩形池的长度比其宽度多10米。池的面积是875平方米。查找池的尺寸。

Draw a sketch:
::绘制草图 :

Define: Let $\begin{aligned} x = \end{aligned}$ the width of the pool
::定义:让 x = 池的宽度

Then $\begin{aligned} x + 10 = \end{aligned}$ the length of the pool
::然后 x+10 = 池的长度

Translate: The area of a rectangle is $\begin{aligned} A = length \times width \end{aligned}$ , so we have $\begin{aligned} x (x + 10) = 875 \end{aligned}$ .
::翻译: 矩形区域是 A= 长xwidth, 所以我们有 x( x+10) = 875 。

Solve:
::解决 :

$\begin{aligned} x^{2} + 10 x & = 875 \\ x^{2} + 10 x - 875 & = 0 \end{aligned}$

::x2+10x=875x2+10x-875=0

Apply the quadratic formula with $\begin{aligned} a = 1, b = 10 \end{aligned}$ and $\begin{aligned} c = - 875 \end{aligned}$
::应用 a=1, b=10 和 c875 的二次公式

$\begin{aligned} x & = \frac{- 10 \pm \sqrt{(10)^{2} - 4 (1) (- 875)}}{2 (1)} \\ x & = \frac{- 10 \pm \sqrt{100 + 3500}}{2} \\ x & = \frac{- 10 \pm \sqrt{3600}}{2} = \frac{- 10 \pm 60}{2} \\ x & = \frac{- 10 + 60}{2} and x = \frac{- 10 - 60}{2} \\ x & = \frac{50}{2} = 25 and x = \frac{- 70}{2} = - 35 \end{aligned}$

::x10(10)2-4(1)(- 875)2(1)x10100+3500x1030.0021030.00210602x10+602和 x10-602x=502=25和x70235

Since the dimensions of the pool should be positive, we want $\begin{aligned} x = 25 m e t e r s \end{aligned}$ . So the pool is $\begin{aligned} 25 m e t e r s \times 35 m e t e r s \end{aligned}$ .
::因为水池的尺寸应该是正的, 我们需要x=25米。所以水池是25米×35米。

Check: $\begin{aligned} 25 \times 35 = 875 m^{2} \end{aligned}$ . The answer checks out.
::检查: 25×35=875 m2. 答案检查出来。

Review
::回顾

Solve the following quadratic equations using the method of your choice.
::使用您选择的方法解决以下二次方程。

$\begin{aligned} x^{2} - x = 6 \end{aligned}$
::x2 - x=6 x2 - x=6

$\begin{aligned} x^{2} - 12 = 0 \end{aligned}$
::x2 - 12=0

$\begin{aligned} - 2 x^{2} + 5 x - 3 = 0 \end{aligned}$
::-2x2+5x-3=0

$\begin{aligned} x^{2} + 7 x - 18 = 0 \end{aligned}$
::x2+7x-18=0

$\begin{aligned} 3 x^{2} + 6 x = - 10 \end{aligned}$
::3x2+6x 10

$\begin{aligned} - 4 x^{2} + 4000 x = 0 \end{aligned}$
::- 4x2+4000x=0

$\begin{aligned} - 3 x^{2} + 12 x + 1 = 0 \end{aligned}$
::3x2+12x+1=0

$\begin{aligned} x^{2} + 6 x + 9 = 0 \end{aligned}$
::x2+6x+9=0

$\begin{aligned} 81 x^{2} + 1 = 0 \end{aligned}$
::81x2+1=0

$\begin{aligned} - 4 x^{2} + 4 x = 9 \end{aligned}$
::- 4x2+4x=9

$\begin{aligned} 36 x^{2} - 21 = 0 \end{aligned}$
::36x2-21=0

$\begin{aligned} x^{2} - 2 x - 3 = 0 \end{aligned}$
::x2-2x-3=0

The product of two consecutive integers is 72. Find the two numbers.
::两个连续整数的产物是 72. 找出这两个数字。

The product of two consecutive odd integers is 1 less than 3 times their sum. Find the integers.
::两个连续奇数整数的产物是其总和的1小3倍。查找整数。

The length of a rectangle exceeds its width by 3 inches. The area of the rectangle is 70 square inches, find its dimensions.
::矩形的长度超过其宽度3英寸。矩形的面积为70平方英寸,找到它的尺寸。

Angel wants to cut off a square piece from the corner of a rectangular piece of plywood. The larger piece of wood is $\begin{aligned} 4 f e e t \times 8 f e e t \end{aligned}$ and the cut off part is $\begin{aligned} \frac{1}{3} \end{aligned}$ of the total area of the plywood sheet. What is the length of the side of the square?
::Angel想从胶合板的长方形片角上切掉一块正方形块。更大的木块是4英尺×8英尺,切掉的部分是胶合板总面积的13。方形侧的长度是多少?

Mike wants to fence three sides of a rectangular patio that is adjacent the back of his house. The area of the patio is $\begin{aligned} 192 f t^{2} \end{aligned}$ and the length is 4 feet longer than the width. Find how much fencing Mike will need.
::Mike想围住他家后院的三边。院子的面积是192平方英尺, 长度比宽度长4英尺。找找Mike需要多少栅栏。

Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
::Sam从25英尺高处直接扔下一个鸡蛋。蛋的最初速度是16英尺/秒。鸡蛋到达地面需要多久?

Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?
::阿曼达和多尔文在同一时间离开他们的家阿曼达向南走和多尔文自行车向东走半小时后他们又相距方圆5.5英里而多尔文的距离比阿曼达多了3英里

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。

比较解决四方形的比较方法

Section outline

Comparing Methods for Solving Quadratics
::比较解决四方形的比较方法

Solving Quadratic Equations
::解析二次赤道

Solving Real-World Problems by Completing the Square
::通过完成广场解决现实世界问题

Real-World Application: Velocity
::现实世界应用:速度

Real-World Application: Distance Between Two Objects
::Real- World 应用程序: 两个对象之间的距离

Solve Applications Using Quadratic Functions by any Method
::使用任何方法的二次函数解决应用程序

Real-World Application: Fencing
::真实世界应用程序:

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)

Section outline

Comparing Methods for Solving Quadratics ::比较解决四方形的比较方法

Solving Quadratic Equations ::解析二次赤道

Solving Real-World Problems by Completing the Square ::通过完成广场解决现实世界问题

Real-World Application: Velocity ::现实世界应用:速度

Real-World Application: Distance Between Two Objects ::Real- World 应用程序: 两个对象之间的距离

Solve Applications Using Quadratic Functions by any Method ::使用任何方法的二次函数解决应用程序

Real-World Application: Fencing ::真实世界应用程序:

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Comparing Methods for Solving Quadratics
::比较解决四方形的比较方法

Solving Quadratic Equations
::解析二次赤道

Solving Real-World Problems by Completing the Square
::通过完成广场解决现实世界问题

Real-World Application: Velocity
::现实世界应用:速度

Real-World Application: Distance Between Two Objects
::Real- World 应用程序: 两个对象之间的距离

Solve Applications Using Quadratic Functions by any Method
::使用任何方法的二次函数解决应用程序

Real-World Application: Fencing
::真实世界应用程序:

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)