Course: 10.5 平根应用 | The Way To Learn

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平方根应用
Square Root Applications
::平方根应用

We can use the methods we’ve learned so far in this section to find approximate solutions to quadratic equations, when taking the square root doesn’t give an exact answer.
::我们可以使用我们在本节所学到的方法来寻找四方形方程式的近似解决方案,

Solving Quadratic Equations
::解析二次赤道

1. Solve the following quadratic equations.
::1. 解决以下四方方程式。

a) $\begin{aligned} x^{2} - 3 = 0 \end{aligned}$
$\begin{aligned} Isolate the x^{2} : x^{2} = 3 \\ Take the square root of both sides : x = \sqrt{3} and x = - \sqrt{3} \end{aligned}$

::a) x2 - 3=0 孤立 x2: x2=3 ,取两侧的平方根:x=3 和 x3

Answer: $\begin{aligned} x \approx 1.73 \end{aligned}$ and $\begin{aligned} x \approx - 1.73 \end{aligned}$
::答复:x1.73和x1.73

b) $\begin{aligned} 2 x^{2} - 9 = 0 \end{aligned}$
$\begin{aligned} Isolate the x^{2} : 2 x^{2} = 9 so x^{2} = \frac{9}{2} \\ Take the square root of both sides : x = \sqrt{\frac{9}{2}} and x = - \sqrt{\frac{9}{2}} \end{aligned}$

::b) 2x2 - 9=0 孤立 x2: 2x2=9 so x2=92 将两侧的平方根:x=92 和 x92 取为平方根

Answer: $\begin{aligned} x \approx 2.12 \end{aligned}$ and $\begin{aligned} x \approx - 2.12 \end{aligned}$
::答复:x____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

2. Solve the following quadratic equations.
::2. 解决以下四方方程式。

a) $\begin{aligned} (2 x + 5)^{2} = 10 \end{aligned}$
:a) (2x+5)2=10

$\begin{aligned} Take the square root of both sides : & 2 x + 5 & = \sqrt{10} and 2 x + 5 = - \sqrt{10} \\ Solve both equations : & x & = \frac{- 5 + \sqrt{10}}{2} and x = \frac{- 5 - \sqrt{10}}{2} \end{aligned}$

::取取两边的平方根: 2x+5=10 和 2x+5\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x\\\\\\\\\\\\\ x\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Answer: $\begin{aligned} x \approx - 0.92 \end{aligned}$ and $\begin{aligned} x \approx - 4.08 \end{aligned}$
::答复:x0.92和x4.08

b) $\begin{aligned} x^{2} - 2 x + 1 = 5 \end{aligned}$
$\begin{aligned} Factor the right-hand-side : & (x - 1)^{2} & = 5 \\ Take the square root of both sides : & x - 1 & = \sqrt{5} and x - 1 = - \sqrt{5} \\ Solve each equation : & x & = 1 + \sqrt{5} and x = 1 - \sqrt{5} \end{aligned}$

::b) x2 - 2x+1=5 右侧方块x- 1)2=5 取两侧的平方根:x-1=5 和 x- 1=5 和 x- 1\%5Solve 每一个方程式:x=1+5 和 x=1-5

Answer: $\begin{aligned} x \approx 3.24 \end{aligned}$ and $\begin{aligned} x \approx - 1.24 \end{aligned}$
::答复:x3.24和x1.24

Solve Applications Using Quadratic Functions and Square Roots
::使用二次曲线函数和平方根解决应用程序

Quadratic equations are needed to solve many real-world problems. In this section, we’ll examine problems about objects falling under the influence of gravity. When objects are dropped from a height, they have no initial velocity; the force that makes them move towards the ground is due to gravity. The acceleration of gravity on earth is given by the equation
::解决许多现实世界问题需要二次方程式。本节将检查受重力影响的物体的问题。当物体从高度下降时, 它们没有初始速度; 促使它们向地面移动的力是由重力决定的。地球重力的加速是由等式决定的。

$\begin{aligned} g = - 9.8 m / s^{2} or g = - 32 f t / s^{2} \end{aligned}$

::g$9.8 m/s2org$32 ft/s2

The negative sign indicates a downward direction. We can assume that gravity is constant for the problems we’ll be examining, because we will be staying close to the surface of the earth. The acceleration of gravity decreases as an object moves very far from the earth. It is also different on other celestial bodies such as the moon.
::负号表示向下方向。我们可以假设重力对于我们将要研究的问题来说是恒定的, 因为我们将停留在接近地球表面的地方。当物体远离地球时, 重力加速度会下降。在月球等其他天体上, 重力也有所不同。

The equation that shows the height of an object in free fall is
::显示对象在自由坠落时高度的方程式是

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

::y=12gt2+y0 y=12gt2+y0

The term $\begin{aligned} y_{0} \end{aligned}$ represents the initial height of the object, $\begin{aligned} t \end{aligned}$ is time, and $\begin{aligned} g \end{aligned}$ is the constant representing the force of gravity. You then plug in one of the two values for $\begin{aligned} g \end{aligned}$ above, depending on whether you want the answer in feet or meters. Thus the equation works out to $\begin{aligned} y = - 4.9 t^{2} + y_{0} \end{aligned}$ if you want the height in meters, and $\begin{aligned} y = - 16 t^{2} + y_{0} \end{aligned}$ if you want it in feet.
::y0 表示对象的初始高度, t 是时间, g 是代表重力的常数。然后, 您可以插入以上 g 的两个值中的一个, 取决于您是否想要以英尺或米表示答案。因此, 如果想要以米表示高度, 公式可以工作到 y4. 9t2+y0 , 如果您想要以米表示高度, 则可以工作到 y 16t2+y0 。

Real-World Application: Object Falling
::Real- World 应用程序: 对象坠落

1. How long does it take a ball to fall from a roof to the ground 25 feet below?
::1. 从屋顶摔到地下25英尺需要多久?

$\begin{aligned} Since we are given the height in feet, use equation : & y & = - 16 t^{2} + y_{0} \\ The initial height is y_{0} = 25 f e e t, so : & y & = - 16 t^{2} + 25 \\ The height when the ball hits the ground is y = 0, so : & 0 & = - 16 t^{2} + 25 \\ Solve for t : & 16 t^{2} & = 25 \\ t^{2} & = \frac{25}{16} \\ t & = \frac{5}{4} or t = - \frac{5}{4} \end{aligned}$

::既然给出了脚的高度, 请使用方程: y16t2+y0 初始高度为 y0= 25 英尺, 所以: y16t2+25 当球击中地面时高度为 y= 0, 所以: 016t2+25Solve , t: 16t2= 25t2= 2516t= 54 或 t54

Since only positive time makes sense in this case, it takes the ball 1.25 seconds to fall to the ground.
::因为在这种情况下,只有积极的时间才有意义,所以需要1.25秒的球落到地上。

2. A rock is dropped from the top of a cliff and strikes the ground 7.2 seconds later. How high is the cliff in meters?
::2. 岩石从悬崖顶上掉下来,然后在7.2秒后撞击地面。

$\begin{aligned} Since we want the height in meters, use equation : & y & = - 4.9 t^{2} + y_{0} \\ The time of flight is t = 7.2 s e c o n d s : & y & = - 4.9 (7.2)^{2} + y_{0} \\ The height when the ball hits the ground is y = 0, so : & 0 & = - 4.9 (7.2)^{2} + y_{0} \\ Simplify : & 0 & = - 254 + y_{0} so y_{0} = 254 \end{aligned}$

::既然我们需要以米计的高度, 请使用方程: y4. 9t2+y0 飞行时间为 t= 7.2 秒: y4. 9( 7.2)2+y0 球撞击地面时的高度为 y=0, 所以: 04. 9( 7.2)2+y0 简化: 0 254+y0 so y0=254

The cliff is 254 meters high.
::悬崖高254米

Example
::示例示例示例示例

Example 1
::例1

Victor throws an apple out of a window on the $\begin{aligned} 10^{t h} \end{aligned}$ floor which is 120 feet above ground. One second later Juan throws an orange out of a $\begin{aligned} 6^{t h} \end{aligned}$ floor window which is 72 feet above the ground. Which fruit reaches the ground first, and how much faster does it get there?
::Victor从10楼的窗户上扔出一个苹果,该楼在地面上120英尺。一秒钟后,Juan从6楼的窗户上扔出一个橙子,该窗户在地面上72英尺。哪一种水果先到达地面,再到多快?

Let’s find the time of flight for each piece of fruit.
::让我们为每一块水果寻找飞行时间。

Apple:
::苹果 :

$\begin{aligned} Since we have the height in feet, use this equation : & y = - 16 t^{2} + y_{0} \\ The initial height is y_{0} = 120 f e e t : & y = - 16 t^{2} + 120 \\ The height when the ball hits the ground is y = 0, so : & 0 = - 16 t^{2} + 120 \\ Solve for t : & 16 t^{2} = 120 \\ t^{2} = \frac{120}{16} = 7.5 \\ \underline{t = 2.74} or t = - 2.74 s e c o n d s \end{aligned}$

::既然我们脚高, 请使用这个方程: y16t2+y0 初始高度为 y0= 120 英尺: y16t2+120 当球击中地面时高度为 y= 0, 所以: 016t2+120Solve for t: 16t2= 120t2= 1202= 12016= 7. 5t= 2. 74_ 或 t2. 74 秒

Orange:
::橙色 :

$\begin{aligned} The initial height is y_{0} = 72 f e e t : & 0 = - 16 t^{2} + 72 \\ Solve for t : & 16 t^{2} = 72 \\ t^{2} = \frac{72}{16} = 4.5 \\ \underline{t = 2.12} or t = - 2.12 s e c o n d s \end{aligned}$

::初始高度为 y0= 72 英尺: 016t2+72Solve, t: 16t2= 72t2= 7216=4.5t= 2.12_ 或 t2.12 秒

The orange was thrown one second later, so add 1 second to the time of the orange: $\begin{aligned} t = 3.12 s e c o n d s \end{aligned}$
::橙子在稍后一秒钟被抛出, 所以在橙子时间增加一秒钟: t=3. 12秒

The apple hits the ground first. It gets there 0.38 seconds faster than the orange.
::苹果先落地,比橙子快0.38秒

Review
::回顾

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

$\begin{aligned} x^{2} = 11 \end{aligned}$
::x2=11xx2=11

$\begin{aligned} 5 x^{2} = 0.01 \end{aligned}$
::5x2=0.01

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

$\begin{aligned} x^{2} - 20 = 0 \end{aligned}$
::x2 - 20=0

$\begin{aligned} 3 x^{2} + 14 = 0 \end{aligned}$
::3x2+14=0

$\begin{aligned} (x - 6)^{2} = 5 \end{aligned}$
:x-6)2=5

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

Susan drops her camera in the river from a bridge that is 400 feet high. How long is it before she hears the splash?
::Susan把摄像机从400英尺高的桥上扔到河里

It takes a rock 5.3 seconds to splash in the water when it is dropped from the top of a cliff. How high is the cliff in meters?
::从悬崖的顶部掉下来时,要花5.3秒才能在水中喷落。悬崖的米数有多高?

Nisha drops a rock from the roof of a building 50 feet high. Ashaan drops a quarter from the top story window, 40 feet high, exactly half a second after Nisha drops the rock. Which hits the ground first?
::尼沙从一座50英尺高的建筑物的屋顶上落下一块石头。阿沙安从顶层故事窗口投下四分之一,40英尺高,正好在尼沙扔下岩石后半秒。谁先落地呢?

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

平方根应用

Square Root Applications ::平方根应用

Solving Quadratic Equations ::解析二次赤道

Solve Applications Using Quadratic Functions and Square Roots ::使用二次曲线函数和平方根解决应用程序

Real-World Application: Object Falling ::Real- World 应用程序: 对象坠落

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Square Root Applications
::平方根应用

Solving Quadratic Equations
::解析二次赤道

Solve Applications Using Quadratic Functions and Square Roots
::使用二次曲线函数和平方根解决应用程序

Real-World Application: Object Falling
::Real- World 应用程序: 对象坠落

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)