Course: 10.8 二次曲线公式

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二次曲线公式
Quadratic Formula
::二次曲线公式

The Quadratic Formula is probably the most used method for solving quadratic equations. For a quadratic equation in standard form , $\begin{aligned} a x^{2} + b x + c = 0 \end{aligned}$ , the quadratic formula looks like this:
::二次方程可能是用来解析二次方程的最常用的方法。对于标准形式的二次方程, x2+bx+c=0, 二次方程看起来是这样:

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

::xbb2 - 4ac2a

This formula is derived by solving a general quadratic equation using the method of that you learned in the previous section.
::此公式是使用上一节所学方法解决一般四方方程式的方法得出的。

We start with a general quadratic equation: $\begin{aligned} a x^{2} + b x + c = 0 \end{aligned}$
::我们从一个一般的二次方程开始:x2+bx+c=0

Subtract the constant term from both sides: $\begin{aligned} a x^{2} + b x = - c \end{aligned}$
::从两边减去常数:x2+bxc

Divide by the coefficient of the $\begin{aligned} x^{2} \end{aligned}$ term:
::除以 x2 术语的系数 :

$\begin{aligned} x^{2} + \frac{b}{a} x = - \frac{c}{a} \end{aligned}$
::x2+bax *ca +bax *ca +bax *ca +bax*ca +bax *ca +bax*ca +bax*ca +bax*ca +bax+bax+bax*ca

Rewrite:
::重写 :

$\begin{aligned} x^{2} + 2 (\frac{b}{2 a}) x = - \frac{c}{a} \end{aligned}$
::x2+2(b2a)xca

Add the constant $\begin{aligned} {(\frac{b}{2 a})}^{2} \end{aligned}$ to both sides:
::将常数(b2a) 2 添加到双方:

$\begin{aligned} x^{2} + 2 (\frac{b}{2 a}) x + {(\frac{b}{2 a})}^{2} = - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} \end{aligned}$
::x2+2(b2a)x+(b2a)2ca+b24a2

Factor the perfect square trinomial :
::将完美的平方三角体乘以:

$\begin{aligned} {(x + \frac{b}{2 a})}^{2} = - \frac{4 a c}{4 a^{2}} + \frac{b^{2}}{4 a^{2}} \end{aligned}$
:x+b2a)24ac4a2+b24a2

Simplify:
::简化 :

$\begin{aligned} {(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a^{2}} \end{aligned}$
:x+b2a)2=b2-4ac4a2

Take the square root of both sides:
::以双方的平方根:

$\begin{aligned} x + \frac{b}{2 a} = \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} and x + \frac{b}{2 a} = - \sqrt{\frac{b^{2} - 4 a c}{4 a^{2}}} \end{aligned}$
::x+b2a=b2-4ac4a2和 x+b2a_b2-4ac4a2

Simplify:
::简化 :

$\begin{aligned} x + \frac{b}{2 a} = \frac{\sqrt{b^{2} - 4 a c}}{2 a} and x + \frac{b}{2 a} = - \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$
::x+b2a=b2-4ac2a和 x+b2a_b2a_b2-4ac2a

$\begin{aligned} x = - \frac{b}{2 a} + \frac{\sqrt{b^{2} - 4 a c}}{2 a} and x = - \frac{b}{2 a} - \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$
::xb2a+b2-4ac2a和 xb2a-b2-4ac2a

$\begin{aligned} x = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and x = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$
::xb+b2 - 4ac2a和 xb - b2 - 4ac2a

This can be written more compactly as $\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ .
::这可以以 xbb2 - 4ac2a 更简明的方式写成。

You can see that the familiar formula comes directly from applying the method of completing the square. Applying the method of completing the square to can be tedious, so the quadratic formula is a more straightforward way of finding the solutions.
::您可以看到,熟悉的公式直接来自应用完成方形的方法。应用完成方形的方法可以变得乏味, 所以二次公式是找到解决方案的更直截了当的方法。

Solve Quadratic Equations Using the Quadratic Formula
::使用二次曲线公式解决二次赤道等量

To use the quadratic formula, just plug in the values of $\begin{aligned} a, b, \end{aligned}$ and $\begin{aligned} c \end{aligned}$ .
::要使用二次方程式, 只需插入 a、 b 和 c 的值。

1. Solve the following quadratic equations using the quadratic formula.
::1. 使用四方形公式解决下列四方形。

Start with the quadratic formula and plug in the values of $\begin{aligned} a, b \end{aligned}$ and $\begin{aligned} c \end{aligned}$ .
::以二次公式开始,插入a、b和c的值。

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

$\begin{aligned} Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = 2, b = 3, c = 1 & x & = \frac{- 3 \pm \sqrt{(3)^{2} - 4 (2) (1)}}{2 (2)} \\ Simplify: & x & = \frac{- 3 \pm \sqrt{9 - 8}}{4} = \frac{- 3 \pm \sqrt{1}}{4} \\ Separate the two options: & x & = \frac{- 3 + 1}{4} and x = \frac{- 3 - 1}{4} \\ Solve: & x & = \frac{- 2}{4} = - \frac{1}{2} and x = \frac{- 4}{4} = - 1 \end{aligned}$

::二次曲线公式: xbb2 - 4ac2a+plug, 数值a=2, b=3, c=1x33(3)2-4(2)(2)(2)简化:x39-84314

Answer: $\begin{aligned} x = - \frac{1}{2} \end{aligned}$ and $\begin{aligned} x = - 1 \end{aligned}$
::答复:x%12和x%1

b) $\begin{aligned} x^{2} - 6 x + 5 = 0 \end{aligned}$
::b) x2-6x+5=0

$\begin{aligned} Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = 1, b = - 6, c = 5 & x & = \frac{- (- 6) \pm \sqrt{(- 6)^{2} - 4 (1) (5)}}{2 (1)} \\ Simplify: & x & = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm \sqrt{16}}{2} \\ Separate the two options: & x & = \frac{6 + 4}{2} and x = \frac{6 - 4}{2} \\ Solve: & x & = \frac{10}{2} = 5 and x = \frac{2}{2} = 1 \end{aligned}$

::二次曲线公式:x=1, b6, c=5x(- 6) (-6) (-6)-2-4(1)(5)2(1) 简化:x=636-202=6162 将两个选项分开:x=6+42和x=6-42Solve:x=102=5和x=22=1

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

c) $\begin{aligned} - 4 x^{2} + x + 1 = 0 \end{aligned}$
:c) -4x2+x+1=0

$\begin{aligned} Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = - 4, b = 1, c = 1 & x & = \frac{- 1 \pm \sqrt{(1)^{2} - 4 (- 4) (1)}}{2 (- 4)} \\ Simplify: & x & = \frac{- 1 \pm \sqrt{1 + 16}}{- 8} = \frac{- 1 \pm \sqrt{17}}{- 8} \\ Separate the two options: & x & = \frac{- 1 + \sqrt{17}}{- 8} and x = \frac{- 1 - \sqrt{17}}{- 8} \\ Solve: & x & = - .39 and x = .64 \end{aligned}$

::二次曲线公式:xbb2-4ac2a+lug, 数值为a4, b=1, c=1x1, 11, 11, 2-4(4)(1) (2-4) 简化:x11+16-8, 117-8) 17-8。

Answer: $\begin{aligned} x = - .39 \end{aligned}$ and $\begin{aligned} x = .64 \end{aligned}$
::答复:x%39和x=.64

Often when we plug the values of the coefficients into the quadratic formula, we end up with a negative number inside the square root. Since the square root of a negative number does not give real answers, we say that the equation has no real solutions. In more advanced math classes, you’ll learn how to work with “complex” (or “imaginary”) solutions to quadratic equations.
::当我们将系数值插入二次公式时,我们往往会在平方根内出现负数。由于负数的平方根没有给出真正的答案,我们说方程式没有真正的解决方案。在更先进的数学课程中,你将学会如何用“复合”(或“想象”式)方法解决四方方程式。

2. Use the quadratic formula to solve the equation $\begin{aligned} x^{2} + 2 x + 7 = 0 \end{aligned}$ .
::2. 使用二次方程式解析公式x2+2x+7=0。

$\begin{aligned} Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = 1, b = 2, c = 7 & x & = \frac{- 2 \pm \sqrt{(2)^{2} - 4 (1) (7)}}{2 (1)} \\ Simplify: & x & = \frac{- 2 \pm \sqrt{4 - 28}}{2} = \frac{- 2 \pm \sqrt{- 24}}{2} \end{aligned}$

::a=1, b=2, c=7x22(2)-2-4(7)2(1)简化:x24-282}2222

Answer: There are no real solutions.
::答复:没有真正的解决办法。

To apply the quadratic formula, we must make sure that the equation is written in standard form. For some problems, that means we have to start by rewriting the equation.
::为了应用二次公式,我们必须确保方程式以标准形式写成。对于某些问题,这意味着我们必须从重写方程式开始。

Finding the Vertex of a Parabola with the Quadratic Formula
::使用 Quadratic 公式查找 Parabola 的顶点

Sometimes a formula gives you even more information than you were looking for. For example, the quadratic formula also gives us an easy way to locate the vertex of a parabola .
::有时一个公式会提供比您想要的更多的信息。例如,二次公式也给我们找到抛物线顶点的简单方法。

Remember that the quadratic formula tells us the roots or solutions of the equation $\begin{aligned} a x^{2} + b x + c = 0 \end{aligned}$ . Those roots are $\begin{aligned} x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}, \end{aligned}$ , and we can rewrite that as $\begin{aligned} x = \frac{- b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} . \end{aligned}$
::记住, 二次方程式告诉我们方程式 x2+bx+c=0 的根或方程式。这些根是 xbb2-4ac2a, 我们可以重写为 xb2a+bx+c=4ac2a 。

Recall that the roots are symmetric about the vertex . In the form above, we can see that the roots of a quadratic equation are symmetric around the $\begin{aligned} x - \end{aligned}$ coordinate $\begin{aligned} \frac{- b}{2 a} \end{aligned}$ , because they are $\begin{aligned} \frac{\sqrt{b^{2} - 4 a c}}{2 a} \end{aligned}$ units to the left and right (recall the $\begin{aligned} \pm \end{aligned}$ sign) from the vertical line $\begin{aligned} x = \frac{- b}{2 a} \end{aligned}$ .
::回顾根与顶端对称。在上表, 我们可以看到四方方方程的根在 x- 坐标- b2a 周围是对称的, 因为它们是从垂直线 x b2a 向左和向右的 b2 - 4ac2a 单位( 重回符号 ) 。

In the equation $\begin{aligned} x^{2} - 2 x - 3 = 0 \end{aligned}$ , the roots -1 and 3 are both 2 units from the vertical line $\begin{aligned} x = 1 \end{aligned}$ , as you can see in the graph below:
::在方程 x2-2x-3=0中,根-1和3是垂直线x=1的2个单位,如下图所示:

Guided Practice
::实践指南实践指南实践指南实践指南实践指南实践指南实践指南

Solve the following equations using the quadratic formula.
::使用二次方程式解决以下方程式。

Example 1
::例1

$\begin{aligned} x^{2} - 6 x = 10 \end{aligned}$

$\begin{aligned} Re-write the equation in standard form: & x^{2} - 6 x - 10 & = 0 \\ Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = 1, b = - 6, c = - 10 & x & = \frac{- (- 6) \pm \sqrt{(- 6)^{2} - 4 (1) (- 10)}}{2 (1)} \\ Simplify: & x & = \frac{6 \pm \sqrt{36 + 40}}{2} = \frac{6 \pm \sqrt{76}}{2} \\ Separate the two options: & x & = \frac{6 + \sqrt{76}}{2} and x = \frac{6 - \sqrt{76}}{2} \\ Solve: & x & = 7.36 and x = - 1.36 \end{aligned}$

::x2 - 6x= 10 以标准格式重写方程式: x2 - 6x - 10= 0 二次曲线式: xbb-2 - 4ac2a+ 数值a=1, b6, c10x ( - 6) ( - 6)2- 4(1)( - 10)2(1) 简化: x= 636+402= 6762 将两个选项分开: x=6+762和 x=6- 762Solve:x=7.36和 x1.36

Answer: $\begin{aligned} x = 7.36 \end{aligned}$ and $\begin{aligned} x = - 1.36 \end{aligned}$
::答复:x=7.36和x1.36

Example 2
::例2

$\begin{aligned} - 8 x^{2} = 5 x + 6 \end{aligned}$

$\begin{aligned} Re-write the equation in standard form: & 8 x^{2} + 5 x + 6 & = 0 \\ Quadratic formula: & x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ Plug in the values a = 8, b = 5, c = 6 & x & = \frac{- 5 \pm \sqrt{(5)^{2} - 4 (8) (6)}}{2 (8)} \\ Simplify: & x & = \frac{- 5 \pm \sqrt{25 - 192}}{16} = \frac{- 5 \pm \sqrt{- 167}}{16} \end{aligned}$

::-8x2=5x+6 以标准格式重写方程式: 8x2+5x+6=0 二次公式: xbb2-4ac2a+8, b=5, c=6x5(5)2-4(8)(6)(6)(8) 简化: x525-19216}516716

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

Review
::回顾

Solve the following quadratic equations using the quadratic formula.
::使用二次方程式解决以下二次方程。

$\begin{aligned} x^{2} + 4 x - 21 = 0 \end{aligned}$
::x2+4x-21=0

$\begin{aligned} x^{2} - 6 x = 12 \end{aligned}$
::x2 - 6x=12

$\begin{aligned} 3 x^{2} - \frac{1}{2} x = \frac{3}{8} \end{aligned}$
::3x2 - 12x=38

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

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

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

$\begin{aligned} 4 x^{2} = x \end{aligned}$
::4x2=xx

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

$\begin{aligned} 5 x^{2} - 2 x + 100 = 0 \end{aligned}$
::5x2-2x+100=0

$\begin{aligned} 100 x^{2} + 10 x + 70 = 0 \end{aligned}$
::100x2+10x+70=0

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。

10.8 二次曲线公式

Section outline

General

二次曲线公式

Quadratic Formula
::二次曲线公式

Solve Quadratic Equations Using the Quadratic Formula
::使用二次曲线公式解决二次赤道等量

Finding the Vertex of a Parabola with the Quadratic Formula
::使用 Quadratic 公式查找 Parabola 的顶点

Guided Practice
::实践指南实践指南实践指南实践指南实践指南实践指南实践指南

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)

Section outline

General

二次曲线公式

Quadratic Formula ::二次曲线公式

Solve Quadratic Equations Using the Quadratic Formula ::使用二次曲线公式解决二次赤道等量

Finding the Vertex of a Parabola with the Quadratic Formula ::使用 Quadratic 公式查找 Parabola 的顶点

Guided Practice ::实践指南 实践指南 实践指南 实践指南 实践指南 实践指南 实践指南

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Quadratic Formula
::二次曲线公式

Solve Quadratic Equations Using the Quadratic Formula
::使用二次曲线公式解决二次赤道等量

Finding the Vertex of a Parabola with the Quadratic Formula
::使用 Quadratic 公式查找 Parabola 的顶点

Guided Practice
::实践指南实践指南实践指南实践指南实践指南实践指南实践指南

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)