Solving Quadratic Equations

Solving Quadratic Equations

Let's examine the example from the previous page:

In that example, we had an equation of the form:

t2 - 10t - 600 = 0

which we factored into two terms as follows:

(t + 20) (t - 30)

The answer would be t = -20 and t = 30. Since -20 has no meaning,

the answer is simply 30.

What if we cannot factor the equation?

Suppose that another project has a cost function of the form:

This equation cannot be factored as in the first example. One way to solve this particular equation is by completing the square. We first move the 7 to the right hand side of the equal sign.

Then we proceed to add a number squared to both sides of the equation to complete the square as follow:

At this point, the number has to be guessed.

When we reduce the equation, we get

The equation yields two answers

If we cannot factor the equation, we can still solve it by the method of completing the square as shown in the above example.

A few simple facts that you should know

Did you know that there are other methods for solving a quadratic equation, such as factoring, completing the square, or using the quadratic formula?

How do I know which method to use?

?Use factoring when the equation is simple and the factors are obvious. Use completing the square when you cannot

factor the equation. When in doubt, use the Quadratic

Formula, shown on the next page, which works for any

quadratic equations.

What is actually meant by solving a quadratic equation?

?Solving a quadratic equation means finding the values of x where the graph cuts the x-axis.

Do you know how the graph for a quadratic equation look? The graph for a quadratic equation is a parabola.

?If the parabola cuts the x-axis at only one point it means that the quadratic equation has two solutions with the same value (the value of x where the parabola touches the

x-axis).

?If the parabola cuts the x-axis at two points it means that it has two solutions (the points where the parabola crosses the x-axis).

?If the parabola doesn't cut the x-axis it means that the quadratic equation doesn't have any real solution.

Try to find the solutions of x2 + 1 , x2 + 2x + 1 , x2 - 1 by using the Graphing workbench .

Quadratic Equations

Quadratic Equations Let's glance at an example: The setup cost for the project to develop a new tank design is 50 thousand dollars. The development cost is a function of time. In other words, the cost increases with time. From previous contracts, the cost function is estimated to be t2 - 10 t where t is the time measured in working days. Suppose that a cost analyst has 650 thousand dollars allocated for the project, and he/she wants to know how long before he/she needs to allocate more resources to complete the project? Let's consider a solution

The total cost for the tank is equal to the sum of the initial cost and the development cost. t2 - 10t + 50 If $650 thousand has been given, we want to know how much time before more resources need to be allocated? t2 - 10t + 50 = 650 If we move 650 to the left of the equal sign, this is what we have t2 - 10t - 600 = 0 To solve the equation above, we just factor the terms as follows: (t + 20) (t - 30) The equation yields two answers: t = -20 and t = 30. Since negative time has no physical meaning, t = 30 is the only valid solution. Thus, it takes 30 working days before the cost analyst needs to allocate more resources to fund the project.

10 Reaction Equations要点

10. REACTION EQUATIONS Clicking the Reaction Equations button in the main menu shows the Reaction Equations Window, see Fig. 1. With this calculation option you can calculate the heat capacity, enthalpy, entropy and Gibbs energy values of a single species as well as of specified reactions between pure substances. See the reference state definitions, valid notations and abbreviations for the description of the chemical formulae in Chapter 28. Databases, chapter 2. 10.1 One Chemical Substance Fig. 1. Reaction Equations Window of HSC Chemistry.

Fig. 2.Thermodynamic data of A12O3 (alumina) displayed by the Reaction Equation option of HSC Chemistry. By entering a single chemical formula into the Formula box you will get similar tables of thermochemical data as presented in many thermochemical data books. HSC will, however, provide the results faster and exactly at those temperatures which you really want. Please follow these steps: 1.Write a chemical formula into the box, Fig. 1. For example: Fe, Na2SO4, Al2O3, SO4(-a), H(+a) or SO2(g). See the valid notation and syntax of chemical formulae in Chapter 21.2. 2.Select the lower limit, upper limit and temperature step. 3.Select the Temperature and Energy Units, by clicking the corresponding buttons. 4.Select the Format of the results. Normal (Absolute scale): H(species), S(species) and C(species) This format is used for example in the famous I. Barin, O. Knacke, and O. Kubaschewski data compilation1. Delta (Formation functions):

Part1OrdinaryDifferentialEquations(常微分方程式微分变数只有一个)

Chapter 1 First-Order ODEs C C h h a a p p t t e e r r 22 S S e e c c o o n n d d --O O r r d d e e r r L L i i n n e e a a r r O O D D E E s s ((二二階階線線性性常常微微分分方方程程式式)) Chapter 3 Higher-Order Linear ODEs Chapter 5 Series Solutions of ODEs Chapter 6 Laplace Transforms ? Ordinary differential equations may be divided into two large classes, linear (線性) and nonlinear (非線性) ODEs. Where nonlinear ODEs are difficult to solve, linear ODEs are much simpler because there are standard methods for solving many of these equations. 22..11 H H o o m m o o g g e e n n e e o o u u s s L L i i n n e e a a r r O O D D E E s s o o f f S S e e c c o o n n d d O O r r d d e e r r ((二二階階線線性性齊齊性性微微分分方方程程式式)) ? A second-order ODE is called linear (線性的) if it can be written as ()()()y p x y q x y r x '''++=. (1) ? 線性：方程式的每一項都不得出現()y x 和其導數(y ', y '',…)的乘積或自乘 In case ()0r x =, the equation is called homogeneous (齊性的). In case ()0r x ≠, the equation is called nonhomogeneous (非齊性的). The functions ()p x and ()q x are called the coefficients of the ODEs. ? Theorem 1 Superposition Principle for the Homogeneous Linear ODE (適用於線性 齊性常微分方程式的疊加原理) If both 1()y x and 2()y x are solutions of the homogeneous linear ODE ()()0y p x y q x y '''++=, (2) then a linear combination (線性組合) of 1y and 2y , say 1122()()c y x c y x +, is also a solution of the differential equation. Proof –

LanguageProblem-Solving1

阅读词汇记忆 heroic /h?'r???k/ adj.英雄的；deliver /d?'l?v?/ v.递送；投递vehicle /'vi??kl/ n.车辆；交通工具burst /b??st/ n.爆发；爆炸emergency /i'm??d??nsi/n.紧急情况wrecked /rekt/ adj. 毁坏的passenger/'p?s?nd??/n.乘客colleague /'k?li?ɡ/n.同事 employer /?m'pl???/n.雇主customer /'k?st?m?/ n. 客户；顾客flame /fle?m/n.火焰；火苗steam /sti?m/n.蒸汽 disabled /d?s'e?bld/adj. 残废的remove/r?'mu?v/ v.移动 abandon /?'b?nd?n/ v.抛弃confirm /k?n'f??m/ v.确证；批准approach /?'pr??t?/ v.接近 still /st?l/adj. 安静的 calm /kɑ?m/ adj.平静的；沉着的explode /?k'spl??d/vi.爆炸forbidden /f?'b?dn/ adj.被禁止的；patience /'pe??ns/n.耐性；忍耐，skill /sk?l/ n.技能；技巧 effort /'ef?t/n. 努力 promise /'pr?m?s/ n.许诺；希望for certain肯定的 for consideration以供考虑 step forward向前进 back off后退 set out出发 in case万一 move on前进 slip away逃走 fall apart崩溃；土崩瓦解；破碎 get hold of把握；抓住；得到 take charge of接管，负责阅读词汇记忆 heroic /h?'r???k/ adj.英雄的；deliver /d?'l?v?/ v.递送；投递vehicle /'vi??kl/ n.车辆；交通工具burst /b??st/ n.爆发；爆炸emergency /i'm??d??nsi/n.紧急情况wrecked /rekt/ adj. 毁坏的passenger /'p?s?nd??/n.乘客colleague /'k?li?ɡ/n.同事 employer /?m'pl???/n.雇主customer /'k?st?m?/ n. 客户；顾客flame /fle?m/n.火焰；火苗steam /sti?m/n.蒸汽 disabled /d?s'e?bld/adj. 残废的remove /r?'mu?v/ v.移动 abandon /?'b?nd?n/ v.抛弃confirm /k?n'f??m/ v.确证；批准approach /?'pr??t?/ v.接近 still /st?l/adj. 安静的 calm /kɑ?m/ adj.平静的；沉着的explode /?k'spl??d/vi.爆炸forbidden /f?'b?dn/ adj.被禁止的；patience /'pe??ns/n.耐性；忍耐，skill /sk?l/ n.技能；技巧 effort /'ef?t/n. 努力 promise /'pr?m?s/ n.许诺；希望for certain肯定的 for consideration以供考虑 step forward向前进 back off后退 set out出发 in case万一 move on前进 slip away逃走 fall apart崩溃；土崩瓦解；破碎 get hold of把握；抓住；得到 take charge of接管，负责

Systems of Linear Equations 4

Systems of Linear Equations 4 In this chapter direct methods for solving systems of linear equations A x =b. 1111 1.n n nn a a b A b a a bn ???? ? ???==???? ???????? Will be presented.Here A is given n n ? matrix. And b is a given vector. We assume in addition that A and b are teal. Althougn this restriction is inessen. Tial in most of the methods. In contrast to the iterative methods(Chatpte 8), the direct methods discussed here produce the solution in finitely many steps. Assuming computations without roundoff errors. This problem is closely related to that of computing the inverse 1A - of the matrix A provided this inverse exists. For if 1A - is known,the solution x of Ax=b can be obtained by matrix vector multiplication, x =1A -b.Conversely. the i th column i a of 1A -=(1,n a a ) is the solution of the linear system i Ax e =,where i e =T (0,,0,1,0,,0) is the i th unit vector. A general introduction to numerical linear algebra is given in Golub and van Loan(1983) and Stewart(1973). ALGOL programs are found in Wilkinson and Reinsch(1971), FORTRAN programs in Dongarra, Bunch,Moler, and stewart(1979)(LINPACH), and in Andersen et al.(1992)(LAPACK). 4.1 Gaussian Elimination. The Triangular Decomposition of a Matrix In the method of Gaussian elimination for solving a system of linear equations (4.1.1) ,Ax b = Where A is an n n ? matrix and n b R ∈. The given system (4.1.1) is transformed in steps by appropriate rearrangements and linear combinations of equations into a system of the form R x c =, 1110 n nn r r R r ?? ? ?= ?????? Which has the same solution as (4.1.1). R is an upper triangular matrix. So that R x c = can easily be solved by “back substitution ”(so long as 0,1,,ii r i n ≠= ): 1n i ik k k i i ii c r x x r =+-= ∑ for ,1,,1i n n =- . In the first step of the algorithm an appropriate multiple of the first equation is subtracted from all of the other equations in such a way that the coefficients of 1x vanish in these equations: hence. 1x remains only in the first equation. This is possible only if 110a ≠. Of course, which can be achieved by rearranging the equations if necessary. As long as at least one 10i a =. Instead of

Solving Quadratic Equations

Solving Quadratic Equations Let's examine the example from the previous page: In that example, we had an equation of the form: t2 - 10t - 600 = 0 which we factored into two terms as follows: (t + 20) (t - 30) The answer would be t = -20 and t = 30. Since -20 has no meaning, the answer is simply 30. What if we cannot factor the equation? Suppose that another project has a cost function of the form: This equation cannot be factored as in the first example. One way to solve this particular equation is by completing the square. We first move the 7 to the right hand side of the equal sign. Then we proceed to add a number squared to both sides of the equation to complete the square as follow: At this point, the number has to be guessed. When we reduce the equation, we get

Solving Polynomial Equations

? Solving Polynomial Equations 3.3Introduction Linear and quadratic equations,dealt within Sections 3.1and 3.2,are members of a class of equations,called polynomial equations .These have the general form: a n x n +a n ?1x n ?1+...+a 2x 2+a 1x +a 0=0 in which x is a variable and a n ,a n ?1,...,a 2,a 1,a 0are given constants.Also n must be a positive integer and a n =0.Examples include x 3+7x 2+3x ?2=0,5x 4?7x 2=0and ?x 6+x 5?x 4=0.In this Section you will learn how to factorise some polynomial expressions and solve some polynomial equations. Prerequisites Before starting this Section you should ...?be able to solve linear and quadratic equations Learning Outcomes On completion you should be able to ...?recognise and solve some polynomial equations

Balancing Chemical Equations化学方程式的配平

Balancing Chemical Equations You may remember that the law of conservation of mass says that matter is neither created nor destroyed during a chemical reaction. This means that all chemical reactions must be balanced—the number of atoms, moles, and ultimately the total mass must be conserved during a chemical process. Here are the rules to follow when balancing equations: 1st. Determine the correct formulas for all the reactants and products in the reaction. 2nd. Begin balancing with the most complicated-looking group. A polyatomic ion that appears unchanged on both sides of the equation can be counted as a single unit. 3rd. Save the elemental (single elements) reactant and products for last, especially if it is hydrogen or oxygen. Keep your eye out for diatomic molecules such as oxygen, hydrogen, and the halogens. 4th. If you get stuck, double the most complicated-looking group and try again. 5th. Finally, make sure that all coefficients are in the lowest-possible ratio. 6th. Know when to quit! None of the reactions you will encounter will be that difficult. If the coefficients are getting wild, double-check what you’ve done since you may have a simple mistake. When balancing reactions, keep your hands off the subscripts! Use only coefficients to balance chemical equations. Now let’s try an example. When you solve it yourself, make sure to follow the steps! Example Write the balanced equation for the reaction between chlorine and sodium bromide, which produces bromine and sodium chloride. Explanation First write the chemical formulas—be on the lookout for the diatomic elements (such as Cl2): Next, find the reagent with the scariest subscripts. In this case, start with Cl2. You need a coefficient of 2 in front of NaCl, which then requires a coefficient of 2 in front of NaBr. The balanced equation becomes

Human problem solving

均是精品，欢迎下载学习！！！ 欢迎下载百度文库资源 资料均是本人搜集 Assignment # 2 RESEARCH PAPERS 1. Newell, A. and H.A Simon, Human Problem Solving, Englewood Cliffs, NJ., Prentice Hall, 1972 (summary paper) 2. Hayes-Roth,F., Rule based systems, Communications of the ACM, vol . 28, no 9, pp., 921-932, september 1985 3. Michie, D., Expert systems, The computer journal, vol 23, no 4, pp 369-376, 1980

H U M A N P R O B L E M S O L V I N G A l l e n N e w e l l&H e r b e r t A.S i m o n REFERENCES This project is an attempt to build the catalogue for Artificial Intelligence books. I would proudly say I used Internet as a least preferable source for getting information, although I know I could have found many more on internet. What I tried to do was to collect all books that exist in the CUNY (City University of NY) library database. So that it would make students and readers to know range of books that exist in CUNY and also provide them exact location to find those books. Other than that what I tried to do, was to get the reference books usually found at the end of each books as references. This was a much better source as it gave me the exact books related to the research topic. In this assignment I tried to avoid the duplication of records, if in case there exist one, I am sorry for that. Enjoy! Prof. Danny Kopec CIS 718 S S u u b b m m i i t t t t e e d d b b y y A A a a l l i i a a R R a a f f i i q q u u e e O O c c t t o o b b e e r r1188,,22000055

Exact_Equations

Exact Equations If we have a di?erential equation in the form M(x,y)+N(x,y)y =0 and we can?nd a continuously di?erentiable functionψ(it is called potential function)such that ?ψ?x (x,y)=M(x,y), ?ψ ?y (x,y)=n(x,y) and such thatψ(x,y)=c de?nes y=φ(x)implicitly as a di?erentiable func-tion of x(To check this,we often Implicit function Theorem.For details, read the textbook for153or254),the equation is said to be an exact di?er-ential equation. Then M(x,y)+N(x,y)y =?ψ ?x + ?ψ ?y dy dx = d dx ψ(x,φ(x)), and we obtain d dx ψ(x,φ(x))=0. Thus,the solution is given implicitly by ψ(x,y)=c where c is an undetermined constant.But usually,this just gives us an inte-gral curve instead of a function,so if there is intial condition,it is necessary to determine the c,and then the branch and the interval of de?nition if needed. 1

问题解决流程(Problem Solving Process)

Problem Solving Process 问题解决流程 Prepared by 拟制沈冰、胡小龙 Date 日期 2012-03-16 Reviewed by 评审人陶春植、尤莎、胡小龙、石 燕、王亚平、崔佳雯、李博 Date 日期 2012-05-24 Approved by 批准Date 日期 yyyy-mm-dd

Revision Record 修订记录

Table of Contents 目录 1Purpose 目的 (5) 2Scope 范围 (5) 3Abbreviations and Acronyms 术语和缩略语 (5) 4Policy 方针 (6) 5Process Description 过程描述 (6) 5.1Roles and Responsibilities 角色和职责 (6) 5.2Entrance Criteria 入口准则 (6) 5.3Inputs 输入 (7) 5.4Activities 活动 (7) 5.5Outputs 输出 (14) 5.6Exit Criteria 出口准则 (14) 6Resources and Tools 资源和工具 (14) 7Configuration Management and Assets 配置管理和资产 (15) 8Training 培训 (15) 9Process Measures 过程度量 (15) 10Tailoring Guidelines 裁剪指南 (15) 11Verification 验证 (15) 12Related Process 相关过程 (15) 13Reference Materials 参考文献 (15)