# Algebraic Expression Applied Math Practice Quiz-Middle Tennessee State University .

Week 1 Instruction 05/26/2020 Well, let us get started with the first week of our remote process. The following lecture is supposed to prepare students to Test 1. It briefly covers all sections of Chapter 1 of our textbook. You are recommended to read this message in detail, and to review Chapter 1, paying special attention to its Examples. Note that we are going to write down in red answers of the problems that will be discussed in this lecture. We will focus, first, on a very important subject of our course. That is the simplification of algebraic expressions. As an illustrative example, let us simplify the following expression ๐2 โ 9 ๐2 โ8๐+16 โ ๐โ4 (1) 5๐ + 15 Recalling the standard algebraic identities (๐ ยฑ ๐)2 = ๐2 ยฑ 2๐๐ + ๐ 2 and ๐2 โ ๐ 2 = (๐ + ๐)(๐ โ ๐), we are able to ultimately simplify (1) as ๐2 โ 9 ๐2 โ8๐+16 โ ๐โ4 5๐ + 15 = (๐+3)(๐โ3) (๐โ4)2 โ ๐โ4 5(๐ +3) = ๐โ3 5(๐โ4) (2) . Another important course issue that is related to Test 1 is the solution of algebraic equations. For illustration, let us solve the following algebraic equation 2(3๐ฅโ1) 4๐ฅโ1 = . (3) 5๐ฅ+1 3๐ฅ+2 To solve this equation means to determine value(s) of the variable ๐ฅ that make it true. To proceed with the solution, we transform (3) into 2(3๐ฅ โ 1)(3๐ฅ + 2) = (5๐ฅ + 1)(4๐ฅ โ 1) which is mathematically equivalent to 18๐ฅ 2 + 6๐ฅ โ 4 = 20๐ฅ 2 โ ๐ฅ โ 1 or 2๐ฅ 2 โ 7๐ฅ + 3 = 0 (4) The equality in (4) represents a quadratic equation. Therefore upon using the quadratic formula, we obtain two solutions for (4) as ๐ฅ1,2 = โ๐ยฑโ๐2 โ4๐๐ 2๐ = โ(โ7)ยฑโ(โ7)2 โ4โ2โ3 2โ2 = 7ยฑ5 4 . Thus, the two distinct solutions of the equation in (3) are ๐ฅ1 = 3 and ๐ฅ2 = 1/2 . Another important type of equations (that Test 1 deals with) is radicals containing equations, as the one below โ๐ฅ + 15 โ โ๐ฅ โ 1 = 4 . (5) just for example. To solve this equation, we rewrite it in the following equivalent form (6) โ๐ฅ + 15 = 4 + โ๐ฅ โ 1 . And then we square both sides of (6), keeping in mind that its left side represents the sum of two additive terms, whose square must be obtained with the aid of the first of the standard identities shown in (2). This results into ๐ฅ + 15 = 16 + 8โ๐ฅ โ 1 + (๐ฅ โ 1) or 8โ๐ฅ โ 1 = 0 , from which it follows that the single solution of the equation in (5) is ๐ฅ = 1 . Another important subject covered in Test 1 is the solution of inequalities. Before we go any further with inequalities, let us point out the formal difference between solutions of equations, on one hand, and inequalities, on the other hand. The point is that solution set of an equation is a limited set of numbers (๐ฅ1 , ๐ฅ2 , ๐ฅ3 , . . . , ๐ฅ๐ ), where n is a number of distinct solutions. Solution of an inequality represents, in contrast, an interval of the ๐ฅ โ axis . As an example, we consider the following inequality |2๐ฅ + 17| โค 9 In order to solve it, we recall that for the inequality |๐| โค ๐ to be true, the following two inequalities must be true ๐ โฅ โ๐ and ๐โค๐. With this in mind, for the inequality in (7) we have 2๐ฅ + 17 โฅ โ9 and 2๐ฅ + 17 โค 9 . From (8), it follows that ๐ฅ โฅ โ13 whereas from (9), we have ๐ฅ โค โ4 . Thus, summarizing, the solution for (7) is โ13 โค ๐ฅ โค โ4 Or, in other words, it is the closed interval [โ13, โ4] on the ๐ฅ โ ๐๐ฅ๐๐ . (7) (8) (9) The last covered in Test 1 subject is the rationalization of radical-containing expressions. To illustrate this subject, we solve a problem. That is, let us rationalize the denominator of the following ordinary fraction type expression 3 โ โ7 (10) 3 + โ7 To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator. This does not, of course, change the quantity of (10), just transforming it to (3 โ โ7)(3 โ โ7) (11) (3 + โ7)(3 โ โ7) In compliance with the second of the standard algebraic identities in (2), the denominator in (11) represents difference of squares 32 โ (โ7)2 = 9 โ 7 = 2 while the numerator in (11) is square of difference 32 โ 2 โ 3 โ โ7 + (โ7)2 = 16 โ 6โ7 Wherefore, the whole fraction in (10) reads as 8 โ 3โ7 . The following is Test 1 (the first of the five tests to be offered during this five week-long remote session). You have to turn it in by 5:00pm on May 29th. Please, send it to me as a scan or a picture of your work via the MTSU D2L email. And the following is just a friendly reminder. Make sure that you provide me with your stepby-step work on each-and-every offered problem in the test. I do not accept just answers. Test 1 1. Simplify the algebraic expression 3๐โ12 ๐2 +8๐+16 โ ๐+4 ๐2 โ16 . 2. Solve the given algebraic equation 2๐ฅโ3 ๐ฅ+2 = 3๐ฅโ2 5๐ฅโ8 . 3. Solve the radical containing equation โ๐ฅ + 4 โ โ๐ฅ โ 5 = 3 . 4. Find solution of the given inequality |5๐ฅ โ 3| โค 8 . 5. Rationalize the denominator 5+โ3 2โโ3 .