September 12, Math Analysis Ch 1 Review Solutions. #1. 8x + 10 = 4x 30 4x 4x 4x + 10 = x = x = 10.

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Avice O’Neal’
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1 #1. 8x + 10 = 4x 30 4x 4x 4x + 10 = x = x = 10 Sep 5 7:00 AM 1

2 #. 4 3(x + ) = 5x 7(4 x) 4 3x 6 = 5x 8 + 7x CLT 3x = 1x 8 +3x +3x = 15x = 15x x = 6 15 Sep 5 7:00 AM

3 #3. y y = y y = y = 5y y +14 = 5y + y y 14 = 3y + 1 = 3y 3 3 y = 4 Find LCD for ALL fractions. LCD = 30 CLT Aug 30 8:3 AM 3

4 # 4. 3( x) x 1 6 3x x 1 4 3x x 1 +3x +3x 4 5x x 5 5 x 1 To graph inequalities: TWO WAYS (1) Coordinate System () Number line For x 1 First think of x 1 as x = 1 Second, what does tell us, solid or dotted line? Third, shade in the solution. In this case all values of x I Remember when graphing the cases: y = # x = # Unfilled >, < Filled, Aug 8 1:34 PM 4

5 #5. y+9 < 5 TWO WAYS OF SOLVING ABSOLUTE VALUE INEQUALITIES: The familiar method of breaking the absolute value inequality into TWO inequalities. (1) y The first inequality stays the same but drops the absolute value bars a. y + 9 < 5 b. y + 9 > 5 y < 4 y > 14 The second inequality has the inequality symbol flipped and the number becomes its opposite. () y + 9 < 5 5 < y + 9 < 5 Solve for y < y < 4 In this second form, what I call the compounded form, the inequality sign moves to the front BUT DOES NOT FLIP. The number becomes its opposite again and is placed at the very front. Whatever you do to the middle part, you also have to do to ALL OTHER PARTS! To Graph on a coordinate system, remember the three steps from before! Don't remember? Look back! y < 4 y > 14 Aug 30 8:33 AM 5

6 # 6. 3 x x x 3 4 x 1 1 x 4 (1) Coordinate System Proper compound Inequalities are written with the inequality pointing to the left. () Number line Aug 30 8:39 AM 6

7 # 7. Standard form of a complex number a + bi a. 9 4i Real term: 9 Imaginary: 4i conjugate: opposite of imaginary term 9 + 4i b. 5i Looking at standard form a + bi Real term: 0 Imaginary: 5i conjugate: opposite of imaginary term 5i c. 10 Real term: 10 Imaginary: 0 conjugate: 10 Aug 30 8:4 AM 7

8 #8a. +4i #8b. ( 3 + 5i) (4 8i) 3 + 5i 4 + 8i i CLT #8c. (1 i)(3 + 4i) FOIL 3 + 4i 6i 8i Note : i = -1 3 i 8( 1) 3 i i Aug 30 8:46 AM 8

9 # 9. x 7 = x = 7 When we have a quadratic equation with bx term missing, then we should convert the remaining terms into ax = ±c x all by itself x = 7 x = ± 7 Aug 30 8:51 AM 9

10 # 10. 5x + 0 = x = x = 4 x = -4 x = i 4 x = ±i ax = ±c Aug 30 8:54 AM 10

11 # 11. x = 4x ax + bx + c = 0 x 4x = 0 When you are missing "c" from ax + bx + c = 0 Just keep it in this form ax + bx = 0 I can factor, but what? What do they have in common? x(x ) = 0 If you aren't sure about your factoring, CHECK IT by distributing. Check. x(x ) = 0 x 4x So what is our solution? x(x ) = 0 Remember this? Ex. x 5x + 6 = 0 (x )(x 3) = 0 We could easily factor that and just say the following, right? x = ; x = 3 But why can we do that? (x )(x 3) = 0 (x ) = 0 => x = (x 3) = 0 => x = 3 Returning to our problem, what is our solution? x(x ) = 0 x(x ) = 0 x = 0 x = 0 (x ) = 0 x = Aug 30 8:56 AM 11

12 # 1. x = 7x - 3 x - 7x + 3 = 0 Solution if you are good at factoring (x - 1)(x - 3) = 0 x = 1/, x = 3 If you are not good at factoring, you will have to use the quadratic formula! x = b ± b 4ac a ALWAYS PUT YOUR a, b, c VALUES IN PARENTHESES! It helps you avoid making silly errors! x = ( 7) ± ( 7) 4()(3) () x = 7 ± x = 7 ± 5 4 x = 7 ± 5 4 x = = 1 4 = 3 x = = 4 = 1 Aug 30 9:01 AM 1

13 # 13. m + m + 1 = 0 If the m's are bothering you, change the variables to something you are more used to, like x's x + x + 1 = 0 x = b ± b 4ac a x = (1) ± (1) 4(1)(1) (1) x = 1 ± 1 4 x = 1 ± 3 x = 1 ± i 3 Aug 30 9:05 AM 13

14 # 14. y = 3 (y+1) y = 3y + 3 If you have a x you want to think quadratic function ax + bx + c = o y - 3y - 3 = 0 Why work with fractions though? There are TWO EASIER WAYS OF SOLVING THIS: (1) Find a common denominator y = 3y + 3 y = 3y + 3 => y - 3y - 3 = 0 () Finding LCD and multiply (You have to remember this trick!) LCD = y = 3y + 3 ( ) y = 3 y + 3 y = 3y + 3 y - 3y - 3 = 0 Now let's find the solution: x = b ± b 4ac a x = ( 3) ± ( 3) 4()( 3) () x = 3 ± 33 4 x = 3 ± or 3 4 ± 33 4 Aug 8 1:44 PM 14

15 #15. 5x 6 x = 0 5x 6 = x ( 5x 6) = (x) 5x 6 = x 5x + 6 5x = x 5x = (x )(x 3) x = ; x = 3 Note: ax + bx + c = 0 try to keep ax positive Aug 30 9:11 AM 15

16 # x The question is when is the expression a real number? Conversely, when do we have imaginary numbers? x 0 ( x) (0) x x x 5 Aug 30 9:13 AM 16

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19 #18. u u = u 6 3u 3 What can we do? Find a common denominator, but what is it? If we had the fractions = 6 3 => LCD = 6 Then it would be easy. So 6 is part of the common denominator, but what about the u's? 3 1 = Factor u u = u 6 3u 3 u u = (u 1) 6 3(u 1) We see two of the three denominator also have (u 1) as part of the denominator. Therefore our LCD should be 6(u 1). Note 6u 6 could also work just as well. 3(u 3) 1 (1 u) 3*(u 1) = (u 1) (u 1) 6 *3(u 1) Remember, whatever you do to the bottom (denominator) you also have to do to the top (numerator). 3(u 3) 1 (1 u) 3*(u 1) = (u 1) (u 1) 6 *3(u 1) 3u 9 ( u) 6(u 1) = (u 1) 6(u 1) 6(u 1) Once you have a common denominator, you can cancel/cross it out and just work with the numerator. 3u 9 = u 1 + u 3u 9 = 3u 3 3u 3u 9 = 3 No Solution Sep 4 8:16 AM 19

20 #19. x + 3 x LCD = 4 Sep 4 7:11 AM 0

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23 #a. d(a, B) = 15 b. d(a, C) = 13 c. d(b, C) = 8 First, "d" means distance Second, distance is always positive There are at least two ways of solving this problem. Either are fine. Our goal should be familiarizing ourselves with the notation d(, ). 1. Since distance is always positive a formula is given written as an absolute value. Either are fine as both will produce the same solution d = b a = a b. The other, simpler "old school", method involves just drawing a number line, plotting the locations of the points, and counting the tick marks Sep 4 8:41 AM 3

24 #3a. (3 + i) (3 + i) i + 3i + i 6 i i + i 3 i 9 + 6i + i 3 i 6 + 4i + i Note : i = i + ( 1) 5 + 4i Be very careful when foiling. Common mistake: (3 + x) => when foiled 9 + i (3 + x) => (3 + x)(3 + x) = 9 + 3x + 3x + x Sep 4 8:46 AM 4

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26 #4a. ( 4) (3 9) ( i 4) (3 i 9) ( i) (3 3i) i 3 + 3i 1 + i #4b i i 4 i 3 + i ( i) (3 i) (3 + i) (3 i) ( i) (3 i) (3 + i) (3 i) Again, when dividing complex numbers, we have to multiply the denominator by the conjugate of the denominator! What is the conjugate? Real part stays the same Imaginary becomes its opposite 6 4i 3i + i 9 + 6i 6i 4i 6 7i + i 9 4i Note : i = i + ( 1) 9 4( 1) 6 7i i 13 Sep 4 8:55 AM 6

27 #4c i 5 i i i Until now, we have used the conjugate of the complex number denominator to help us divide complex numbers into each other. There is an exception to the rule: When there is no REAL term of the complex number, just the IMAGINARY term, we multiply by Let's try it out (4 + 5i) (i) STOP! Are we done yet? NO! We shouldn't have i's in the denominator. How do we get rid of the i? One more note: Our answers should always be in form of * i i i i 4i + 5i i 4i + 5( 1) ( 1) 4i 5 or i + 5 a + bi Note : i = -1 with the real part first and the bi, or imaginary part, second True solution can either be 5 4i or 5 i Sep 4 9:05 AM 7

28 #5. ( y ) = 0 (y ) = 0 y = 0 y = y = or y = = 4 * 5 = 4 * 5 = 5 Sep 4 9:11 AM 8

29 # u = u u ( u = ) u u + 3 = u u u + 3 = 0 u = b ± b 4ac a LCD = u Perform the Trick: Find LCD and multiply Just in case you don't see why u is the common denominator and not u = 4 With numbers its easy. You would say 16 is the LCD, but isn't 16 really just 4? = 4 u = ( ) ± ( ) 4(1)(3) (1) u = ± 4 1 u = ± 8 u = ± i 8 u = ± i 8 = 4 * = 4 * = u = 1 ± i Sep 4 9:16 AM 9

30 #7. x x x 6 x 3 = 3 When looking at this problem, you might want to think finding a common denominator is a good way to start But... before we multiply the two denominators together, we also might be interested in factoring x x 6 Factoring we have (x 3)(x + ) x (x 3)(x ) x 3 = 3 Notice, we do actually have a factor in common. LCD = (x 3)(x + ) Using our trick of finding the LCD and then multiplying, we have (x 3)(x + ) ( ) x (x 3)(x + ) x 3 = 3 x (x + ) = 3(x 3)(x + ) x x 4 = 3(x x 6) x 4 = 3x 3x 18 +x +4 +x +4 0 = 3x x 14 We have a Quadratic Function, Looks Like we will need the Quadratic Formula x = b ± b 4ac a x = ( ) ± ( ) 4(3)( 14) (3) x = ± x = ± 17 6 x = ± 43 6 x = 1 ± or x = ± 3 17 = 4 * 43 = 4 * 43 = 43 Sep 4 1:51 PM 30

31 1 x (x 3 ) = x = x Sep 4 1:01 PM 31

32 #9. m 4 + 5m 36 = 0 If the m's bother you, we can also change this to x 4 + 5x 36 = 0 Before we solve this, let's take a quick look at a standard quadratic function :1 ax + bx 1 + c = 0 x 4 + 5x 36 = 0 Notice in a quadratic function the powers share a :1 ratio. The same as in the original problem. When this happens, we have a special quadratic function case. This means, as long as there is a :1 ratio between the highest power and the next highest power in a function, we can think of it as a quadratic function. :1 Therefore we have one of two options: (1) We "pretend" for a split second that we really do have a quadratic function and either factor or use the quadratic formula to find our solutions. THERE'S ONLY ONE CATCH!!!! Our solutions are not x =..., but rather x =... Factoring x 4 + 5x 36 = 0 (x + 9)(x 4) = 0 Our solutions are: x + 9 = 0 x = 9 x = ± 9 x = ±i 9 x = ±3i x 4 = 0 x = 4 x = ± 4 x = ± () We could use our newly learned u substitution. Let u = x Our problem can then be rewritten as u + 5u 36 = 0 Either factoring or using quadratic formula we would have (u + 9)(u 4) = 0 Our solutions according to u are: u + 9 = 0 u 4 = 0 u = 9 u = 4 If we choose to use the u substitution method, we just have to remember to sub back in our original parts u = 9 u = 4 x = 9 x = 4 x = ± 9 x = ±i 9 x = ± 4 x = ± x = ±3i Sep 6 8:36 AM 3

33 when put together, the other needed factor FINAL SOLUTION: y = 9 4 y = 3 Sep 6 8:43 AM 33

34 #31..15x 3.73(x 0.930) = 6.11x.15x 3.73x = 6.11x 1.58x = 6.11x +1.58x +1.58x = 7.69x x = Sep 6 9:05 AM 34

35 # x Solve for x x x x.4 Note about inequality notation: The inequality signs, WHEN WRITTEN AS A COMPOUND INEQUALITY LIKE ABOVE, should always point left.4 x 1.1 Sep 6 9:10 AM 35

36 This is a good time to talk about INTERVAL NOTATION Until now, we have written intervals as inequalities like in the last problem.4 x 1.1 We said as a compound inequality, the inequality signs should point left. And we are used to graphing these types of inequalities on a number line using and and as or lines on a coordinate system. As we are getting closer to the college level, we also have to begin using "college" notation. This isn't something new or completely different. We are just going to modify what we already know. Let's take a look at the last solution, we had:.4 x 1.1 The way I think about the new notation is simple. Look at the symbol, it has a bar underneath the < sign. For me, the bar Which kind [ of looks like Brackets, [ ], tell us the interval we are working with includes the point. So our solution,.4 x 1.1, can be written as: [.4, 1.1] Sep 6 9:10 AM 36

37 # x > 5.48 We can rewrite the absolute value inequality as a compound inequality. the number becomes its opposite x > 5.48 the inequality sign stays the same 5.48 > x > 5.48 To solve for x, we can start by subtracting Whatever we do to one part, we have to do for all parts > x > x < < < x < 1.17 Notice the "solution" may be pointing to the left as we want it to, but the statement tells us that x is greater than 4.0 but less than It would be better if we wrote our solution as: x > 4.0 or x < 1.17 With the x's first Sep 6 9:16 AM 37

38 # t t 1 Now we can perform our trick: LCD = ( t 1 ) t t t 4 4 Find the LCD and multiply it into ALL PARTS t True solution when written in compound form. 4 4 Sep 6 9:0 AM 38

39 # x x = 0 x = b ± b 4ac a x = ( 4.57) ± ( 4.57) 4(6.09)( 8.86) (6.09) x = 4.57 ± x = 4.57 ± x = 4.57 ± x = 1.64 x = Your solutions may vary based on how you rounded. That's ok for now. As long as your solutions are close to those numbers, its fine. Sep 1 9:40 AM 39

Chapter 1 Review of Equations and Inequalities

Chapter 1 Review of Equations and Inequalities Part I Review of Basic Equations Recall that an equation is an expression with an equal sign in the middle. Also recall that, if a question asks you to solve