University of Phoenix MTH 209 Algebra II Week 4 The FUN FUN FUN continues! A jump to Radicals! Section 9.1 What is a radical? If you have 22=4 Then 2 is the root of four (or square root) If you have 23=8 Then 2 is the cube root of eight. Definitions nth Roots If a=bn for a positive integer n, then : b is the nth root of a Specifically if a = b2 then b is the square root And if a=b3 then b is the cube root More definitions If n is even then you have even roots.
The positive even root of a positive even number is called the principle root For example, the principle square root of 9 is 3 The principle fourth root of 16 is 2 If n is odd, then you find odd roots. Because 25 = 32, 2 is the fifth root of 32. Enterthe The Radical Symbol We use the So we can define : symbol for a root n a where if n is positive and even and a is positive, then that denotes the principle nth root of a
If n is a positive odd integer, then it denotes the nth root of a If n is any positive integer then n 0 = 0 The radical radicand We READ n a as the nth root of a The n is the index of the radical The a is the radicand Example 1 pg 542 Find the following roots 2 = 25, the answer is 5 a) because 5 25
b) c) 3 6 3=-27 the answer is 3 because (-3) 27 6 =64 the answer is 2 because 2 64 d) because 4 ** Ex. 7-22**
=2, 4= -2 4 A look into horror! Whats up with these? 9 4 81 6 64
What two numbers (or 4 or 6) satisfy these? They are imaginary they are NOT real numbers and will be NOT be dealt with in section 9.6 it was removed in this version of the class. Roots and Variables Definition: Perfect Squares x2 ,x4 ,x6 ,x8 ,x10, x12 , EASY to deal with 2 x x 4 x x 2
6 x x 3 Cube roots and Variables Definition: Perfect Cubes x3 ,x6 ,x9 ,x12 ,x15, x18 , EASY to deal with with CUBE roots 3 3 x x 3 6 x x
2 3 9 x x 3 Example 2 Roots of exponents page 544 a) x b) 3
c) 5 t 22 18 s x t 30 11 6
s ** Ex. 23-34** 6 The totally RADICAL product rule for radicals What if you have 2 3 and you want to square it? 2 2 2 ( 2 3 ) ( 2 ) ( 3 ) 2 3 6
The Product Rule for Radicals The nth root of a product is equal to the product of the nth roots. Which looks like: n n n ab a b provided all of these roots are real numbers. Example 3 Using said product rule page 545 a) 4 y 4 y 2 y 2 y b)
8 8 4 3 y 3 y 3 y y 4 By a convention of old people, we normally put the radical on the right 3 Ex 3c c) 3
2 3 3 2 3 125w 125 w 5 w ** Ex. 35-46** 2 Example 4 pg 545 a) b) c) 12 4 3 4 3 2 3
3 4 54 3 27 2 3 27 3 2 33 2 80 4 16 5 4 16 4 5 24 5 5 d) 64 5 32 2 5 32 5 2 25 2 ** Ex 47-60** Example 5 pg 546 a) 20 x 3 4 x 2 5 x 4 x 2 5 x 2 x 5 x b) 3
c) 4 40a 8 3 8a 6 5a 2 3 8a 6 3 5a 2 2a 2 3 5a 2 4 11 4 4 11 4 4 4 11
48a b 48a b 48a b 2ab 5 7 5 5 2 5 5 5 2 5
d) w w w w w w w ** Ex 61-74** 24 2 3b 3 What about those Quotients? The nth root of a quotient is equal to the quotient of the nth roots. In symbols it looks like: n n a
a n b b Remember, b cant equal zero! And all of these roots must be real numbers. Example 6 pg 547 Simplify radically a) 25 25 5 9 9 3 c)
3 3 3 b b b 3 125 5 125 ** Ex. 75-86 ** b) d)
15 15 3 3 3 21 3 3 21 7 x x x
2 6 6 3 y y y Example 7 pg 547 Simplify radically with prod. & quot. rule a) b) 50 25 2 5 2 49 7 49
c) 4 a 5 4 a 4 4 a a 4 a 2 8 4 b b b8 ** Ex. 75-86 ** 5 3 3
5 3 2 3 x x x x x 3 8 2 8 2 Example 8 page 548 Domani
a) x 5 You just dont want the negative _ number So x 5 0...x 5 b) 3 c) 4 x 7
So all xs are ok! 2 x 6 This time all x 3 are allowed Section 9.1 Riding the Radical Definitions Q1-Q6 Find the root in numbers Q7-22 Find the root in variables Q23-Q34 Use the product rule to simplify Q35-Q74 Quotient rule Q75-98 The domain Q99-106 Word problems Q107-117
9.2 Rati0nal Exp0nents This is the rest of the story! Remember how we looked at the spectrum of powers? 23=8 22=4 21=2 20=1 2-1=1/2 2-2=1/4 Defining it If n is any positive integer then a Provided that 1/ n n a
n a is a real number 9.2 Ex. 1 pg 553 More-on Quadratic Equations a) 3 b) 4 c) d) 1/ 3
35 35 1/ 4 xy ( xy ) 1/ 2 5 1/ 3 3 5 a a **Ex. 7-14**
Example 2 pg 55 Finding the roots **Ex. 15-22** a) 41/ 2 4 2 1/ 3 b) ( 8) 1/ 4 8 2 4 c) 81 81 3 1/ 2 d) ( 9) 1/ 2 e) 9
3 9not _ real 9 3 The exponent can be anything! The numerator is the power The denominator is root 2 3 Another definition m/n a or a
m/n 1/ n m m 1/ n (a ) (a ) n m n ( a ) a m And negatives? Just upside down.
a m/n 1 a m/n or a m/n 1 n m ( a) Example 3 pg 554
Changing radicals to exponents a) 3 2 2/3 a x 1 1 3/ 4 b) 3 / 4 m 3 4 m m ** Ex. 23-26**
Example 4 page 554 Exponents going to radicals a) b) 5 2/3 3 2 5 1 2/5 a 2
5 a **Ex. 27-30** Cookbook 1 a m/n Reciprocal Power Root Cookbook 2 1. Find the nth root of a 2. Raise your result to the nth power 3. Find the reciprocal
Example 5 page 555 Rational Expressions a) b) c) d) 27 2 / 3 (271/ 3 ) 2 32 9 1 1 1 3/ 2 4 1/ 2 3 3 (4 ) 2 8 1 1 1 3/ 4 81
1/ 4 3 3 (81 ) 3 27 1 1 1 1 5/3 ( 8) 1/ 3 5 5 (( 8) ) ( 2) 32 32
**Ex. 31-42** Everything at a glance (remember these from 2x before?) Example 6 pg 556 Using product and quotient rules a) b) 1/ 6 27 1/ 2 27 3/ 4 1 / 6 1 / 2
27 27 2/3 9 5 3 / 4 1/ 4 2/ 4 1/ 2 5 5 5 5
1/ 4 5 **Ex 43-50** Example 7 pg 557 Power to the Exponents! **Ex.51-60** a) b) 1/ 2 3 12 10 1 / 2 (3 ) 6
c) 1/ 2 2 9 3 1/ 2 36 2 3 (3 12) 1/ 2 6
5 3 6 1/ 3 2 9 3 1/ 3 2 3 27 3 2
3 2 4 Square roots have 2 answers! 2 1/ 2 (x ) x ( same _ as ) 2 x x Ex 8 pg 558 So you use absolute values with roots a) b)
8 4 1/ 4 (x y ) 9 x 8 1/ 3 **Ex. 61-70** 2
2 x y _ or _ x y 3 x 2 Ex 9 pg 558 Mixed Bag **Ex. 71-82** a) x 2 / 3 y 4 / 3 x 6 / 3 x 2 1/ 2 a 1/ 2 1/ 4 1/ 4 a a b) a1/ 4
1/ 2 c) 3 1/ 2 1/ 2 3 1/ 2 ( x y ) ( x ) ( y ) 2 x 1/ 3 d) y 1/ 2 1/ 3
y 2 x 1/ 2 1/ 4 x y y1 / 6 x 3/ 2 x1/ 4
3/ 2 y Section 9.2 Radical Ideas Definitions Q1-Q6 Rational Exponents Q7-42 Using the rules of Exponents Q43-Q60 Simplifying things with letters Q61-Q82 Mixed Bag Q83-126 Word problems Q127-136 Section 9.3 Now adding, subtracting and multiplying You treat a radical just like you did a variable.
You could add xs together (2x+3x = 5x) And ys together (4y+10y=14y) So you can add 5 2 5 3 5 and 10 2 2 2 12 2 Ex 1 page 563 Add and subtract a) 3 5 4 5 7 5 b) 4 w 6 w 5 w c) 3 5 4 3 6 5 3 3 7 5
d) 4 3 4 3 3 3 3 3 3 6 x 2 x 6 x x 4 6 x 3 x ** Ex. 5-16**
Ex 2 pg 563 Simplifying then combining 8 18 4 2 9 2 2 2 3 2 5 2 a) note 8 18 26!!! 3 2x b) 2 3 2 2
4 x 5 18 x x 2 x 2 x 5 9 x 2 x x 2 x 2 x 15 x 2 x 16 x 2 x 2 x c) 3 16 x 4 y 3 3 54 x 4 y 3 3 16 x 4 y 3 3 2 x 2 xy 3 2 x 3xy 3 2 x xy 3 2 x **Ex. 17-32** 3 54 x 4 y 3 3 2 x
Ex 3 Multiplying radicals with the same index pg 564 **33-46** a) 5 6 4 3 5 4 6 3 20 18 20 3 2 60 2 b) 3a 2 6a 18a 3 9a 2 2a 3a 2a c) d) 3 4 3 4 3 16 3 8 3 2 23 2 4 x 3 4 x 2 4 x 5 4 x 4 4 x x 4 x 4 2 8
16 2 16 Ex 4 Multiplying radicals pg 565 ** Ex. 47-60** a) b) c) 3 2 (4 2 3 a (3 a 2 3 ) 3 2 4 2 3 2 3 12 2 3 6 24 3 6 a 2 ) 3 a 2 3
3 a 3 3 a 2 a 3 5 3 3 2 5 FOIL 2 3 3 3 2 3 2 5 5 3 3 5 2 5 18 4 15 3 15 10 8 15 d) 3
2 x 9 32 2 3 x 9 ( x 9 ) 2 9 6 x 9 x 9 x 6 x 9 One of those special products reminder Weve looked a lot at: (a+b)(a-b) = a2-b2 So if you multiply two things like this with radicals, just square the first and last and subtract them! Example 6 Multiplying Conjugates pg566 **Ex. 61-70** 2 2
a) (2 3 5 )(2 3 5 ) 2 (3 5 ) 4 45 41 b) ( 3 2 )( 3 2 ) 3 2 1 c) ( 2 x y )( 2 x y ) 2 x y Example 7 Why not even mix the indiceswhy the heck not? Remember aman=am+n a) 3 2 4 2 21/ 3 21/ 4 27 /12 12 27 12 128 b) 3 2 3 21/ 3 31/ 2 2 2 / 6 33 / 6 6 2 2 6 33 6 2 2 33 6 108 Section 9.3 Adding and stuff Definitions Q1-Q4 Addin and Subtractin Q5-32
Multiplying Q33-Q60 Conjugates Q61-Q70 Multiplying different indices Q71-78 Everything Q79-110 Word problems Q127-136 9.4 Quotients and Denominator Problems Remember: 2 2 2 3 3 3 2 2 2 2 Example 1 pg 570
**Ex 1-8** Fix the denominator no radicals! a) b) 3 3 5 15 5 5 5 5 3 3 3 4 33 4 33 4 3 3 3 3 2 2 2 4 8
Step by Step help A radical expression of index n is in simplified form if it has: 1. No perfect nth power as factors of the radicand 2. No fractions inside the radical, and 3. No radicals in the denominator Example 2 Simplifying Radicals a) 10 10 6 60 4 15 2 15 15
6 6 6 3 6 6 6 b) 5 3 5 3 5 3 3 3 15 3 3 3 3 3 3 9 3 9 9 3 27
3 **Ex. 9-18** Example 3- Doing the same with letters. **Ex. 19-28** a) a a b ab b b b b b) y x xy x xy x3
x3 x2 x x x x x 2 2 2 3 5 y y y y y y y y y y y5 y4
c) x 3 x 3 x 3 y 2 3 xy 2 3 xy 2 2 3 3 3 3 y y y y 3 y y 3
Dividing Radicals n n a n a a b n b b n Example 4 Dividing w/same index a) 10 10 10 5 2
5 5 b) 3 2 2 3 3 2 3 2 3 3 6 6 2 2 3 2 3 3 2 3 2 2 10 x 10 x c) 3 10 x 2 3 5 x 3 3 2x
3 5x 5x 3 **Ex. 29-36** Ex 5 Or simplify BEFORE you divide a) b) 4 3 2 3 3 2x 6x 12 72 x
6x 36 2 x 6 2 x 3 2 x 2 x 4 4 16 4 a 4 2 16a a 4 a 4 4 a 4 a 4 a 4 5 ** Ex 37-44**
16 Ex 6 Simplifying radical expressions a) 4 12 4 2 3 2(2 3 ) 2 3 4 4 2 2 2 6 20 6 2 5 2(3 5 ) b)
3 2 2 2 **Ex. 45-48 5 Ex 7 Rationalizing the denominator using conjugates a) 2 3 4 b) (2 3 ) (4 3 (4 3 ) (4
3 ) 8 6 3 3 11 6 3 13 13 3) 5 5 ( 6 2) 30 10 4 6 2 6 2 ( 6 2) ** Ex. 49-58** All the tricks againnow things
to powers Ex. 8 pg 575 a) b) c) d) 3 3 3 (5 2 ) 5 ( 2 ) 125 8 125 2 2 250 2 3 4 (2 x ) 2 3 3 4
3 4 12 ( x ) 16 x 16 x 3 3 3 3 3 6 (3w 2 w ) 3 w ( 2 w ) 27 w (2 w) 54w 4 3
3 3 4 3 (2t 3t ) 2 t ( 3t ) 8t **Ex. 59-70** 34 27t 3 4 Section 9.4 Dividing Radical Stuff
Rationalize the Denominator Q1-8 Simplifying Radicals Q9-28 Dividing Radicals Q29-Q48 Using Conjugates Q49-Q58 Powers of Powers Q59-70 Everything Q71-Q108 Word problems Q109-110 Quantum Leap again to Section 9.5 this time Now we do the SAME thing but we are solving equations and working with word problems with the RADICALS making a return. The odd root Property
(arent they ALL odd?) Remember (-2)3 = -8 and 23 = 8 So the solution of x3= 8 is 2 and the solution of x3 = -8 is 2 Because there is only one real odd root of each real number. (Flash card time) The Odd-Root Property If n is an odd positive integer n x k x = k is equivalent to n for any real number k Try it on for size Example 1 page 579
a) 3 x 27, x 3 27, x 3 b) x 5 32 0, x 5 32, x 5 32, x 2 c) ( x 2) 2 24, x 2 3 24, x 2 2 3 3 **Ex. 5-12** The Even-Root Property Oooh spooky.
If you have x2=4 the answer is 2. Right? Bzzzt! We know (2)2=4. Great! BUT (-2)2=4 also. So the solution is x={2,-2} Another way to write this is x=2 So in x4=16, x=4 And in x6=5 is x= 6 5 Leaving the book for a page Remember Johns fractional exponent trick? The solution of x2 is x Lets solve it: we have Thats why 2 x 1
2 x 1 2 1/ 2 (x ) x 1 2 2 (1) 1 x 1 1/ 2 2
( x ) (1) x 1 And the last one from the previous slide was And in x6=5 is x= 6 5 x 6 5 x 5 ( x 6 )1/ 6 (5)1/ 6 6 x
1 6 6 6 (5) x 5 1 6 1 6 6 6
x 5 x 6 5 The Even root problem In short, if the number inside the even root is positive, you have a + and answer. If its zero, the answer it just zero. If its negative, you have no solution (in this universe it is an imaginary number). Technically the Even Root Property looks like If k > 0, then xn = k is equivalent to x n k If k = = then xn = 0 is equivalent to x=0 If k<0, then xn =k has no real solution Example 2 page 580 using the EVEN root property a)
2 x 10, x 10 so _ the _ solution _ set _ is { 10, 10}or{ 10} b) w8=0 so w=0 c) x4= -4 has no real solution (to the physicists = 2i, to engineers it is 2j ) **Ex. 13-18** Example 3 page 581 Using this same property a) (x-3)2=4 x-3=2 or x-3=-2 x=5 or x=1 The solution set is {1,5} Example 3b b) 2( x 5) 2 7 0 2( x 5) 2 7
7 2 7 7 x 5 , or __ x 5 2 2 ( x 5) 2 multiply _ 7 2 by 2 2 7 2 14
14 2 2 4 2 14 14 , or __ x 5 2 2 10 14 10 14 x , or __ x 2 2 2
2 10 14 10 14 x , or __ x 2 2 x 5 Example 3c x4-1=80 x4=81 x= 3 81= 3 So the solution set is {-3,3} ** Ex 19-28** Nonequivalent solutions or Extraneous Solutions When you solve an equation by squaring both sides, you can get answers that DONT satisfy the equation you are working with.
These are extraneous. Throw them out! Example 4 Solve: a) 2 x 3 5 0, 2 x 3 5 2 ( 2 x 3) 5 2 x 3 25 2 x 28 2 x 14 _ or _{14} b) 3 3 x 5 3 x 1, 3
3x 5 3 3 x 1 3x 5 x 1 2 x 6 x 3 _ or _{ 3} 3 3 x 18 x 2
( 3 x 18) x 3 x 18 x 2 2 Ex 4c **Ex. 29-48** 2 x 3 x 18 0 ( x 6)( x 3) 0 x 6 0 __ or __ x 3 0 x 6 _ or _ x 3 Checking 3( 3) 18 3 ________ 3(6) 18 6 9 3 _______________ 36 6 3 3 __ bad _ answer !____ 6 6 __ good !!
5x 1 x 2 1 5 x 1 1 x 2 5x 1 2 1 x2 2
5 x 1 1 2 x 2 x 2 5 x 1 3 x 2 x 2 And sometimes you need to square both sides twice EX5 **Ex. 49-84** 4 x 4 2 x 2 Checking 2x 2 x 2 2x 2 2
x2 2 4 x 2 8 x 4 x 2 4 x 2 9 x 2 0 (4 x 1)( x 2) 0 4 x 1 0 _ or _ x 2 0 1 x ___ or ___ x 2 4 1 x ___ or ___ x 2 4 Plug _ it _ in _ to _ 5 x 1 1
5 1 4 1 1 2 4 4 5 2 1 22 9 x 2 1 9 1 3 1 4 2 2 4 3 2 1
Example 6 page 584 **Ex. 65-76** 2 3 a) x 4 2 3 b) 3 3 x 4 x 2 64 x 8 ___ or ___ 8
Both _ work , so { 8,8} ( w 1) 2 5 4 2 5 ( w 1)
5 4 5 1 ( w 1) 2 1024 w 1 1 1024 1 1 __ or __ w 1 32 32 33 31
w __ or __ w 32 32 31 33 Both _ work , so { , } 32 32 w 1 Ex 7 page 585 Not all good things have a solution (2t-3)-2/3=-1 [(2t-3)-2/3]-3=(-1)-3 (2t-3)2=-1 Error! We cant take the square root of this! There is no solution in this universe.
**Ex 77-78** Aid in Solving these things 1. In raising each side of an equation to an even power, we can create equations that give extraneous solutions. Check em! 2. When applying the even-root property, remember that there is a positive and a negative root for any positive real number. 3. For equations with rational exponents, raise each side to a positive or negative integral power first, then apply the even- or odd- root property. (Postive fraction raise to a positive power; negative fraction raise to a negative power.) And back to a few applications The distance formula If you have a triangle with points (x 1,y2) and (x2,y2) you can use the Pythagorean theorem
to get the distance a2+b2=c2 becomes 2 d ( x2 x1 ) ( y2 y1 ) 2 Example 8 page 585 Looking Figure 9.1 we want to know the distance from first to third base when the bases are 90feet apart. x 2 902 902 x 2 8100 8100 x 2 16, 200 x 16, 200 90 2 Then _ we _ ignore _ the _ negative answer. Putting out fires in section 9.5
Some definitions Q1-Q4 Solving things with radials with one answer Q5-Q12 Solving for two or no answers Q13-28 Solving and checking for extraneous answers Q29-64 Solving Q65-98 Word problems Q99-124 Changing Chapters Chapter 10.1-10.3 Putting it all together, factoring to graphing. The big sum up. Section 10.1 Factoring and Completing the Square This is MORE of the same. Nothing new
EXCEPT square roots may show up. If you keep your head about you, this will go down like castor oil. Review of Factoring ax2+bx+c=0 Where a,b,c are real and a isnt equal to 0 (or thats cheating). Review of the Cookbook 1. Write the equation with 0 on the right hand side ( stuff=0) 2. Factor the left hand side. 3. Use the zero factor property to set each factor equal to zero. 4. Solve the simplest equations. 5. Check the answers in the original equation. Example 1 page 610 Solving a quadratic equation by factoring **Ex. 5-14**
2 3x 4 x 15 0 (3 x 5)( x 3) 0 3x 5 0 ___ or ___ x 3 0 3x 5 ____ or ___ x 3 5 x 3 5 So _ the _ solution _ set _ is _ ,3 3 Example 2 pg 611 Review of the Even-Root Property This should also go down quickly since youve done it sooooo much! If you solve (a-1)2=9 you get We _ know _ x 2 k _ is _ also _ x k (a 1) 2 9 a 1 9
So _ a 1 3 _ or _ a 1 3 a 4 _ or _ a 2 { 2, 4} **Ex. 15-24** Completing the Square (making polynomials the way YOU want them) Can you make factorable polynomials if you are only given the first two terms? What about x2+6x ? To find the last term, remember that you start with two of the things added together that make that middle term that when multiplied together equal that last term. Or, in other words, (b/2)2 is your last term. The rule for finding the last term x2+bx has a last term that makes the entire
polynomial look like: x2+bx+(b/2)2 Ex 3 pg 612 Raiders of the Lost Term 2 a) x 8 x __ 2 (8 / 2) 4 4 16 2 So _ x 8 x 16 b) x 2 5 x __ 25 ( 5 / 2) ( 5 / 2)( 5 / 2)
4 25 2 So _ x 5 x 4 2 Ex 3 continued 4 x x __ 7 1 4 2 2 2 4 ( ) ( )( ) 2 7 7 7 49 4 4 2
So _ x x 7 49 2 c) d) **Ex. 25-32** 3 x x __ 2 1 3 2 3 3 9 ( ) ( )( ) 2 2 4 4 16 3
9 2 So _ x x 2 16 2 Example 4 pg 612 Remember the perfect square trinomials? Were looking for things in the form a2+2ab+b2=(a+b)2 a) x2+12x+36 = (x+6)2 b) y2-7y+49/4 = (y-7/2)2 c) z2-4/3z + 4/9 = (y-2/3)2 **Ex 33-40** If a=1 then we can complete the squares Example 5 pg 613 Given x2+6x+5=0 The perfect square whos first two terms are x2+6x
is x2+6x+9 So we just add 9 to both sides to FORCE this to be a perfect square! x2+6x+5+9=0+9 x2+6x+9=9-5 (x+3)2=4 Now we solve it Solving Ex 5 2 ( x 3) 4 x 3 4 2 x 3 2 _ or _ x 3 2 x 1 _ or _ x 5 **Ex. 41-48** If the coefficient of a isnt 1 Too bad. To make this work, you have to MAKE it = 1!! So divide both sides in their entirety by
whatever is before the a For example if you have 2x2+4x+10=8 Then divide EVERYTHING by 2 Making it x2+2x+5=4 then work on The cookbook for these critters: Solving Quadratic Equations by Completing the Squares 1. 2. 3. 4. The coefficient of x2 must be 1 Get only the x2 and x terms alone on the RHS Add to each side the coefficient of x Factor the left hand side as the square of a binomial 5. Apply the even root property (plus or minus the square root of the remaining number) 6. Solve for x
7. Simplify Example 6 pg 614 a isnt 1 ** Ex.49-50** 2x2+3x-2=0 2 x 2 3x 2 0 2 2 3 x 2 x 1 0 2 3 2 x x __ 1 2 3 9 9
2 x x 1 2 16 16 2 3 25 x 4 16 3 25 4 16 3 5 3
5 x _ or _ x 4 4 4 4 2 1 8 x _ or _ x 2 4 2 4 x 2 x 3x 6 0 2 x 3x __ 6 9 9 2
x 3x 6 4 4 Example 7 pg 615 x2-3x-6=0 **Ex. 51-60** 2 3 33 x 2 4 3 33 x 2
4 3 33 x 2 4 3 33 3 33 3 33 x _ or _ , 2 2 2 Now well disguise dishwashing liquid as hand lotion Ex 8 pg 615 (square first then solve) ** Ex. 61-64** Can we deal with square roots in the problem? x 3 153 x
x 3 2 153 x 2 x 2 6 x 9 153 x x 2 7 x 144 0 ( x 9)( x 16) 0 x 9 0 _ or _ x 16 0 x 9 _ or _ x 16 Checking ... Root _ x 9?
9 3 ? 153 9 12 144 _ good ! Root _ x 16 16 3 ? 153 13 169 13 _ BAD ! Extraneous _ root 1 3 5 x x 2 8 The _ LCD _ is 8 x ( x 2) 1 3 5 8 x( x 2) 8 x( x 2) 8 x( x 2) x
x 2 8 8 x 16 24 x 5 x 2 10 x Example 9 pg 616 LCD then complete the squares 32 x 16 5 x 2 10 x 5 x 2 42 x 16 0 (5 x 2)( x 8) 0 So _ 5 x 2 0 _ or _ x 8 0 2 x _ or _ x 8 5 2 Both _ work _ so _ ,8 5 ** Ex. 65-68**
Toying with the dark side imaginary solutions! Example2 10 pg 616 **Ex. 69-78** x 4 x 12 0 2 x 4 x ___ 12 2 x 4 x 4 12 4 ( x 2) 2 8 x 2 8 x 2 i 8 2 i 4 2 x 2 2i 2 Section 10.1 Try a completed square on for size!
Definitions Q1-4 Review solve by factoring Q5-14 Use the even root property Q15-24 Finish the perfect square trinomial Q25-32 Factor perfect square trinomials Q33-40 Solve by completing the square Q41-58 Potpourri of problems Q59-66 Complex Answers Q67-90 Check answers Q91-94 Word Problems Q95-106 Section 10.2 The Return to the Temple of the Quadratic Formula
Or how to get the answer without doing ANYTHING that is hard as what we have already done!! The scientific term for it is: PlugnChug Remember the standard form? ax2+bx+c=0 We can solve for x and always find out what x (or the xs) are. Developing it Well just look at it like one would look at the Grand Canyon. You can enjoy it and get into it if youre in good shape. (where was I going with this slide?) ax 2 bx c 0, ax 2 bx c 0 , a
a b c x 2 x 0, a a b c x 2 x , a a 2 b b2 b the _ square _ of _ 1/ 2 _ of _ _ is _ 2 a 4a 2a b
b2 c b2 x x 2 2 , a 4a a 4a 2 2 b b 2 c 4a , x 2 2 a 4 a a 4
a 2 b b 2 4ac , x 2 2a 4a b b 2 4ac x , 2 2a
4a b b 2 4ac x , 2a 4a 2 b b 2 4ac x 2a The Quadratic Formula ax2+bx+c=0 where a isnt 0 2 b b 4ac x 2a Example 1 pg 623
(become the numbers) **Ex. 7-14** x2+2x-15=0 a=1 b=2 c=-15 2 b b 4ac x 2a 2 22 4 1 15 x , 2(1) 2 4 60 x , 2 2 64 x , 2
2 8 x , 2 2 8 2 8 So, x , or , , 2 2 x { 5,3} Example 2 pg 623 Only solution **Ex. 15-20** 4x2-12x+9=0 a=4 b=-12 c=9 2 b b 4ac
x 2a 2 12 ( 12) 4 4 9 x , 2(4) 12 144 144 x , 2 12 0 12 3 x 8 8 2 Example 3 pg. 624 Two irrational solutions **Ex. 21-26**
2x +16x+3=0 a=2 b=16 c=3 2 2 b b 4ac x 2a 6 (6) 2 4 2 3 x , 2(2) 6 36 24 6 12 x , 4 4 6 4 3 2 3 2 3
x , 4 2 2 2 3 2 3 2( 3 3) x , 2 2 2 2 3 3) x 2 Example 4 pg 625 Two imaginary solutions (they are in elsewhere) **Ex. 27-32** x2+x+5=0 a=1 b=1 c=5
2 b b 4ac x 2a 2 1 (1) 4 15 1 19 x , 2(1) 2 1 i 19 x , remember _ 1 i 2 The big picture Use the quick reference guide on
PAGE 538 for all the different ways to solve ax2+bx+c=0 How many solutions? Look to the discriminate. From the earlier examples, you get two answers when the b2-4ac is positive You get one answer when b2-4ac is = 0. And no real answers, only imaginary ones, when b2-4ac is negative. Example 5 pg. 626 **Ex. 33-48** a) x2-3x-5=0 b2-4ac = (-3)2-4*1*(-5)=9+20=29 two real ans. b) x2=3x-9 x2-3x+9 =0 b2-4ac = (-3)2-4*1*9= 9-36=-27 two imag. answers c) 4x2-12x+9=0 b2-4ac = (-12)2-4*4*9= 144-144=0 One real ans.
Ex 6 - Application pg 626 **Ex. 77-96** If the area of a table is 6 sq ft. And one side is 2 feet shorter than the otherwhat are the dimensions? The Setup x(x-2)=6 Or x2-2x-6=0 a=1,b=-2,c=-6 2 ( 2) 2 4 1 6 2 28 x , 2(1) 2 2 28 2 4 7 2 2 7 x
, 2 2 2 x 1 7 Danger _ 1 7 _ is _ negative toss _ it So _ x 1 7 The _ sides _ are _ x _ and _ x 2 Side1:1 7, and 7 1 or _ 3.65 feet _ and _ 1.65 feet Section 10.2 The Quadratic Formula
Definitions Q1-Q6 Solve using the formula Q7-32 How many solutions? Q33-48 Solve it the way you want.. Q49-66 Using a calculator Q67-76 Word Problems Q77-106 10.3 More-on Quadratic Equations We just wont get this far this classsorry! You can email me and work through it if you want to! The material below is no longer part of MTH 209 Go back, there are dragons ahead! Section 10.4 I see Quadratic Functions
Definition, If Y is determined by a formula with X in it, we say y is a quadratic function of x y=ax2+bx+c Example 1 Given a number, whats the other (more plugging in) a) y=x2-x-6; given (2, ), ( ,0) The ()s are (x,y) y=22-2-6=4-2-6=-4 So the first is (2,-4) The other one makes us factor x2-x-6=0 Which is (x-3)(x+2)=0 so x=3 or 2
This one gives us two answers (3,0) and (-2,0) example 1b s=-16t2+48t+84 given (0, ), ( ,20) This time its (t,s) inside the ()s The first is s=-16(0)2+48(0)+84 = 84 Its ordered pair is (0,84) The second is 20 =-16t2+48t+84 =-16t2+48t+64 =t2-3t+4 =(t-4)(t+1) so t=4 or 1 Giving us (-1,20) and (4,20) as answers. Graphing. Plug in all values in the universe for x, and see what y is A parabola. They look like this!
Example 2 Graphing y=x2 We can go back to the old try a few numbers method. x -2 -1 0 1 2 y=x2 4 1
0 1 4 Can you picture that? It looks like this! With a positive a the U shape opens upward. Domain and Range The domain is the extent of the graph in X The range is the extent of the graph in Y In this graph X = , (domain) Y is only above and including 0 : [0, )
(range) Example 3 : y=4-x2 y=4-x2 x -2 -1 0 1 2 y=4-x2 0 3 4 3 0 Figuring out more quickly where is the VERTEX? The Vertex is the minimum point (if the
thing opens upward) or maximum point (if the thing opens downward). We can find the vertex by using the front part of: 2 b b 4ac x 2 a b Mainly x 2a The vertexs above For y=x2 The vertex is: (0,0) for y=4-x2 here it is (0,4)
Example 4 Using Graph y=-x2-x+2 b ( 1) 1 x 2a 2( 1) 2 y _ there _ is... y x 2 x 2 2 1 1 9 1 1 y 2 2 4 4 4 2 2 1 9 The _ Vertex _ is _( x, y ) ,
2 4 b x 2a Example 4 continued Plug them numbers in x -2 -1 -1/2 0 1 y=-x2-x+2 0 2 9/4
2 0 Example 5 Do it some more a) y=x2-2x-8 b Using x give us x=1, then y=12=2*1-8 = -9 2a The vertex then is (1,-9) To find the y-intercept, we can plug in x=0 and find y=022*0-8 = -8 So its (0,-8) To find the x-intercept(s) we can plug in y=0 x2-2x-8=0 or (x-4)(x+2)=0 so x=4 or x=-2 Now we have sleuthed out some points and can plot it Example 5a, the graph
Example 5b a) s=-16t2+64t us b t=2, then s=-16*22+64(2) = 64 Using give t The vertex then 2isa(2,64) (since it is (t,s)) To find the s-intercept, we can plug in t=0 and find s=-16(0)2+64*0 = 0 So its (0,0) To find the t-intercept(s) we can plug in s=0 -16t2+64t=0 or -16t(t-4)=0 so 16t=0 or t-4=0 t=0 or t=4 the t intercept(s) will be (0,0) and (4,0) Now we have sleuthed out some points and can plot it Example 5b the picture Graph that quadratic soldiers!
Section 10.3 Definitions Q1-6 Complete the ordered pairs Q7-10 Graph the equations Q11-30 Find the max or min Q31-38 Word Problems Q39-48 and beyond New 10.4 Section 10.5 Alas, Jean Luc. All good things must come to an end. This time we put much of the earlier material together to do your favorite thing! Well graph quadratic INEQUALITIES on the
number line. Then you can go run in the beautiful spring air and feel the joyful burden of learning algebra fall off your mathematical shoulders. Again its just a small step Quadratic Inequalities They look like: ax2+bx+c > 0 where a, b,c are real numbers and a isnt 0 We can use , , , Example 1 x2+3x-10 > 0 (x+5)(x-2)>0 The product is positive so both may be negative or both may be positive Value x+5=0 x+5>0
Where if x= -5 if x>-5 On the number line Put a 0 above 5 Put + signs to the right of 5 x+5<0 if x<-5 Put signs to the left of -5 Example 1 continued Value x-2=0 x-2>0
Where if x= 2 if x>2 On the number line Put a 0 above 2 Put + signs to the right of 2 x-2<0 if x<2 Put signs to the left of 2 Example 2 2 2 x 5 x 3
2 2 x 5 x 3 0 (2 x 1)( x 3) 0 SOOoooo one is neg, one is pos. or the opposite. 2x-1=0 if x=1/2 2x-1>0 if x>1/2 2x-1>0 if x<1/2 x+3=0 if x=-3 x+3>0 if x>-3 x+3<0 if x<-3
The cookbook 1. Write the inequality with 0 on the right 2. Factor the quadratic polynomial on the left 3. Make a sign graph showing where each factor is positive, negative or zero. 4. Use the rules for multiplying signed numbers to determine which regions satisfy the original equations. A reminder about ratios and inequalities x2 2x 3 2 1 2, 0, x 3 x 5 x 4 x 1
You need an LCD to add fractions If you multiply by 1 to solve for x, you must reverse the inequality sign (but you dont have to do anything to the inequality sign if you divide or multiply by a positive number) Example 3 A rational inequality x2 2, x 3 x 2 2( x 3) 0, x 3 x 3 x 2 2x 6 0, x 3 x 3 x 2 2x 6 0,
x 3 x 8 0 x 3 Ex 3 continued x-3=0 if x=3 x-3>0 if x>3 x-3<0 if x<3 -x+8=0 if x=8 -x+8 > 0 if x>8 -x+8<0 if x<8 Example 4 Now put a ratio on both sides 2 1 0,
x 4 x 1 2( x 1) 1( x 4) 0, ( x 4)( x 1) ( x 1)( x 4) 2( x 1) ( x 4) 0, ( x 4)( x 1) 2x 2 x 4 0, ( x 4)( x 1) x 2 0 ( x 4)( x 1) The cookbook for rational inequalities with a sign graph 1. Rewrite the equation with a 0 on the right hand side 2. Use only addition and subtraction to get an
equivalent inequality 3. Factor the numerator and denominator if possible 4. Make a sign graph showing where each factor is positive, negative and zero. 5. Use the rules for multiplying and dividing signed numbers to determine the regions that satisfy the original inequality. Getting your + and regions correct. Using a test point. Sometimes you cant factor the portions of the quadratic equation. Are you stuck? Heavens no! Why not just use the quadratic equationthen test a few points? Example 5 x2-4x-6>0 2 b b 4ac
x , 2a 4 162 4(1)( 6) x 2( 6) 4 40 4 2 10 2 10 2 2 Example 5 continued We can use these points to divide the line by , 2 Note:
10 , 2 10, 2 10 , 2 10, 2 10 5.2, and , 2 10 1.2 So well choose 2, 0 and 7 as test points. We plug those into the first equation and see which are true Example 5 finishing it up Test Point Value of x2-4x-6 at test point
Sign of x2-4x-6 in interval of test point -2 0 7 6 -6 15 Positive Negative Positive The quadratic inequalities using Test Points Cookbook 1. Rewrite the inequality with 0 on the right 2. Solve the quadratic equation that results from replacing the inequality symbol with the equals
symbol 3. Locate the solutions to the quadratic equation on a number line 4. Select a test point in each interval determined by solving the quadratic equation 5. Test each point in the original quadratic inequality to determine which intervals satisfy the inequality. Example 6 After setting up the problem we have the equation P=-x2+80x-1500 For what x is her profit positive (x = magazine subscriptions) -x2+80x-1500>0 x2-80x+1500<0 (x-30)(x-50)<0 The Final Practice! Section 10.5
Definitions Q1-4 Solve each inequality Q5-16 Do it with rational inequalities Q17-36 Solve each inequality using interval notation Q37-60 Word Problems Q61-66 Wrap it up with a final Go forth and multiply. And factor. And find roots. etc
Example 7 Writing in simplified form Simplify a) 10 10 6 60 4 15 2 15 15 6 6 6 3 6
6 6 b) 3 5 3 5 3 5 3 3 3 15 3 15 3 3 3 3 9 3 9 9 3 27 Rationalize YOUR denominator These square roots (like 2 3 and 5 are irrational numbers. It is customary to rewrite the fraction with a rational number in the denominator. That is rationalize it. Remember we can always do this:
2 2 2 since 2 2 4 2 Ex 5 Lets rationalize some denominators! a) Rationalize: b) 3 3 5 15 5 5 5 5 3 3 3 2 3 2 33 4 3 3 3 3 2
2 2 2 2 Simplified Radical Form for Radicals of Index n A radical expression of index n is in simplified form if it has: 1) no perfect nth powers as factors of the radicand 2) no fractions inside the radical, and 3) no radicals in the denominator Example 7 Of course, we can insert variables into this! Simplify a) 6 (look for even things you can work with!) 6
6 12 x 4 x 3 4 x 3 2 x 3 3 b) 5 9 4 8 2 98 x y 49 x y 2 xy 7 x y
4 2 xy Example 8 Working with denominators and radicals We (traditionally remember) want to get rid of the in the denominators a) a a b ab b b b b
b) x3 x3 x2 5 y y5 y4 y x xy x xy x x x x 2 2 2 3 y y y y y
y y y y x Of course, why not also complicate things with cube roots and 4th roots a) 3 8 3 6 3 12
4 2 40 x 8 x 5 x 2 x 23 5x 2 b) 4 c) 12 5
3 3 3 x y x y y x y 4 y 4 3 x x x y 3y 3y 3 3
4 y 2 y 2 3 3 xy y 2
3 xy y 3 2