Sect Definition of the nth Root

Transcription

1 Cocept #1 Sect Defiitio of the th Root Defiitio of a Square Root. The square of a umber is called a perfect square. So, 1,, 9, 16, 2, 0.09, ad 16 2 are perfect squares sice 1 = 12, = 2 2, 9 = 2, 16 = 2, 2 = 2, 0.09 = (0.) 2 ad 16 2 = ( ) 2. The reverse operatio of squarig a umber is fidig the square root of a umber. We ca use the idea of perfect squares to simplify square roots. The square root of a umber, a, asks what umber times itself is equal to a. For example, the square root of 2 is sice times is 2. Defiitio of a Square Root Let a represet a o-egative real umber. The, the square root of a, deoted a, is a umber whose square is a. a is called the positive or pricipal square root of a. a is called the egative square root of a. Simplify: Ex. 1a 6 Ex. 1b 0.2 Ex. 1c 9 81 Ex. 1d 6.2 Ex. 1e 0 Ex. 1f 1 Ex. 1g 2 Ex. 1h Solutio: a) Sice 8 2 = 6, the 6 = 1 6 = 8 b) Sice (0.) 2 = 0.2, the 0.2 = 0. c) Sice ( ) 2 = , the 9 81 = 9 d) Sice (2.) 2 = 6.2, the 6.2 = = 2. e) Sice 0 2 = 0, the 0 = 0 f) Sice 1 is ot a perfect square, the eed to use a calculator to get approximatio: 1 = g) 2 is ot a real umber. h) 1 = 1 2 1

2 8 Cocept #2 Defiitio of a th Root The square root of a umber is the reverse process of squarig a umber. We ca also do this with other powers as well. The cube root of a umber is the reverse process of cubig a umber. The fourth root of a umber is the reverse process of raisig a umber to the fourth power ad so forth. Defiitio of the th root Let a be a real umber ad let be a atural umber greater tha 1 1) If is eve ad a 0, the a is the pricipal th root of a ad a is the egative th root of a. 2) If is odd, the a is the pricipal th root of a ad a is the egative th root of a. I the radical a, the symbol is called the radical sig, the umber a is called the radicad, ad is called the idex. For the square root, the idex is 2 ad we usually do ot write the idex. Keep i mid that the eve root of a egative umber is udefied i the real umbers. Here is a list of perfect powers i the followig table: Perfect Cubes Perfect Fourth Powers Perfect Fifth Powers 2 = 8 2 = 16 2 = 2 = 2 = 81 = 2 = 6 = 26 = 102 = 12 = 62 = 12 6 = = = 6 = = 201 = 16,80 8 = 12 8 = = 2,68 9 = 29 9 = = 9,09 10 = = 10, = 100,000 Simplify the followig: Ex. 2a 10,000 Ex. 2c 16 Ex. 2e 9,09 Ex. 2b 6 6 Ex. 2d 1 Ex. 2f 10,000,000

3 9 Solutio: a) Sice 10 = 10,000, the 10,000 = 1 10,000 b) Sice ( ) = 6, the 6 = c) Sice 2 = 16, the 16 = 2 6 d) 1 is ot a real umber. Cocept # e) Sice ( 9) = 9,09, the 9,09 = 1 ( 9) = 9 f) Sice 10 = 10,000,000, the 10,000,000 Simplify the followig: Roots of Variable Expressios = 10 = 1 9,09 = 10 Ex. a ( ) 2 Ex. b ( ) Ex. c ( ) Solutio: a) ( ) 2 = 9 = b) ( ) c) ( ) = = 201 = = Ex. d ( ) d) ( ) = 16,80 = Notice that there is a patter occurrig. If the power ad the idex are both the same ad a odd umber, we get the same umber for the aswer as the umber we started with: ( ) = ( ) = Same Same If the power ad the idex are both the same ad a eve umber, we get the absolute value of the umber we started with for the aswer : ( ) 2 = ( ) Absolute Value Absolute Value I geeral, a = a if is odd ad a = a if is eve. This illustrates how to take the th root of variable expressios. =

4 60 Defiitio of Let be a atural umber greater tha oe. The 1) If is odd, the a = a. 2) If is eve, a = a. a Simplify the followig: Ex. a (r s) Ex. b (a b) Ex. c 9x 2 0x+2 Ex. d a 1 b c 10 Ex. e 6 Solutio: 6a 6 b 2 c 12 a) (r s) b) (a b) = (r s) = r s. = (a b) = a b. c) Sice 9x 2 0x + 2 = (x ) 2, the 9x 2 0x+2 = (x ) 2 = (x ) = x d) a 1 b c 10 e) 6 6a 6 b 2 = c 12 c 2 = (a ) b (c 2 ) a 6 (b ) 6 2ab = (c 2 ) 6 c 2 = a bc 2.. But 2, b, c 2 are ot egative, so they ca be pulled out of the absolute value: 2ab = 2b a = 2 a b c 2 c 2. Notice that with part d ad e, we rewrote the powers as perfect th powers before applyig the th root. Here are some patters to look for: Perfect Squares Perfect Cubes Perfect th Powers Perfect th Powers (x 2 ) 2 = x (x 2 ) = x 6 (x 2 ) = x 8 (x 2 ) = x 10 (x ) 2 = x 6 (x ) = x 9 (x ) = x 12 (x ) = x 1 (x ) 2 = x 8 (x ) = x 12 (x ) = x 16 (x ) = x 20 (x ) 2 = x 10 (x ) = x 1 (x ) = x 20 (x ) = x 2 (x 6 ) 2 = x 12 (x 6 ) = x 18 (x 6 ) = x 2 (x 6 ) = x 0 (x ) 2 = x 1 (x ) = x 21 (x ) = x 28 (x ) = x ( x 2 )2 = x ( x ) = x ( x ) = x ( x ) = x

5 61 Simplify: Ex. a t 6 v 8 Ex. b Ex. c x y 21 z 1 Solutio: 62a 8 81b 16 Ex. d 216x 9 y a) Sice 6 2 = ad 8 2 =, the t 6 v 8 = (t ) 2 (v ) 2 = t v. But v 0, so t v = v t. b) Sice 8 = 2 ad 16 =, the 62a 8 = 81b 16 a = 2. But all the powers are o-egative, so b a 2 b = a 2. b c) Sice 21 = ad 1 = 2, the x y 21 z 1 = x (y ) (z 2 ) = xy z 2. d) Sice 9 =, the 216x 9 y = 1( 6x y) = 6x y. (a 2 ) (b ) = 1 ( 6) (x ) y Cocept # Pythagorea Theorem I a right triagle, there is a special relatioship betwee the legth of the legs (a ad b) ad the hypoteuse (c). This is kow as the Pythagorea Theorem: Pythagorea Theorem I a right triagle, the square of the hypoteuse (c 2 ) is equal to the sum of the squares of the legs (a 2 + b 2 ) c 2 = a 2 + b 2 c a b Keep i mid that the hypoteuse is the logest side of the right triagle.

6 62 Fid the legth of the missig sides (to the earest hudredth): Ex..8 m Ex. 6 1 ft 10.2 m 6 ft Solutio: Solutio: I this problem, we have I this problem, we have oe leg the two legs of the triagle ad the hypoteuse ad we are ad we are lookig for the lookig for the other leg: hypoteuse: c 2 = a 2 + b 2 c 2 = a 2 + b 2 (1) 2 = (6) 2 + b 2 c 2 = (.8) 2 + (10.2) 2 22 = 6 + b 2 (solve for b 2 ) c 2 = = 6 c 2 = = b 2 To fid c, take the square To fid b, take the square root root* of 16.88: of 189: c = ± = ± b = ± 189 = ± 1... c 12.8 m b 1. ft * - The equatio c 2 = actually has two solutios 12.8 ad 12.8, but the legths of triagles are positive, so we use oly the pricipal square root. Solve the followig: Ex. 8 Leroy leaves St. Philip s College i his car drivig east at 0 mph o Marti Luther Kig Drive. Juaita leaves at the same time drivig south at 0 mph o New Braufels Aveue. How far apart are they after twelve miutes? Solutio: Twelve miutes is 12 Sice d = rt, the: 60 = 1 of a hour. Leroy traveled = 0( 1 ) = 6 miles Juaita traveled = 0( 1 ) = 8 miles South 8 miles East 6 miles

7 Now, draw a picture: We have a right triagle with two legs ad we are lookig for the hypoteuse: c 2 = (8) 2 + (6) 2 c 2 = c 2 = 100 To fid c, take the square root* of 100: c = ± 100 = ± 10 Agai, we used oly the pricipal square root. c = 10 miles Leroy ad Juaita are te miles apart. Cocept # Radical Fuctios If is a atural umber greater tha oe, the f(x) = x is a radical fuctio. The domai will deped o the idex. If is odd, the domai will be a real umbers. If the idex is eve, the the domai will be all values of x that make the radicad greater tha or equal to zero. Fid the domai of the followig: Ex. 9a f(x) = 6 x Ex. 9b g(x) = + 2x Ex. 9c r(x) = 6 1 x 2 Ex. 9d h(x) = x 8 Solutio: a) Sice the idex is eve, the radicad has to be o-egative: Solve: 6 x 0 (switch the iequality sig whe x 6 dividig by a egative umber) x 1. The domai is (, 1.], b) Sice the idex is odd, the radicad is defied for all real umbers. The domai is (, ). c) Sice the idex is eve, the radicad has to be o-egative. But the radicad is i the deomiator, so it also caot be zero: Solve: x 2 > 0 x > 2 x > 2 Thus, the domai is ( 2, ). 6

8 6 d) Sice the idex is odd, the radicad is defied for all real umbers. But the radicad is i the deomiator, so it also caot be zero: x 8 0 x 8 x 1.6 So, the fuctio is defied for all real umbers except 1.6. Thus, the domai is (, 1.6) (1.6, ). Match the fuctio with the graph: Ex. 10a f(x) = x 2 Ex. 10c g(x) = x+2 Ex. 10b h(x) = x 1 i) ii) iii) Solutio: a) Sice the idex is odd, the domai is all real umbers. The oly graph that matches it is #ii. b) Sice h(x) is the egative square root, the y-values are egative. This matches #iii. c) Sice g(x) is the positive square root, the y-values are positive. This matches #i.