0.9 Radicals and Equations

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1 0.9 Radicals and Equations 0.9 Radicals and Equations In tis section we eview simlifying exessions and solving equations involving adicals. In addition to te oduct, quotient and owe ules stated in Teoem 0. in Section 0., we esent te following esult wic states tat n t oots and n t owes moe o less undo eac ote. Teoem 0.. Simlifying n t owes of n t oots: Suose n is a natual numbe, a is a eal numbe and n a is a eal numbe. Ten ( n a) n a if n is odd, n a n a; if n is even, n a n a. Since n a is defined so tat ( n a) n a, te fist claim in te teoem is just a e-woding of Definition 0.8. Te second at of te teoem beaks down along odd/even exonent lines due to ow exonents affect negatives. To see tis, conside te secific cases of ( ) and ( ). In te fist case, ( ) 8, so we ave an instance of wen n a n a. Te eason tat te cube oot undoes te tid owe in ( ) is because te negative is eseved wen aised to te tid (odd) owe. In ( ), te negative goes away wen aised to te fout (even) owe: ( ) 6. Accoding to Definition 0.8, te fout oot is defined to give only non-negative numbes, so 6. Hee we ave a case wee ( ), not. In geneal, we need te absolute values to simlify n a n only wen n is even because a negative to an even owe is always ositive. In aticula, x x, not just x (unless we know x 0.) We actice tese fomulas in te following examle. Examle Pefom te indicated oeations and simlify.. x +. t 0t x x + Solution. ( x ) (x) 6. 8x. q ( 8y 8y) +( 0 L 8 80). We told you back on age tat oots do not distibute acoss addition and since x + cannot be factoed ove te eal numbes, x + cannot be simlified. It may seem silly to stat wit tis examle but it is extemely imotant tat you undestand wat maneuves ae legal and wic ones ae not. See Section 5. fo a moe ecise undestanding of wat we mean ee. If tis discussion sounds familia, see te discussion following Definition 0.9 and te discussion following Extacting te Squae Root on age 8. You eally do need to undestand tis otewise oible evil will lague you futue studies in Mat. If you say someting totally wong like x +x + ten you may neve ass Calculus. PLEASE be caeful!

2 Peequisites. Again we note tat t 0t t 0t + 5, since adicals do not distibute acoss addition and subtaction. In tis case, oweve, we can facto te adicand and simlify as t 0t + 5 (t 5) t 5 Witout knowing moe about te value of t, we ave no idea if t so t 5 is ou final answe. 5 5 is ositive o negative. To simlify 8x, we need to look fo efect cubes in te adicand. Fo te cofficient, we ave To find te lagest efect cube facto in x, we divide (te exonent on x) by (since we ae looking fo a efect cube). We get wit a emainde of. Tis means +, so x x + x x (x ) x. Putting tis altogete gives: 8x 6 (x ) x (x ) 6x x 6x Facto out efect cubes Reaange factos, Poduct Rule of Radicals. In tis examle, we ae looking fo efect fout owes in te adicand. In te numeato is clealy a efect fout owe. Fo te denominato, we take te owe on te L, namely, and divide by to get. Tis means L 8 L (L ). We get L L (L ) L Quotient Rule of Radicals Poduct Rule of Radicals Simlify Witout moe infomation about, we cannot simlify any fute. Howeve, we can simlify L. Regadless of te coice of L, L 0. Actually, L > 0 because L is in te denominato wic means L 6 0. Hence, L L. Ou answe simlifies to: L L 5. Afte a quick cancellation (two of te s in te second tem) we need to obtain a common denominato. Since we can view te fist tem as aving a denominato of, te common denominato is ecisely te denominato of te second tem, namely ( x ). Wit Let t and see wat aens to t 0t + 5 vesus t 0t In geneal, t 5 6 t 5 and t 5 6 t + 5 so watc wat you e doing!

3 0.9 Radicals and Equations common denominatos, we oceed to add te two factions. Ou last ste is to facto te numeato to see if tee ae any cancellation ootunities wit te denominato. x x + ( (x) x x + x ) ( (x) Reduce x ) x x + x ( x ) Mutily (x x ) ( x ) ( x ) + x ( x ) Equivalent factions x( x ) x ( + x ) ( Multily x ) x(x ) ( x ) + x ( x ) Simlify x(x ) + x ( x ) Add x(x + ) ( x ) Facto x(x ) ( x ) We cannot educe tis any fute because x is ieducible ove te ational numbes. 6. We begin by woking inside eac set of aenteses, using te oduct ule fo adicals and combining like tems. q ( 8y 8y) +( q 0 80) ( 9 y y) +( 5 6 5) q ( 9 y y) +( 5 6 5) q ( y y) + ( 5 5) q ( y) +( 5) q y +( ) ( 5) y + 5 y + 0 To see if tis simlifies any fute, we facto te adicand: y + 0 (y + 0). Finding no efect squae factos, we ae done.

4 Peequisites Teoem 0. allows us to genealize te ocess of Extacting Squae Roots to Extacting n t oots wic in tun allows us to solve equations 6 of te fom X n c. Extacting n t oots: If c is a eal numbe and n is odd ten te eal numbe solution to X n c is X n c. If c 0 and n is even ten te eal numbe solutions to X n c ae X ± n c. Note: If c < 0 and n is even ten X n c as no eal numbe solutions. n Essentially, we solve X n c by taking te n t oot of bot sides: X n n c. Simlifying te left side gives us just X if n is odd o X if n is even. In te fist case, X n c, and in te second, X ± n c. Putting tis togete wit te ote at of Teoem 0., namely ( n a) n a, gives us a stategy fo solving equations wic involve n t and n t oots. Stategies fo Powe and Radical Equations If te equation involves an n t owe and te vaiable aeas in only one tem, isolate te tem wit te n t owe and extact n t oots. If te equation involves an n t oot and te vaiable aeas in tat n t oot, isolate te n t oot and aise bot sides of te equation to te n t owe. Note: Wen aising bot sides of an equation to an even owe, be sue to ceck fo extaneous solutions. Te note about extaneous solutions can be demonstated by te basic equation: x. Tis equation as no solution since, by definition, x 0 fo all eal numbes x. Howeve, if we squae bot sides of tis equation, we get ( x) ( ) o x. Howeve, x doesn t ceck in te oiginal equation, since, not. Once again, te oot 7 of all of ou oblems lies in te fact tat a negative numbe to an even owe esults in a ositive numbe. In ote wods, aising bot sides of an equation to an even owe does not oduce an equivalent equation, but ate, an equation wic may ossess moe solutions tan te oiginal. Hence te cautionay emak above about extaneous solutions. Examle Solve te following equations.. (5x + ) 6. (5 w) 7 9. t + t +6. y x + x 6. n ++n 0 Fo te emaining oblems, assume tat all of te vaiables eesent ositive eal numbes. 8 6 Well, not entiely. Te equation x 7 as seven answes: x and six comlex numbe solutions wic we ll find using tecniques in Section.7. 7 Pun intended! 8 Tat is, you needn t woy tat you e multilying o dividing by 0 o tat you e fogetting absolute value symbols.

5 0.9 Radicals and Equations 5 7. Solve fo : V (R ). 8. Solve fo M : M M 9. Solve fo v: m m 0. Assume all quantities eesent ositive eal numbes. v c Solution.. In ou fist equation, te quantity containing x is aleady isolated, so we extact fout oots. Since te exonent ee is even, wen te oots ae extacted we need bot te ositive and negative oots. (5x + ) 6 5x + ± 6 Extact fout oots 5x + ± 5x + o 5x + x 5 o x We leave it to te eade tat bot of tese solutions satisfy te oiginal equation.. In tis examle, we fist need to isolate te quantity containing te vaiable w. Hee, tid (cube) oots ae equied and since te exonent (index) is odd, we do not need te ±: (5 w) 7 9 (5 w) 7 8 Subtact (5 w) 56 Multily by 7 5 w 56 Extact cube oot 5 w ( 8)(7) 5 w 8 7 Poduct Rule 5 w 7 w 5 7 Subtact w w 5+ 7 Divide by Poeties of Negatives Te eade sould ceck te answe because it ovides a eaty eview of aitmetic.. To solve t + t + 6, we fist isolate te squae oot, ten oceed to squae bot sides of te equation. In doing so, we un te isk of intoducing extaneous solutions so cecking

6 6 Peequisites ou answes ee is a necessity. t + t + 6 t + 6 t Subtact t ( t + ) (6 t) Squae bot sides t + 6 t + t F.O.I.L. / Pefect Squae Tinomial 0 t t + Subtact t and 0 (t )(t ) Facto Fom te Zeo Poduct Poety, we know eite t 0 (wic gives t ) o t 0 (wic gives t ). Wen cecking ou answes, we find t satisfies te oiginal equation, but t does not. 9 So ou final answe is t only.. In ou next examle, we locate te vaiable (in tis case y) beneat a cube oot, so we fist isolate tat oot and cube bot sides. y + 0 y + Subtact y + Divide by y + Poeties of Negatives! ( y + ) Cube bot sides y + ( ) y + 7 y 7 y 7 Subtact 7 7 y 7 7 y 7 5 Common denominatos Subtact factions Divide by multily by Since we aised bot sides to an odd owe, we don t need to woy about extaneous solutions but we encouage te eade to ceck te solution just fo te fun of it. 9 It is wot noting tat wen t is substituted into te oiginal equation, we get If te + 5 wee 5, te solution would ceck. Once again, wen squaing bot sides of an equation, we lose tack of ±, wic is wat lets extaneous solutions in te doo.

7 0.9 Radicals and Equations 7 5. In te equation x + x, we ave not one but two squae oots. We begin by isolating one of te squae oots and squaing bot sides. x + x x x Subtact x fom bot sides ( x ) ( x) Squae bot sides x x + ( x) F.O.I.L. / Pefect Squae Tinomial x x + ( x) x x + 8x Distibute x 5 8x x Gate like tems At tis oint, we ave just one squae oot so we oceed to isolate it and squae bot sides a second time. 0 x 5 8x x x 6 x Subtact 5, add 8x (x 6) ( x) Squae bot sides x x + 6 6( x) x x x x x Subtact 6, add x (6x 8x + 5) 0 Facto (x )(8x 5) 0 Facto some moe Fom te Zeo Poduct Poety, we know eite x 0 o 8x 5 0. Te fome gives x 5 wile te latte gives us x 8. Since we squaed bot sides of te equation (twice!), we need to ceck fo extaneous solutions. We find x 5 8 to be extaneous, so ou only solution is x. 6. As usual, ou fist ste in solving n ++n 0 is to isolate te adical. We ten oceed to aise bot sides to te fout owe to eliminate te fout oot: n ++n 0 n + n Subtact n ( n + ) ( n) Raise bot sides to te t owe n + n Poeties of Negatives 0 n n Subtact n and 0 (n )(n + ) Facto - tis is a Quadatic in Disguise At tis oint, te Zeo Poduct Poety gives eite n 0 o n + 0. Fom n 0, we get n, so n ±. Fom n + 0, we get n, wic gives no eal solutions. 0 To avoid comlications wit factions, we ll foego dividing by te coefficient of x, namely. Tis is efectly fine so long as we don t foget to squae it wen we squae bot sides of te equation. Wy is tat again?

8 8 Peequisites Since we aised bot sides to an even (te fout) owe, we need to ceck fo extaneous solutions. We find tat n woks but n is extaneous. 7. In tis oblem, we ae asked to solve fo. Wile tee ae a lot of lettes in tis equation, aeas in only one tem:. Ou stategy is to isolate ten extact te cube oot. V (R ) V (R ) Multily by to clea factions V R Distibute V R Subtact R V R Divide by R V Poeties of Negatives R V Extact te cube oot Te ceck is, as always, left to te eade and igly encouaged. 8. Te equation we ae asked to solve in tis examle is fom te wold of Cemisty and is none ote tan Gaam s Law of effusion. As was mentioned in Examle 0.8., subscits in Matematics ae used to distinguis between vaiables and ave no aitmetic significance. In tis examle,,, M and M ae as diffeent as x, y, z and 7. Since we ae asked to solve fo M, we locate M and see it is in a denominato in a squae oot. We eliminate te squae oot by squaing bot sides and oceed fom tee. M M M M Squae bot sides M M M M Multily by M to clea factions, assume, M 60 M M Divide by, assume 60 As te eade may exect, cecking te answe amounts to a good execise in simlifying ational and adical exessions. Te fact tat we ae assuming all of te vaiables eesent ositive eal numbes comes in to lay, as well. including a Geek lette, no less!

9 0.9 Radicals and Equations 9 9. Ou last equation to solve comes fom Einstein s Secial Teoy of Relativity and elates te mass of an object to its velocity as it moves. We ae asked to solve fo v wic is located in just one tem, namely v, wic aens to lie in a faction undeneat a squae oot wic is itself a denominato. We ave quite a lot of wok aead of us! m m m m v m 0 v c c m 0 Multily by v to clea factions c! v c m 0 Squae bot sides v c m 0 Poeties of Exonents m m v c m 0 Distibute m v c m 0 m Subtact m m v c (m 0 m ) Multily by c (c 6 0) m v c m 0 + c m Distibute v c m c m 0 m Reaange tems, divide by m (m 6 0) c m c m 0 v m Extact Squae Roots, v > 0 so no ± c v (m m 0 ) Poeties of Radicals, facto m v c m m 0 m v c m m 0 m c > 0 and m > 0 so c c and m m Cecking te answe algebaically would ean te eade geat ono and esect on te Algeba battlefield so it is igly ecommended Rationalizing Denominatos and Numeatos In Section 0.7, tee wee a few instances wee we needed to ationalize a denominato - tat is, take a faction wit adical in te denominato and e-wite it as an equivalent faction witout See tis aticle on te Loentz Facto.

10 0 Peequisites a adical in te denominato. Tee ae vaious easons fo wanting to do tis, but te most essing eason is tat ationalizing denominatos - and numeatos as well - gives us an ootunity fo moe actice wit factions and adicals. To el efes you memoy, we ationalize a denominato and ten a numeato below: and In geneal, if te faction contains eite a single tem numeato o denominato wit an undesiable n t oot, we multily te numeato and denominato by wateve is equied to obtain a efect n t owe in te adicand tat we want to eliminate. If te faction contains two tems te situation is somewat moe comlicated. To see wy, conside te faction 5. Suose we wanted to id te denominato of te 5 tem. We could ty as above and multily numeato and denominato by 5 but tat just yields: 5 ( 5 5) We aven t emoved 5 fom te denominato - we ve just suffled it ove to te ote tem in te denominato. As you may ecall, te stategy ee is to multily bot numeato and denominato by wat s called te conjugate. Definition 0.7. Congugate of a Squae Root Exession: If a, b and c ae eal numbes wit c > 0 ten te quantities (a + b c) and (a b c) ae conjugates of one anote. a Conjugates multily accoding to te Diffeence of Squaes Fomula: (a + b c)(a b c)a (b c) a b c a As ae (b c a) and (b c + a): (b c a)(b c + a) b c a. Tat is, to get te conjugate of a two-tem exession involving a squae oot, you cange te to a +, o vice-vesa. Fo examle, te conjugate of 5 is + 5, and wen we multily tese two factos togete, we get ( 5)( + 5) ( 5) 6 5. Hence, to eliminate te 5 fom te denominato of ou oiginal faction, we multily bot te numeato and denominato by te conjugate of 5: 5 ( ( + 5) ( + 5) 5)( + 5) ( 5) ( + 5) Wat if we ad 5 instead of 5? We could ty multilying 5by+ 5 to get ( 5)( + 5) ( 5) 6 5, Befoe te advent of te andeld calculato, ationalizing denominatos made it easie to get decimal aoximations to factions containing adicals. Howeve, some (admittedly moe abstact) alications emain today one of wic we ll exloe in Section 0.0; one you ll see in Calculus.

11 0.9 Radicals and Equations wic leaves us wit a cube oot. Wat we need to undo te cube oot is a efect cube, wic means we look to te Diffeence of Cubes Fomula fo insiation: a b (a b)(a + ab + b ). If we take a and b 5, we multily ( 5)( + 5+( 5) ) ( 5) ( 5) So if we wee caged wit ationalizing te denominato of, we d ave: 5 5 ( ( + 5+( 5) ) 5)( + 5+( 5) ) Tis sot of ting extends to n t oots since (a b) is a facto of a n b n fo all natual numbes n, but in actice, we ll stick wit squae oots wit just a few cube oots town in fo a callenge. 5 Examle Rationalize te indicated numeato o denominato:. Rationalize te denominato: Solution. 5 x. Rationalize te numeato: 9+. We ae asked to ationalize te denominato, wic in tis case contains a fift oot. Tat means we need to wok to ceate fift owes of eac of te factos of te adicand. To do so, we fist facto te adicand: x 8 x x. To obtain fift owes, we need to multily by x inside te adical. 6 x 5 x 5 x 5 x 5 x 5 x 5 x x Equivalent Factions Poduct Rule 5 x x 5 5 x x 5 Poety of Exonents Poduct Rule 5 x x 5 8 x x 5 x x Poduct Rule Reduce Simlify 5 To see wat to do about fout oots, use long division to find (a b ) (a b), and aly tis to 5.

12 Peequisites. Hee, we ae asked to ationalize te numeato. Since it is a two tem numeato involving a squae oot, we multily bot numeato and denominato by te conjugate of 9+, namely Afte simlifying, we find an ootunity to educe te faction: 9+ ( 9+ )( 9+ + ) ( Equivalent Factions 9+ + ) ( 9+) ( 9+ + ) (9 + ) 9 ( 9+ + ) ( 9+ + ) ( 9+ + ) 9+ + Diffeence of Squaes Simlify Simlify Reduce We close tis section wit an awesome examle fom Calculus. Examle Simlify te comound faction ten ationalize te numeato of te esult. (x + )+ x + Solution. We stat by multilying te to and bottom of te big faction by x + + x +. (x + )+ x + x + + x + x + + x + x + + x + ((((((( x + + x + ((((((( x + + x + + x + x + + x + x + x + + x + + x + x + + x + x + Next, we multily te numeato and denominato by te conjugate of x + x + +,

13 0.9 Radicals and Equations namely x ++ x + +, simlify and educe: x + x + + x + + x + ( x + x + + )( x ++ x + + ) x + + x + ( x ++ x + + ) ( x + ) ( x + + ) x + + x + ( x ++ x + + ) (x + ) (x + + ) x + + x + ( x ++ x + + ) x + x x + + x + ( x ++ x + + ) x + + x + ( x ++ x + + ) x + + x + ( x ++ x + + ) Wile te denominato is quite a bit moe comlicated tan wat we stated wit, we ave done wat was asked of us. In te inteest of full disclosue, te eason we did all of tis was to cancel te oiginal fom te denominato. Tat s an awful lot of effot to get id of just one little, but you ll see te significance of tis in Calculus.

14 Peequisites 0.9. Execises In Execises -, efom te indicated oeations and simlify.. 9x. 8t. 50y 6. t +t + 5. w 6w x c v. z +z c 8. 5 L 9. x +. t +t x z In Execises - 5, find all eal solutions.! ( ). x + (x) q ( x s " 8 t x) + ( t) x. (x + ) ( y) t 7. x t + 9. x + x y + y + 0. t + 6 9t. x x +. w w. x + x 5 5. x ++ x In Execises 6-9, solve eac equation fo te indicated vaiable. Assume all quantities eesent ositive eal numbes.! () 6. Solve fo : I b L 8. Solve fo g: T g. 7. Solve fo a: I 0 5 a 6 9. Solve fo v: L L 0 v c. In Execises 0-5, ationalize te numeato o denominato, and simlify x 7. x x c c. x + + x +. x + x 7 5. x + x

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