Ics Square Roots

8/8/2019 Ics Square Roots

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102. As before stated, it is customary to use some kindof a table i ns tead of ext ract ing the roots directly, s till thesquare root is so frequently required t ha t i t is well to learnhow to extract square root directly. There a re several goodmethods , and none are much harder than long. division. Themet hod g iv en her e i s the. s impl es t and eas ie st t o r ememberand app ly o f any . Fo r cube root and higher roots, all exactmethods are long and laborious. Th e method g iven in ~ h pages that follow is a general one, applicable to any ront, thei ndex of which i s a n i nte ge r, a nd is ver y e as y to remember

and raise the new number to the same power ; but'if the resul tso obtained is greater than the given number, subtract 1 frointhe right-hand figure of the root and r ai se t he new numbe rto same power. Th e result of these operations is to obtainpowers of two consecutive numbers having five significantfigures each, one of the powers being a little greater and theother a little l ess than the given number. Then proceedingexactly as previously described, divide the second differenceby the first difference and obtain four more figures of the root.Consider example 1, Art. 92 . Th e square root of 31,416

to five significant figures is 177.25. 177.252 = 31,417.5625,which is a little greater than 31,416; hence, subtracting 1from 177.25, 177.242 = 31,414.0176, which is a little l ess than31,416. Th e first difference is 31,417.5625 ~ , 3 1 , 4 1 4 . 0 1 7 6= 305449. Th e second difference is 31,416 - 31,414.017'6= 1.9824; 1.9824 -;- 3.5449 = .55922, or .5592 to four figures.Therefore, "'31,416 = 177.245592 to nine significant figures.Suppose it has been found that the cube root of 37,267

is 33.402 to five significant figures, and it is des ir ed t oobtain more figures. Proceed exactly as befor e. 33.4023= 37,266.397760808, which is a little less th an 37,267;hence, adding 1 to 33.402, 33.4033 = 37,269.744941827.Th e first difference is 37,269.744941827 - 37,266.397760808= 3.347181019; the second differe.nce is 37,267 - 37,266.397760808 = .602239192; .602239192 -;- 3.347181019 = .17992+,or .1799 to four figures. Therefore, ~ 3 ,267 = 33.402179911 .t o n ine significant figures.

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(e)

ARITHMETIC

(a) 55

(d) 1006106

61120

11121

1112203

SOLUTION.-

40-8

11 223EXPLANATloN.-Firstpoint off i nt o per iods of two figures

each. Now, find the largest single number whose square isless than or equal to 31, the first period. This is evidently 5,since 62 = 36, which is greater than 31. Write it to ther ight , a s i n l ong division, and also t o t he lef t , a s shown at (a).'I'his is the f irst f igure of the root. Now, multiply the 5 .at (a) b y th e 5 in the root, and write the result under t hefirst period, a s s hown at (b). Subtract and obt ai n 6 as aremainder.Add t he root' already found to th e 5 at (a), getting 10, and

annex a cipher to this 10, thus making it 10o_, as shown at (d),which call the f ir st t ri a l d iv i so r . Bring down the nextperiod, 50, and annex it to t he remainder 6, as shown at (c),which'call the f i r s t d iv idend . Divide the first dividend (c)b y the first trial divisor (d) and obt ai n 6, which is probably

SQUARE ROOT-EXACT METHOD103. The method is best explained by giving several

examples with full exp lana ti ons of each s tep. In order tomake t h work c le ar er t o t he student a nd e as ie r to follow,the figures i n t he r oo t and, the success ive numbers resul tingfrom their us e are pri nt ed in light and heavy-face typealternately.

EXAMPLE I .-Find the square root of 31,505,769. root3 1'5 0'5 7'6 9 ( 5 6 1 3 Ans.(h) 25(e) 65063 6

145711213366933669

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EXAMPLE 2.-W;hat is the square root of .000576?

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root3.00'00'00'00'00 ( 1 .1 3 205+ Ans.120 0189110010297100692417600001732025

27975

ARITHMETIC

1120

'221'l340

'3343

334602346223464005

SOLUTION. -

EXAMPLE I . -What is t he square roo t of 3? Find the result to fivedecimal places.

346405EXPLANATION.-Annex five periods of two ciphers each

to the right of t he dec imal point . The fi rs t f igure o f therO'ot is f ound to be 1. To ge t t he second figure, we findthat, on dividing 200 by 20, it is 10. This is evidently toolarge.

104. If the number is no t a perfect power, 'the root willconsist of an interminable nunlber of decimal places. Th er esul t m ay b e carried to any required number of decimalplaces by annexing periods of two ciphers each to thenumber.

EXPLANATION .-Beginning at the decimal' point andpointing off the number into periods of two figures each,it is seen that the first period is cotIiposed of ciphers; hence,the first figure of the root must be a cipher. Th e remainingportion of the solution should be per fect ly c lear f rom wha thas' preceded.

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root.00'05'76 ( .024 Ans.417 617 6

ARITHMETIC

224044t

SOLUTION.-

Read ve r y ca r ef u l ly t h a t pa r t of Ar t . 96 whi ch ispr in t ed in I ta l ics . '

the next figure o the roo t. Write 6 in the root , as shown, andalso add it to 100, t he t ri al divisor, making it 106. This iscal led the f i r s t com.plete div i sor .Multiply the first complete divisor, 106, by 6, t he secondfigure i n t he r oo t, .and sub tr ac t t he result f rom the first divi

dend (c); the remainder i s 14. Add the second figure of theroot t o t he complet e d iv isor , 106, 'and annex a cipher, thusgetting 1120, which call the second t r ia l div i sor . Annexthe next per iod, t o th e rema in der in the second' columnmaking it 1457, as sh own at (e), which cal l the seconddiv idend . Dividing 1457 by 1120, we ge t 1 as the nextfigure of the root. Adding t hi s l as t figure of the rootto 1120, the result is 1121, the second com.plete divisor .Multiplying the second c o ~ p l e t e divisor by t he t hi rd figureof the root and sub tr ac ti ng f rom the second dividend, 1457,the remainder is 336.Now, adding the last figure o f t he root to 1121 and annex-

ing a cipher as before) the resul t is 11220, the t h ir d t r ia ldivisor . Annexing the next and last period, 69, to theremainder in the second' column the r es ul t is 33669, theth i rd div idend . Dividing 33669 by 11220, the resultis 3, the fourth figure of the 'root. Adding the fourthfigure of the root to 11220, t he resul t is 11223, the t h i rdcom.plete ( l iv isor . lVlultiplying the third complete divisorby t he f ou rt h figure of the r oo t, t he result is 33669. Subt ra ct ing , th e p roduct f rom t he t hi rd dividend, there is noremainder; hence, "'31,505,769 = 5,613.

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Trying 9, we a dd 9 to 20 and multiply 29 by 9; the resultis 261, a result which is considerably larger than 200; hence,9 i s t oo l arge . In t he s ame way, it i s f ou nd that 8 i s a ls otoo large. Trying 7, 7 times 27 is 189, a result smaller than200; therefore, 7 is the second figure o f t he r oo t. Th e nexttwo figures, 3 and 2, ar e easily found. Th e fifth f igure inthe root is a cipher, s ince t he t ri al divisor, 34640, is greaterthan the' new dividend, 17600. In a case of t hi s k ind weannex ano ther c iphe r ~ 34640, thereby making it 346400,and b ri ng down t he next period, making the 17600, 1760000.Dividing the fourt h dividend, 1760000, by the four th t rialdivisor, 346400, the result is 5.0+. Hence, the next figureof the root is 5, ~ n d , as we now have five decimal places,w'e stop.The square root of 3 is , then, 1.73205+ . .If the second figure o f t he quo ti en t last obtained, 5.0+,had been 5 or a greater digit , the figure' in th e fifth decimal

place would have been increased by 1.E ~ A M P L E 2.-What is the square root of.3 to five decimal places?

ARITHMETIC 47

Ans.

ARITHMETIC

11206266S20013201

SOLUTION . - -EXAJ.\iPLEc-What is the square root of 258.2449? root2'5 8.2 4'4 9 (1 6.0 'l1

15 81562244922449

EXPLANATION.-In the above example, since 320 is greaterthan 224, we place acipher for the$ i rd figure of the rootand annex a cipher to 320, making it 3200. Then, bringingdown the next period, 49, 7 is found to be the fourth figureof the root. Since there is no remainder , the square root of258.2449 is 16.07.

105. If it is required to find the square root of a mixednumber, begin a t t he decimal point and point off the periodsbo th to the right and to the lef t. The manner of finding theroot will then be exactly t he s ame a s i n t he p revi ou s cases.

106. PROOF .-To prove square r oo t, s quar e th e resultobtained. If th e n umber is an e xa ct power, the squ are ofthe root will equal it; if it is no t an exact power, the squareof the root will very nearly equal it .107. Rule . - I . Begin at units place and separate the

number into periods of two figures each, proceeding from lef tto right with the decimal part, i f there be any.

I I . Find the greatest number whose square is contalnedIn thefirst, or leftoohand, period. Write this number as the flrst figure

. in the root; also, write it at the lef t of the given number.

EXPLANATION.-In the above example we annex a cipherto .3, making the first period .30, since every perio'd of adecimal, as was mentioned before, must have two figures init. Th e retnainder of t he work should be per fect ly clear.

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root.30'00'00'00'00 ( .5 4 712+ Ans.25500416840076097910016629247100219084-----2801 ..

109542

109411----1095402

55lOGt1044108071087

710940

1

SOLUTION.-

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MultiPly this number at t h left ,by the first figure of the root,and subtract the result fronz the first period.

I I I . Ad d the first figure of the root to the number n thef ir st column on the lef t and annex a cipher to t ~ result/ this isthe first trial divisor. Annex the secondperiod to the remainderin the second c o l u 1 n n ~ this is the first dividend. Di'vide thediv idend by the trial divisor for the second figure in the rootand add this figure to the trial divisor to form the completedivisor.. Multiply the conzplete divisor by the second figure inthe root and subtract this result fronz 'the dividend. (If thisresult is larger than the dividend, a snzal ler number must betried for the second figure of the root.) Ad d the second fzgure ofthe root to the complete divisor. Annex a cipher for a new trialdivisor, and bring down the third period and annex it to thelast renzainder for the second dividend.

IV . Continue in this manner to the last period, after which,i f any additional places in .the root are required, bring downcipher periods and continue the operation.

V. I f at any ti1ne the trial div isor i s larger than thedividend, place a cipher in the root, annex a cipher to the trialdivisor, and bring down anotherperiod.

VI. I f the root contains an intermz'nable decimal and t isdesired to termin.ate the operation at sonze point, say, t h fourthdecimal place, carry the operation one place farther, and i f thefifth f igure is 5 or greater, increase the fourth f igure by 1 andomit the szgn +.

1.08. Short Methodo-If the number whose root is to beextracted is no t an exact square, t he root will be an interminable decimal. It js the n usual to extra ct the root to acertain nUlnber of significant figures. In such cases thewor k may be greatly s h o r t e ~ e d as follows: Determine tohow many significant figures t he wor k is t o be c ar ri ed, s ayseven, for example; divide this number by 2 and take the nexthigher number. In t he above case, we have 7 -;- 2 =3 t ;hence, we take 4, the next higher number. Now extract the,root i n t he u sual manner unt il four significant figures have

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(a) 432(b) 1,432(c) 5,464(d) .10815-(e) 14.0761Ans. (f ) 997li) 1.5411(h) 1.275(i ) .55(j ) .75593-(k) .88388+

ARITHMETIC

EXAMPLE S FOR PRACT IC E~ ,~ i n d the square root of:

(a) 186,624.(b) 2,050,624.(c) 29,855,296.(d) .. 0116964.(e) 198.1369.(I) 994,009.(g) 2.375 to four decimal places.(h) 1.625 to three decimal places.(i ) .3025.(1") .571428.(k ) .78125.

been obtained. Then form the trial divisor in the usualmanner, bu t omi tt ing to ann ex t he cipher; divide t he l as tremainder by the t rial divisor, as in long division, obtainingas many figur.es of the quo ti en t as there are remain ingf igures of t he roo t, i n thi s case 7 - 4 = 3. The quotient soobtained is the remaining figures o f the roo t.

Consider example 2, Art. 104. Here there are fivefigures in the root. We there fore extract the root to threefigures in the usual manner, obtaining .547 for the firstthree root figures. The nex t t ri al d iv isor i s 1094 (with thec ipher omi tt ed ) and the last remainder is 791. Then, 791-;- 1094 = .723, and the next two figures o f the root are 72,t he who le root being .54.772+. Always carry-the divisionone place farther than desired, and if the last f igu re i s 5 ora greater d ig it , i nc rease the p receding figure by 1. Thismethod should not b e u sed unless the root i s to contain f i . v ~'or more figures.

NOTE.-If the last figure of the roo t found in the regu1a:r . ~ a n n e ris a cipher, carry t he process one p lace farther before dIVIdIng as"described above.

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2RITHMETIC8

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Ics Square Roots