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1334025 1155

1155 934 221

or x == 48841

or -1, and 221

221 the other two roots will be found from the equation

r+a+2r)x+6+2ar+3r=9, or 240-289, r being one of the equal roots before obtained.

And if the equal roots have contrary signs, we shall have albe-ad)=co, and 2=EN-, as before ; the other two roots -)

x= being found from the equation 2 + ax + 6 + g = 0.

4. Also, if the general biquadratic equation 2 + ax +bx'ta td=1, has three equal roots, with the same sign, each of them will be a root of the equation 6x?+3ax=_b. And if two of the three equal roots have contrary signs, we shall have

albc — ad)z=co, and =EN- as in the last case.*

a 1. Supposing that two of the roots of the equation and + 2 + 33 + 63 = = 0, are equal to each other, and have the same sign, it is required to determine them. Here a being =1, 33, and c=63, we shall have, by the

=b= rule above laid down, 3x+2ax+b=3x+2x433=0; or 2fit=11;whence —+1(+11)=-Niges-1+4=

-$ 3, or =- -*-=-, the former of which values (3) being substituted for 2, in the original equation, is found to succeed.

Whence two of the three roots are each = 3; and the remaining root -7.

2. Supposing that two of the roots of the equation *+1242 +2+ =, are equal to each other, and have the same sign; it is required to determine them.

* The mode of solution here given might have been easily extended to equations of the 5th, 6th, &c. power ; but in those above the 4th degree, the expressions for the equal roots, when they have the same sign, are too complicated to be of any practical Rules might have been also introduced for determining, a priori

, by meams of the relations of the coefficients alone, whether or not any given equation has equal roots ; but, except in the case where they have contrary signs, this may in general be more readily done, by finding the root of the quadratic equation to which it is reducible, and then seeing whether the root, thus obtained, be a root of the original equation or not.

In all cases of this kind, however, it may be of use to know, that an equation cannot have equal roots when the last term, and the coefficient of the last but one, are prime to each other ; but the reverse of this is not always true.

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Here a being =,b-, ==, and das, we shall have, by the rule, (3a-86)2 +2(ab6c)x+ac-16d=22-182=0. Or, by transposition and simplifying the terms, 2-1f=; the two roots of which equation are s and =186.

And by trial, the first of these, ß, will be found to be a root of the original equation. Whence the 2 roots required are each=f.

And, since r = , we shall have, by the rule, za+la+2r)x+6+2ar +3/+ (+1)x-*+*+1=1, or 2+1=-; the roots of which latter equation are - $, Hence the four roots of the given equation are }, }, - }, and

3. It is required to determine whether the equation x+112+ 1922—992-252—1, has equal roots, and if so, what they are. Here, since a=11, b=19, =-99, and d=259,

: we shall have, by the rule asbc-ad)=11(-19X99+11 X 252) 11-1881+2772)=9801=99c;whence the equation has two equal roots, with contrary signs ; and, consequently, =EN-=EVH= 19= 3, and — 3, which two x +፡ +

-3 roots will be found to answer the conditions of the question. Also, since =+3, we shall have, by the rule, 29+ ax6+p = 22 + 11x + 28 = 0,

or 2 + 112 =

28; the roots of equations are -4 and —7 Whence the four roots of the given equation are 3, -3, -4, and -7.

1. Supposing that two roots of the equation _48x_-128=0, are equal, and have the same signs ; it is required to determine them.

Ans. -4, -4, and 8. 11. A cubic equation of the general form 28+px?+qxtr=0, has two equal roots with the same signs. Let its 3 roots be de. noted by a, b, c; then we shall have the two following equations; {a+pa+qatr=0}-{2+pb+qb+==~—69+pla' -—62) tala—6=0; and when divided by ab, will give a' tab + 5* tplató+e=0. If we now take b=0, the reduced equationin

+b)q this case will be 3a + 2pa +9=0, or =3x?+2p=+=0, where one of the values of x, taken twice with the same sign, will give the two equal roots of the proposed equation.

2. Given the equation 2+5464-782-1=0, to find whether it has equal roots, and if so, what they are. Ans. 4,-,&-51.

3. Supposing the equation 24 +3a3–14x2—12x+40=0, to have three equal roots, two of which have the same sign ; it is required to determine them.

Ans. 2, 2, -2, and -5. 4. Supposing the equation 24-X-toxtoto=0, to have two equal roots, with the same sign ; it is required to determine all its four roots.

Ans. -4,-, and 1%.

-1, 3

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Of Exponential Equations. 111 An exponential quantity is that which is raised to some unknown power, or which has a variable quantity for its indes, as a, a, z®, or ał, &c.

a*
*

See Logarithms page 302. And an exponential equation is that which is formed between any expression of this kind and some other quantity whose value is known ;: asazh, x=a, &c. ; where it is to be observed, that the first of these equations a*=b, when converted into logarithms,

log. 6 is the same as z log. a=log. b; or x= ;

and the second

log. a x=d, when so converted, is the same as x log, x=log. a; in the last of which cases the approximate value of 2 may be determined as follows:

Rule. Find, by trial, two numbers nearly equal to the value of x, and substitute them in the given equation, x log. x= = log. ar instead of the unknown quantity, noting the results obtained from each, as in the rule of Double Position, before laid down, in art. 31.

Then, by means of a certain number of successive operations, performed in the same manner as is there described, the value of & may be found to any degree of accuracy required.

1. Given x=100, to find an approximate value of x. Here, by the above formula, we have a log. x= log. 100 = 2,

And since x is easily found, by a few trials, to be nearly in the middle between 3 and 4, but rather nearer the latter than the for. mer, let 3.5 and 3.6 be taken for the two assumed numbers. Then log. 3.55.5440680; which, being multiplied by 3.5, gives 1.904238 = first result; And log. 3.6 3.5563025; which being multiplied by 3.6, gives 2.002689 for the second result

. Whence 2.002689....3.6....2.002689 1.904238.

...3.5....2. 098451 :

.002689 : .00273 for the first correction ; which, taken from 3.6, leaves 2= 3.59727, nearly. And as this value is found, on trial, to be rather too small, let 3.59727 and 3.59728 be taken as the two as. sumed numbers. Then log. 3.59727 = .5559731 ;which, being multiplied by 3.59727, gives 1.9999854 = first result.

And log. 3.59728 = .5559743; which, being multiplied by 3.59728, gives 1.9999953 = second result; whence

1.9999953. ...3.59728......2
1.9999854. .3.59727....1.9999053

0000099 : ..00001 :: .0000047 : .00000474747 for the second correction ; which, added to 3.59728, gives, 27 3.59728474747, extremely near the truth.

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And in the same way may the value of the unknown quantity be determined, in any other equation of this kind.

1. Given 2*=2000, to find an approximate value of 2.
Here x log. I

log. 2000

3.301030, and 2 is found, by a few trials, to be rather less than 5; let therefore 4.8 and 4.9 be taken for the assumed numbers ; then log. 4.80.681241, which, being multiplied by 4.8, gives 3.269956 = the first result; and

= log. 4.9 = 0.690196, which being multiplied by 4.9, gives 3.381960 = the second result. Whence 3.381960 4.9 3.381960 3.269956 4.8 3.301030 .112004 :. .1 :: .080930 : 00722 for the first correction, which, being taken from 4.9, leaves 1.8278, the answer nearly.

2. Given (6x)=96, to find the approximate value of x.

Here x log. 6x log. 96, and x is easily found,, by a few trials, to be rather less than 2 ; let therefore 1.8 and 1.9 be taken for the two assumed numbers; then log. 60(=10.8) 1.033424, which, multiplied by. 1.8, gives 1.860363 = the first result ; and log. 6x1 = 11.4) = 1.056905, which, multiplied by 1.9, gives 2.008119 the second result. Whence

2.008119 ... 1.9... 2.008119
1.860363 1.8 .. 1.982271

.147756 : .1 :: .025848 :.0174 for the first cor. 3 rection, which, taken from 1.9, leaves 1.8826 for the answer,

3. Given x=123456789, to find the approximate value of .

Here, after a few trials, or from inspection in a table of powers we find x is between 8 and 9, but nearer the latter than the form

Assume therefore x=8.6 and x=8.7. Then log 123456789 = - 8.0915149 Log.8.6=0.9344935 L.8.7=0.9395193 8.17381 8.7 8.09151 8.6

8.7 8.03664 8.6 8.03664 56069610

65766351 .13717 : 0.1 :: .05487 7.4759480 7:5161544

: 04 8.03664410 Res 8:17381791

Whence x = 9.6 + :04 = 8.640 nearly; And repeating the operation by assuming x equal to 8.64 and 8.641; x is found = 8.6400268

4. Given (22—æ+)*, to find the approximate value of x. Ans.x=1.47933.

log. 3m/log. a"
5. Given amatne na+m, to find x. Ans. X=
6. Given arý =m

log. am-log. Za sty=n

to find the values of x and y. ; See Index. 7. Given x=48, (k)=1, and w=100, to find x. Ans. x = 3.26, x= 3,8 and 2 = 3.5972.

a

er.

Mult. by

I

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; a=bEN (B2_4)

Here x is found, by a few trials, to be nearly equal to 1,7; let therefore 1.7 and 1.8 be taken for the assumed numbers : 1st Supposition.

2d Supposition. By logs. X= 2.46471

27

2.88072 0.76471

1.08072 2x = 93529

2x -x= .71929 log. (x-3-1.883496 log. (x-2)= 0.033713 log. (2x— *)=-1.982909 Llog. (22—2+)-1.864943 0.033713

-1.982909

results -1.864943

-1.883496 .168770

.098413 Therefore .168770 : .1::.098413 :.058. Whence 1.8 - .058 =1.742, the value fox, nearly; and, assuming 1.74 and 1.75 for the numbers, and repeating the operation, we get x=1.74793.

1 5. Given at ax=b, to find

the value of &.
ar

62
atba
51; q*_ba'+
4 4

2
b+ (12-4)
is 2 •log. a=log.

log. {bEN (6-4)-log: 2; log. {bEx(62—4)-log, 2 .

log. a 6. Given a tank, to find the value of x.

ab ams tram-b; or came + "; .. am*

6

at1 Take the log. and mz. log. a = log. (ab) --log: (a+1);

a log.(ab)-log (1+1)

milog. a 112. In considering the nature of an exponential of the form 2017 it must be observed that it means a to the power of 6*, and not al to the power of x, which would be abr.

7. Given ea, to find the value of x.

Find by trial a number, n, as near to the true value of x as can be done. Let n+y=x, then by taking the log. of the given equation,

Y (n+y).log.(n+y)= log. a; i. e. (n+y)•log. n:(1+=log. a,

y

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Or (n+y)-log. n+(n+y)M{

у

int&c.} =log. a;

Then by multiplying (n+y) into the series, and rejecting all the terms into which y, y, &c. enter, we obtain (n+y) log. nt

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