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3. Involve now the whole root, and subtract and divide as before ; and so on till the whole is finished.*
Examples. 1. Required the square root of a*—20*x+3aRx?—2ax' +r".
2. Extract the cube root of ro +6r-40x® +961-64.
I+6r'-40x3 +96r-64(1? +21-4
3. Required the square root of a +2ab + 2ac+1 +2bc+c%.
Ans. a +b+c. 4. Required the cube root of zo — 6% +157'-20ro #15r'-6r+1.
Ans. x2 - 2x+1. 5. Required the biquadrate root of 16a7-96aor+216a?r? — 216urs +8174.
Ans. 2a-37. 6. Required the fifth root of 32r» - 80x* + SOP3 – 40x2 + 101-1.
Ans. 2x-I. As this method, in ligh powers, is generally thought too laborious, it may not be improper to observe, that the roots of compound quantities may sometimes be easily discovered ilus :
1. Extract the roots of some of the most simple terms, and connect them together by the sign + or -, as may be judged most suitable for the purpose.
2. Involve the compound root, thus found, to the proper power, and, if it be the same with the given quantity, it is the root required.
3. But if it be found to differ only in some of the sigus, change them from t to —, or from to +, till its power agrees with the given one throughout.
Thus, in the fifth example, the root 22-3, is the difference of the roots of the first and last terms; and in the 3d example, the root a +b+c is the sum of the roots of the 1st, 4th, and 6th term. The same may also be observed of the 6th example, where the root is found from the first and last terms.
(59.) Equations have received different names, according to the power of the unknown quantity, or quantities, which they contain
Such equations, for example, as contain only the first, or simple power of the unknown quantity, or quantities, are called siMPLE EQUATIONS ; those which contain the square, QUADRATIC EQUATIONS ; those which contain the cube, CUBIC EQUATIONS, &c. Thus,
* # 6y = 40-10} are simple equations ;
6x +4 10.r Or,
10r + r = 47-17 Or, 61° + 4y = 10y-3 ?
are quadratic equations ; 10r - 2r?=6r ti
are cubic equations. Or, xy + y'=41--Or -3b8. There are four processes, or rules, which are to be applied in the same way, and in the same order, for the resolution of all equations, and wlich will, as it were mechanically, determine the value of the mknown quantity. These processes are grounded upon the following
AXIOMS. (60.) 1. If equal quantities be added to equal quantities, their sums will be equal.
2. If equal quantities he subtracted from equal quantities, their remainders will be equal.
3. If equal quantities be multiplied by equal quantities, their products will be equal.
4. If equal quantities be divided by cqual quantities, their quotients will be equal.
(61.) Rule I. Any quantity may be transferred from one side of the equation to the other, by changing its sign. Nole.-This rule is founded on Axioms I. and II.
EXAMPLE I. If r + 8 = 15, by subtracting 8 from each side of the equation, then will 2 + 8 8 - 15 -- 8; but 8 - 8 = 0; therefore r = 15 8.
Corol. Hence it appears, that + 8 may be transforred to the other side of the equation by changing it to 8; and we obtain, by this means, the value of x, which is 15 -- 8. or 7.
2. If I + 3 = 7, tlien will r = -7 34. 3. If
xa+b=0 d, then will x =(-d + Q . 4. If 41 8= 3.7 + 20, then will 4.1 37 = 20 + 8, Or I 28.
Examples for Practice. 5. 3.1 5 2x + 9 required the value of x. Ans. I 14. 6. 2.1 + 3 I +17....
Ans. = 14. 7. 5.r 4 : 4.0 + 25
Ans. I = 29. 9 = 6r-3...
Ans. I 6. 9. 4.1 + 2a= 3.1+ gb
96--2a. (62.) Rule II. If the unknozun quantity has a coefficient, then its value may be found by dividing both sides of the equation by that coefficient. Note. This rule is founded on the 5th Axiom.
Examples. 1. Let 4r=28, then, by dividing each side of the equation by 4, 40 28
57; therefore, r=7. 4
+ 2. If br=20—4r, then will x=2. 3. If 12x+20=6r+44, then will x=4. 4. If 25x+10=20x+50, then will x=8.
Examples for Practice. 5. 10.r = 150.
Ans, r=15. 6. 30.r+8=68...
Ans. r= 2. 7. Tôr +14=12r+54
Ans. I=10. 8. gr-3=4r+22....
Ans. r= 5. 9. 175-47+9=3x +39.
Ans. x= 3. 10. 12.r+6.r +18=9r+36.
Ans. = 2. (63.) Rule III. An equation may be cleared of fractions by multiplying each side of it by the denominators of the fractions in succession. Or, . all the denominators may be taken away in one operation, by multiplying each term by their least common multiple. Note.—The foundation of this rule is found in Axiom 3d.
Examples. 1. Let
= 12; then, by multiplying each side of the equation 6 6x
бr by 6, we shall have
12 X 6; but
=x, and 12 x 6=72; 6
6 therefore, I = 72.
. 2. If
+ =7, then will r = 10.
10r Again, multiply by 5, and 5x +
= 70. 5
70 Or, 5x+21=70; that is, 7x=70, or I=
7 which is the answer, az above.
s. If + + 4
= 13, then will £* 12.
We have 6x +4x+30=156;
(64.) Note. In the solution of the following questions, the learner must observe the following rules :-
Ist. To clear the equation of fractions by Rule 3d.
2d. To collect the unknown quantities on one side of the equation, and the known on the other, by the First Rule; and,
3d. To find the value of the unknown quantity by dividing each side of the equation, by its co-efficient, as in Rule 2d.
Examples for Practice.
4. Let 29+ = 22 ; it is required to find ? Ans. s=24.
(65.) QUESTIONS WHICH INVOLVE EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY,
Note: — These questions are designed to shew the method of applying the foregoing roles, previously to our entering upon the management of equations with two naknown quantities.
QUESTION 1. There are two numbers whose difference is 12, and their sum 20; what are the numbers ?
Obs. As their difference is 12, the greater aamber must evidently exceed the less pumber by 12.
Solu. Let x = the less number; then x+12 will be the greater
But by the equation, the sum of the two numbers is 20; or, the greater + the less number = 20. Hence, by addition, (x + 12) + x = 20. That is, 2x + 12
= 20. therefore, 2x
= 20 12 x: 8.
8 and ..
= 4 the less No.
2 Hence the greater number=m+12=4+12 = 16.
QUESTION 2. There are two numbers whose difference is 9, and if three times the greater' be added to five times the less, the sum will be 35; what are those two numbers?
Solu. Let x = the less ; then x+9 = the greater number.
5r. But, by the conditions of the question, 3 times the greater + 5 times the less number = 35.
Hence, by addition, (31 + 27)+5x = 35;
::====the less number;
x x +10=3x x + 10
and +9=1+9=10 the greater number. QUESTION 3. What number is that to which if we add 10, three-fifths of the sum shall be 66 ?
Solu. Let r=the number sought, then x+10 = the number + 10. Now2ths of x+ 10 3
5 3 But, by the question, 5ths of 3 + 10 = 66.
3x + 30 Hence, by substitution, Multiplying by 5, we have 3r+ 30
:. 37 =330-30=300, and x = =100=the number required.