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4. Reduce (c) and (1)3 to equivalent surds having
and (54) 1's.
a common index. Ans. (a)'s ar
5. Reduce (2 (2%) and (4) to equivalent surds hav
ing a common index. Ans.
88. To add or subtract surds.
RULE. Reduce the surds to their simplest form; then if they have the same radical quantity in each, the sum of the coefficients prefixed to this radical will be their sum; and the difference of the coefficients prefixed to the radical will be their difference. But if they have different radi. cal quantities, their sum can only be indicated by placing the sign plus between them, and their difference by placing the sign minus between them.
The reason of this rule is obvious, for the radical quantity may be represented by a letter, and then the rule will be identical with that of addition and subtraction in alge bra.
EXAMPLE. What is the sum and difference of √288 and 128. Here √288=√144×2=12√2, and 128 =√64×2=8√2; hence their sum is 20√2, and their difference is 42.
1. Find the sum and difference of 3/32 and 2/162.
Ans. sum 30/2, diff. 6/2
2. Find the sum and difference of 3,3/54 and 3/250.
Ans. sum 143/2, diff. 43/2.
3. Find the sum and difference of 3/24a4 and/192a.
Ans. sum (2a+4)/3a, diff. (2a—4)/3a. 4. Find the sum and difference of 80 and 45.
Ans. sum 75, diff. 5.
5. Find the sum and difference of (363) and (98a). Ans. sum (6a√a+7√2a), diff. (6a√a_7√2a.) 6. Find the sum and difference of (1000) and (300a3).
Ans. sum (10a2/10a +10a√3a), and diff. 10a2/10a -10a√3a.
TO MULTIPLY SURDS.
89. RULE. Reduce the surds, if necessary, to a common index, then multiply the coefficients together for a coefficient, and the surd quantities together for the surd, over which place the common index.
EXAMPLE. Multiply 3/10 by 2 3/12.
3x2=6, the coefficient, and 3/10/12/120.
.. 6/120 is the result; which, however, can be simplified; for /120-3/8x15-23/15; hence the quantity in its simplest form is 123/15.
EXAMPLE 2. Multiply ✔a by 3/b.
Here √a=a3=a&=(a3)3, and‚3⁄4/b=(b)3=b8=(b2)& √ã×2/5=(a3)3× (b2)3=(a3b2)ð.
1. Multiply 5 √5 by 3√8. 2. Multiply (18) by 53/4. 3. Multiply 10 by 15.
Ans. (233255) or 225000.
90. RULE. Reduce the quantities, if necessary, to a common index, then divide the coefficients and the surd quantities separately as in rational quantities.
EXAMPLE. Divide abac by b3/bc.
Here ab÷ba, the coefficient, and ac÷be the surd
.. the quotient is a .
EXAMPLE 2. Divide 3 ac by 23/bc.
Here the quantities reduced to a common index become 3(a33), and 2 (622).. the coefficient of the quotient
simplest form is (abc), and hence the quotient is
91. RULE. Raise the coefficient of the surd to the required power, and then multiply the exponent of the surd by the exponent of the power.
EXAMPLE. Find the third power of 2√/ac.
Here we raise 2 to the third power, which gives 8, and then multiply, the exponent of the surd, by 3, the exponent of the power, which gives .. the third power of 2√ac is 8(ac)13—8(a2c2 × ac)=8ac(ac) or 8ac√ ac. 1. Raise 2(ac)3 to the second power.
2. Raise 4(bcx2) to the third power.
6. Raise (3+25) to the second power.
EVOLUTION OF SURDS.
92. RULE. Extract the required root of the coefficient, and then multiply the fractional exponent of the surd by the fractional exponent of the root.
EXAMPLE. Extract the square root of 9 ab.
Here the square root of 9 is 3, and the fractional exponent of the surd is, which we are to multiply by the exponent of the root, which gives; hence the quantity sought is 3(ab).
1. Extract the square root of 93.
5. Extract the fourth root of 64/4. Ans. 2× (256)TM1⁄2
EQUATIONS CONTAINING SURDS, ETC.
93. In equations containing surds, before the solution can be effected, the surds must be cleared away; to effect this, transpose all the terms which do not contain surds to one side of the equation, and the surds to the other, then raise both sides to a power denoted by the index of the surd, and if there was only one term containing a surd, the surd will be cleared away, if there be more than one, the operation must be repeated.
If an equation appear under the form xa√xb, or x2axc, it can be solved as an adfected quadratic, by solving first for the power in the second term, and then for the quantity itself.
EXAMPLE. Given √x+9=√x+1.
Squaring both we have x+9=x+2√x+1.
EXAMPLE 2. Given x3+x=72, to find the value of x. Here the equation comes under the form x2+x=c, since the exponent of the first term is double its power in the second; hence we must solve for x. The operation will be as follows: x3+x2=72.
2+x+1=209, by completing square.
+7, by extracting the root.
x2-8 or-9, by transposition. and x64 or 81, by squaring..
hence x=4 or 33/3, by extracting the cube root.
Ans. x 6.
4. Given /2+5=7; to find æ.
5. Given √4x+17+6√x+2=8√x+3; to find x.
6. Given √x+√x—7
11. Given √√√√ax, to find x. Ans. x=
12. Given √x+a+√a—x=b, to find x. b
Ans. x= 1⁄2 (4a—b2) §.
13. Given1+x√x2+12=1+x, to find x. Ans. x=2.
-; to find x.
9 or —
; to find x.
Ans. x 8.
to find x. Ans. x=
to find x. Ans. x=9.