for the third; and 3x ק×√ √ for the fourth. And

continuing the same process, we will find the next term and all after become 0; .. the power sought will evidently be 1+3+3x+x√√x.

6. By (Art. 29), (3+2√/5)2=32+(2,√5)2+2×3× 2√5=29+12√5.

(92.) EVOLUTION OF Surds.

1. The square root of 9 is 3, and the square root of 33 is—33×36, or 3; hence the quantity sought is 3/3.

2. The square root of 36 is 6, and the square root of 3/2—23×4—26, or √/2; hence the quantity sought is 66/2.

9

3. The cube root of 8 is 2, and the cube root of 3/5 _5§×§—5—2/5; hence the quantity sought is 25. 4. The cube root of 27 is 3, and the cube root of √7

is 7×3—7—/7; hence the quantity sought is 3×76, or 3/7.

=

5. The fourth root of 64 is (64)4=(64)15—(218)ik — 2(26)13, and the fourth root of 3/4 is 4×41-(22)12; hence the quantity sought is 2(26)12 (22) 122 (28) 1'4 — 2(256)12.

(94.) EXERCISES IN EQUATIONS.

1. √3x+4=5. Given equation.
3x+4=25, by squaring both sides.
x=7, by transp., and dividing by 3.

..

2. √4+5x=2+3x. Given equation.

4+5x=4+4√3x+3x, by squaring both sides.

2x=4/3x, by transposition.

4x2-48x, by squaring both sides.
..x=12, by dividing by 4x.

3. 3/2x+10+4=8, given equation, /2x+10=4, by transposition,

2x+10=64, by cubing both sides, .x= =27, by transposition and division by 2. 4.2+5=7, given equation,

2=4, transposing and squaring,

.. x=6, by dividing by 3.

5. √4x+17+6√x+2=8√x+3, given equation, √4x+17=2√x+1, by transposition, 4x+17=4x+4x+1, by squaring,

x=4, by transposing and dividing by 4, .. 16, by squaring.

√x2-7x+x-7-21, by mult. by x-7,
√x2-7x=28-x, by transposition,
x2-7x-784-56x+x2, by squaring,
49x=784, by transposition,
.. x=16, by dividing by 49.

8x+4

7.

==

√5+x

4/5+x, given equation,

8x+4=20+4x, by multiplying by √5+x,
.. x=4, by transposing and dividing by 4.

√7x+x2+1=13, by multiplying by √x,
x2+7x=144, by transposing and squaring,

3+2√1+x=1+x, by multiplying by 3/1+x, 4+4x=x2-4x+4, transposing and squaring, 8x=x2, by transposing,

.x=8.

2a

10. √x+√x+a=] given equation,

√x2+ax+x+a=2a, by multiplying by √x+a, x2+ax=a2—2ax+x2, transposing and squaring, 3ax-a, by transposition,

11. √x−√α=✔ax, given equation,
(1—√α) √x=√a, by transposition,

Ja

.. X=

a

by dividing by 1—✔a.

(1—√āja› by squaring.

12. √x+a+√a—x=b, given equation,

x+a+2√a2—x2+a—x=b2, squaring both sides,
2/a2 —x2-b2-2a, by transposition,
4a2-4x2-b4-4ab2+4a2, by squaring,

4x2-be (4a-62), transp. & changing signs,

b

.. a—2 (4a—b2), div. by 4 and ext. root.

13. √1+x√x2+12=1+x, given equation, 1+x√x+12=1+2x+x2, squaring both sides,

√x2+12=2+x, subt. 1 and dividing by x, x2+12=4+4x+x2, by squaring,

..x=2, transposing and dividing by 4.

√16x+x2+16+x=40, by mult. by ✔]6+x,

√/16x+x2=24−x, by transposition,
16x+x2-576-48x+x, by squaring,
64x-576, by transposition,

.. x=9.

64√√x+1024=48x+1056, by mult. both sides by 768, 16x=32, by transposition,

x=4, by dividing by 16 and squaring.

E

15/6x=16/6x-6, clearing of fractions,

6x-6, transposing and changing the signs, .x=6, by squaring and dividing by 6.

·18. x+2x=24, given equation,

x+2x+1=25, completing the square,
x+1=5, by extracting the root,
..x=16 or 36, by transposing and squaring.

19. x2xx, given equation,

1

x-2=x, by dividing by x2,

x-x-2, by transposition,

x-x+1=1, by completing the square,

—=—, extracting the root,

x-2 or -1, by transposition, .. x 4 or 1, by squaring.

20. x2+x3-6, given equation,

+x+1=25, (Art. 94), by completing square, +1=2, by extracting the root,

x-2 or 3, by transposition,

..x=32 or -243, by raising to the fifth power.

21. 3x-2x-133, given equation,

Solving for x2 by formula, (Art. 80), we have

22. (x+12)*+(x+12)a—6, given equation.

This being of the form of an adfected quadratic, in which the unknown quantity is (x+12), we must first solve for (x+12), and then raising both sides to the fourth power, solve the result again for x.

(x+12)*+(x+12)*+1=, completing the square,

(x+12)+=+, extracting the root, (x+12)=2 or -3, by transposition, x+12=16 or 81, raising to the fourth power, ..x=4 or 69.

1

2

(95.) ARITHMETICAL PROGRESSION.

Proof of Theorems, page 64.

l=a+n-ld

n Given theorems.

s={2a+n―ld 2

1. Since la+n―ld, by equation (1),
l-n-ld-a, by transposing n-ld,

..a—l—n—ld.

2. Since s=

Q. E. D.

{2a+n-1d}, by equation (2.)

2s=2an+nxn-ld.

2'

2an-2s—nxn―ld, by transposition.

d.

3. From (1)n-ld—l—a, and

Q. E. D.