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Examples in the Calculus of Radical Quantities.

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4. Find the continued product of a, b, and ✔c.

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Ans.abc.

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5. Find the continued product of a ✔x, b ŵ y, c ↓ z.

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. 10. Find the continued product of a-, a3, a-t.

11. Multiply 6-2a-3 by abc.

Ans. a ᄒ ava

Ans. aTM1⁄2 b−‡c=

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Ans. 241.

17. Multiply—5—√ by −5+v}.

Examples in the Calculus of Radical Quantities.

18. Multiply 9+2 10 by 9-210.

Ans. 41.

19. Multiply 28+3√5−7 √ 2 by √ 72 — 5 √20

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21. Multiply a+b by va-b. Ans. a-b.

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22. Multiply ✔a+cb by ✔a—cb.

Ans, a-c2 b2.

23. Multiply ↓a3+b2 by ↓ a3 — ↓ b2.

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Ans. aab.

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To free an Equation from Radical Quantities.

204. Problem. To free an equation from radical quantities.

First Method of Solution. Free the equation from fractions, as in art. 112.

Bring all the terms multiplied by either of the radical quantities, whether they contain other radical quantities or not, to the first member, and all the other terms to the second member of the equation. Raise both members of the equation to that power, which is of the same degree with the root of the radical factor of the first member, and this radical factor will be made to disappear; and by performing the same process on the new equation thus formed, either of the other radical quantities may be made to disappear, and in most cases which occur in practice it will be found that the equation can, in this way, be freed from radical quantities.

205. EXAMPLES.

1. Free the equation

(a + x)* = (b + x)* — (c + x) *

from radical quantities.

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Solution. The square of this equation is

a+x=b+x−2 (b+x)* (c + x)*+c+x;

hence, by transposition,

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square

2 (b + x)* (c + x) * = x—a+b+c,

of which is

4(b+x)(c+x)=(x−a+b+c)2.

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from radical quantities.

Ans. x=1.

Ans. 4x3-17 x2+34x=57.

206. Scholium. There are cases, however, in which the preceding method of solution is inapplicable on account of the new radical quantities which are introduced by raising the second member to the required power; but in all cases the following method will be found successful.

207. Second Method of solving the problem of art. 204. Place each of the radical quantities equal to some letter not before used in the equation, and raise the equations thus formed to that power which is of the same degree with the root of its radical quantity, and substitute in the given equation for each radical quantity the corresponding letter. If, then, each letter, thus introduced, is considered to represent ɑ

To free an Equation from Radical quantities.

new unknown quantity, the new equations, thus formed, are of the same number with that of their unknown quantities; and, since they are free from radical quantities, all their unknown quantities but one can be eliminated by the method of art. 155.

208. EXAMPLES.

1. Free the equation

(x2 + x + 1)3 — (x2 — x + 1)3 = 1

from radical quantities.

Solution. Place

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y and z are eliminated between these three equations, the resulting equation is

27x48x339 x2-6x+28 = 0.

2. Free the equation

(x + x2)3 + (1 + x2)* = 1

from radical quantities.

Ans. x6+5x1+4x3 +7x2+8x=0.

209. When, in an equation, the same quantity is affected by different radical signs, these radical signs, expressed by fractional exponents, may be reduced

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