Transposing one of the radicals, we obtain

Squaring, we have

√a+x=c−√b+y.

a+x=c2-2c√b+y+b+y.

Transposing, so as to bring the radical to stand alone, we

have

2c√b+y=c+b+y-a-x,

which may be freed from radicals by squaring a second time. Sometimes the two radicals may be of such a form that it is best to bring both to the same member of the equation before involution.

When an equation contains several radical quantities, it may generally be freed from them by successive involutions, but the best mode of procedure can only be determined by trial. Ex. 4. Free the equation

√2x+7+√3x-18=√7x+1

from radical quantities.

Ans. 6x-15x-126-x2+12x+36. When an equation contains a fraction involving radical quantities in both numerator and denominator, it is sometimes best to render the denominator rational by Problem XI.

Ex. 5. Free the equation

√x+ √x a an2

√x-√x -a

x-a

from radical quantities.

Multiply both terms of the first fraction by √x+√x−a, and we have

(177.) The preceding rules for the reduction of radicals, are exact so long as we treat of absolute numbers, but require some modifications when we consider imaginary expressions, such as √3, √—a, &c.

Let it be required, for example, to determine the product of √-a by √-a.

By the rule given in Art. 164,

√-ax√-a=√-ax-a

= √+a2.

Now, √+a2±a, so that there is apparently a doubt as ic the sign with which a ought to be affected in order to answer the question. However, the true result is -a, because any quantity must be equal to the square of its square root.

That is, vaxa is the same as (-a)', and, consequently, is equal to -a.

Again, let it be required to determine the product of √-a by v-b.

The result, however, is not properly ambiguous, and should be-Vab; for we have, according to Art. 148,

In the same manner we shall find for the different powers of V-1 the following results.

Since the four following powers will be found by multiplying +1 by the first, the second, the third, and the fourth powers, we shall again find for the four next powers

so that all the powers of ✔-1 will form a repeating cycle of these four terms.

Whenever the student is at a loss to determine the product of two imaginary quantities, it is best to resolve each of them into two factors, one the square root of a positive quantity, and the other -1, Art. 148.

EXAMPLES.

1. Let it be required to multiply ✔—9 by √—4.

SECTION XII.

EQUATIONS OF THE SECOND DEGREE.

(178.) According to Art. 96, quadratic equations, or equa tions of the second degree, are those in which the highest power of the unknown quantity is a square.

Quadratic equations are divided into two classes.

I. Equations which involve only the square of the unknown quantity and known terms. These are called pure quadratics. Of this description are the equations

ax2-b; 3x+12=150-x', &c.

They are sometimes called quadratic equations of two terms, because, by transposition and reduction, they can always be exhibited under the general form

ax'=b.

II. Equations which involve both the square and the first power of the unknown quantity, together with a known term. These are called affected or complete quadratics. Of this description are the equations

They are sometimes called quadratic equations of three terms, because, by transposition and reduction, they can al ways be exhibited under the general form

ax2+bx=c.`

PURE QUADRATIC EQUATIONS.

(179.) The equation

ax2=b

is easily solved. Dividing each member by a, it becomes

b

a

Whence

b

If be a particular number, either integral or fractional, we

a

can extract its square root either exactly or approximately by the rules of arithmetic.

It is to be remarked, that since the square both of +m and

-m is +m2, so, in like manner, the square of +

✓ and that

are both + Hence the above equation is suscep

+

a

tible of two solutions, or has two roots; that is, there are two quantities which, when substituted for x in the original equation, will render the two members identical. These are

For, substituting each of these values in the original equation ax=b, it becomes

2

ax(+1

and

ax(−√√/-)'=b, or ax2=b; i. e., b=b.

Hence the two values of x are +4 and -4, and they may both be verified by substitution in the original equation. Thus, taking the first value, we have