It will be shown hereafter, (Art. 396), that every equation has as many roots as there are units in the exponent of the highest power of the unknown quantity. We do not, therefore, by this method, in all cases, obtain all the values of the unknown quantity. Thus, in the preceding example, there are four values of x not determined.

2. Given 5x-4√x=33, to find the value of x.

3. Given, √x+12+1/x+12=6, to find the value of x.
Assume, x+12=y; then √x+12=y2, and

Or, without introducing a new letter y, we may consider the whole expression under the radical as the unknown quantity, and proceed to complete the square thus,

√x+12+√/x+12+1=6+1=35 ;

Extracting the root, /x+12+1=±1⁄2;

The principle of both operations is the same, but the use of the new letter renders the process more easily understood by beginners.

4. Given 3x2+√3x2+1=55, to find the value of x.
Adding 1 to each member, the equation becomes

The equation may now be solved like the preceding.

The values of x are +4, 4, +√21, and —√21.

Find the values of x in each of the following examples.

ART. 243. In the preceding examples the form of the trinomial equation, is either given or easily ascertained; but it sometimes happens that questions are given, in which the compound term is not presented to view, but which may be reduced to the form of a trinomial equation by the following method:

If the greatest exponent of the unknown quantity is not even, it must be made even, by multiplying both members of the equation by the unknown quantity. Then extract the square root to two or three terms, and if we find a remainder (omitting known terms if necessary), which is any multiple or any part of the root already found, the given equation may be reduced to a trinomial. of which the compound term will be the root already found.

Example. Given, x3—4ax2—2a2x+12a3—16a, to find x.

Multiplying both sides by x, and transposing, we have

xa—4ax3—2a2x2+12a3x—16a1=0.

Proceeding to extract the square root, we have the following

Hence, the given equation may be written thus:

(x2-2ax)2-6a2(x2-2ax)-16a1=0.

Assuming x2-2ax=y, we find y=8a2, or -2a2; then from the equation x2-2ax-8a2, or -2a2, we find

6. x1—8x3+10x2+24x=-5. Ans. x=5, -1, or 2±√5..

SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE CONTAINING TWO OR MORE UNKNOWN QUANTITIES.

ART. 244. Equations of the second degree, containing two or more unknown quantities, may be divided into two classes.

1st. Pure Equations.

2nd. Adfected Equations.

The first class embraces those equations that may be solved without completing the square; the second, those in the solution of which it is necessary to complete the square. The same equations, however, may sometimes be solved by both methods.

ART. 245. PURE EQUATIONS.-Pure equations may in general be reduced to the solution of one of the following forms, or pairs

We shall explain the general method of solution in each of

these cases.

1. To solve x+y=a (1), and xy=b (2), we must find x-y. Squaring Eq. (1), x2+2xy+y2=a2;

Multiplying Eq. (2) by 4, 4xy,

=4b;

x=1a±1⁄2√a2—4b.

Adding and dividing by 2,

Subtracting and dividing by 2, y=a√a2-4b.

The pair of equations (2) is solved in the same manner, except that in finding x+y we must add 4 times the second equation to the square of the first.

The pair of equations (3) is solved merely by adding and subtracting, then dividing by 2 and extracting the square root.

EXAMPLES IN PURE EQUATIONS.

1. Given, x2+y2=d (1), and x+y=a (2), to find x and y. Squaring Eq. (2), x2+2xy+y2=a2;

Whence, xa√2d—a2, y=ža‡ž√2d—a2.

2. Given, x2+xy+y2=91(1), and x+√xy+y=13(2), to find

By adding,

x+y=10.

Squaring, (5), x2+2xy+y2=100:

Squaring, (4), 4.xy = 36;

(5)

x2-2xy+y2=64, .. x—y=±8.

But, x+y=10, whence, x=9 or 1, and y=1 or 9.

Equations of higher degrees than the second, that can be solved by simple methods, are usually classed with pure equations of the second degree.

3. Given, x*+y3—6, and x2+y3—126, to find x and y.

In all cases of fractional exponents, it renders the operations more simple to learners, to make such substitutions as will render

the exponents integral. In this example, let x*=P, and y3=Q;

then x=P3, and y=Q3. The given equations then become,

Dividing Eq. (2) by (1), P2— PQ+Q2=21;

Squaring Eq. (1),

Subtracting,

P2+2PQ+Q2=36;

3PQ=15,.. PQ=5.

Having P+Q=6, and PQ=5, by the method explained in form (1), we readily find P=5 or 1, and Q=1 or 5.

Whence, x=625 or 1, and y=1 or 3125.

(x+y)(x2+y2)=580 (2), to find x and y.
x3—x2y—xy2+y3=160 (1), by multiplying.
x3+x2y+xy2+y3=580 (2), "

66

2x2y+2xy2=420 (3), by subtracting.

Add (3) to (2), x3-3x2y+3xy2+y3=1000.

Extract cube root,

From (3),

x+y=10.

xy(x+y)=210; .. xy=21.

From x+y=10, and xy=21, we readily find x=7 or 3, and y=3, or 7.

Let the following examples be solved by the preceding or similar methods.

5. x-y=2,
x2+y2=394.

6. x2+y2=13.

xy= 6.

Ans. x=15, or -13;

y=13, or -15.

Ans. x=3;

y=±2.