Required the 5a1d — 6c4n = 5 a d ±√5.a2d-3c2n.

It is manifest that, by strictly observing the nature of the indices, there is but little difficulty to be met with.

To find the square root of compound quantities.

RULE.

1st.-Range the quantities, as in Division; the square root of the first term will be the first term of the quotient set its square under the first term, and subtract.

Then bring down two terms to the remainder, for a dividend; set down double the root found, for the next divisor; next divide the first term of the new dividend by it, and place the result in both the quotient and the divisor; then multiply the whole divisor by that last term of the quotient, subtracting as usual, and so on until nothing remains, always doubling the quotient for a new divisor.

Find the square root of a2 + Qab + b2.

a2+2ab+b2(a + b root required.

Find the square root of a6 + 4a5 + 2a1 + 9a2 4a + 4.

Find the square root of x2 + 2xb + b2 + 2xc + 2bc + c2.

Ans...x+b+c.

The cube root may be extracted as follows:- Find the cube root of the first term, which root place in the quotient; subtract its cube from the first term of the dividend, and bring down the next three terms; divide the first term of the new dividend by three times the square of the quotient, and place the result in the quotient; then, to three times the square of the first term of the quotient, add three times the first term of the quotient into the second, together with the square of the last term, and it forms the divisor, which multiply by the last term, subtract, and thus proceed until nothing remains.

Find the cube root of a3 + 3a2b + 3ab2 + b3.

a3+3a2b+3ab2 + b3(a + b

a3

3a23ab+b2)3a2b+3ab2 + b3

3a2b+3ab2 + 63

Otherwise.

Having found the root of the first term, place it in the quotient, and subtract its cube from that term, bringing down the next term for a new dividend. Next involve the root found to its next lower power, multiplying it by the index of the given power, for a divisor; divide, and place the result in the quotient; then involve the whole root, subtract, and proceed as before.

a3 + 3a2y + 3ay2 + y3(a + y required root.

Find the cube root of a6 + 6a5 + 15a1 + 20a3 + 15a2 + 6a+ 1.

3a1)

3a1)

(a2 + 2a + 1

a6+6a5 +15a4 +20a3 + 15a2 + 6a+ 1

a6

6a5

a6+6a5 +12a4+8a3 = 3rd power of a2 + 2a = remainder.

3a4

a66a5+15a4 + 2a3 + 15a2 + 6a+ 1 =

3rd power of the quotient.

Any root out of any given quantity can be extracted by this Rule.

EQUATIONS.

It will be requisite here to briefly state the nature of ratio and proportion, as far as regards their utility in connexion with Equations.

When two quantities are compared together, that is to say, ab or 10: 5, the first is called the antecedent, and the second the consequent, and the quotient found by dividing the antecedent by the consequent is called the ratio.

If four quantities are proportional, the product of the means is equal to the product of the extremes; in the proportion a: b::c: d, a and d are the extremes, b and c the means. Wherefore, in order to prove the above numerically, they will stand thus,

10: 5 equivalent to = 2, the ratio.

10
5

Suppose 10: 5 :: 12 : 6, the product of the extremes will be 60, and the product of the means will be also 60; therefore the proposition is evident. To prove that they are in proportion, it will be simply

10

12

necessary to place them thus, = and, by clear5 6'

ing fractions, we have, 60 = 60; or the ratio of each = 2. An equation is the equality of two quantities. The assemblage of quantities at the same side of the sign is called the member or side; an equation has

two sides.

That which is at the left is called the first side, and the other is called the second.

In the equation, 2x + b = a, the first side is 2x+b, and the second side is a.

The quantities which compose the same side, when they are separated by + or—, are called terms. Thus, the first side of the equation 2x + b = a contains two terms, namely, 2x and b.

The equation x + 7 = 8x each of its sides, namely,

12 has two terms in

Although I may have taken the equation x+7= 8x 12, by chance, for to serve for example, it ought to be considered (likewise all those of which I shall speak) as coming from a problem of which we can always find an expression, by converting the equation into ordinary language. That of which we have spoken lastly shows how we may find a number such, that in adding 7 to of itself, the sum will be equal to 8 times itself, minus 12.

Similarly, the equation ax + be cx = ac — bx, (in which the letters a, b, and c, stand to represent known quantities, and x to represent unknown) answers to the following question :

To find a number (x) such, that in multiplying it by a given number (a), then adding the product of the two given numbers (b and c), and taking away from this sum the product of the given number (c) by the unknown term (x), there may be a result equal to the product of the numbers a and c, minus that of the numbers b and x.

To draw from an equation the value of the unknown quantities, or to have the unknown terms only on one side of the equation, and the known terms on the other, is what we call solving an equation.

The known and unknown quantities can be expressed by any letters we wish, but the unknown are usually expressed by x and y, and the known by the other letters of the alphabet.