first and third, shall be equal to 28; and of the sum of the three numbers equal to the third minus 1.

Ans, 40, 36, and 24.

QUADRATIC EQUATIONS.

89. The definition of a quadratic equation has been already stated in No. 76.

of

Quadratic equations are of two kinds, Pure and Adfected. A pure quadratic is that which contains only the square the unknown quantity. Thus, x2 = 49; x2+7=43, &c. are pure quadratics. An adfected quadratic is that which contains both the first and second power of the unknown quantity. Thus, x2+ax=b; x2-6x=16, &c. are adfected quadratics.

On the Solution of Pure Quadratic Equations.

90. For the solution of pure quadratics, the following rule may be given: Transpose the terms of the equation in such a manner, that those which contain x2 may stand on one side of the equation, and the known quantities on the other; divide by the coefficient of x2, if it have one, then extract the square root of both sides of the equation, and the result will give the value of x.

Ex. 1. Given x2+11=20, to find the value of x.

Ex. 2. Given 5x2 + 7 = 3x2 + 39, to find the value of x.

And hence, x =±√16=± 4.

*The signs prefixed, indicate that the value of x is either + 3 or 3. That this is the case, in every such instance, is manifest from the consideration, that the square root of a positive quantity is either + or -. Thus, 3 x + 3 = +9, and -3 X -3, also + 9. Hence it follows, that the unknown quantity in every quadratic equation will have two values.

-

Ex. 3. Given ax2+bc, to find the value of x.

When the value of the square of the unknown quantity is not a complete power, the value of the simple quantity can only be obtained by approximation.

On the Solution of Adfected Quadratic Equations.

91. Since the square of a binomial is equal to the squares of its two terms, together with twice their product, (p. 82,) it follows, that if the square of either of the terms, and

twice the product of the two, be given, the other, or its square, may be found. For it is manifest, that the coefficient, or quantity which multiplies the given term in twice the product, is equal to twice the other term; and hence the half of this coefficient will be the other term itself. Thus, if x2+8x be the square of the first term, and twice the product of the two terms of a binomial, then the other = 4, and hence its square = 16.

term will be equal to

8
2

=

The complete square, therefore, of the binomial x+4 is x2+8x+16. In the same manner may the square be completed in every instance.

This process of adding the third term to the two first terms of the square of a binomial is very appropriately called completing the square, and is of the most essential service in the solution of adfected quadratics.

92. From what is stated in the preceding number, we are furnished with the following rule for the solution of adfected quadratics.

Arrange the terms, containing the unknown quantity, on one side of the equation, beginning with that involving the highest power, and transpose all the known quantities to the other side. Then, if that term containing the highest power of the unknown quantity, has any coefficient, either positive or negative, let all the terms be divided by it. If now the square of half the coefficient of the second term be added to both sides of the equation, that side which involves the unknown quantity will become a complete square. Lastly, extract the square root of both sides of the equation, and a simple equation will be obtained, from which the values of the unknown quantity may be easily determined.

EXAMPLES.

Ex. 1. Given x2+8x=48, to find the values of x.

Ex. 2. Given x2 - 6x + 7 = 23, to find the values of x.

Ex. 3. Given x2 + ax = b, to find the values of r.

Ex. 4. Given 3x2 + 12x· 15165, to find the values of x.

The equation must first be cleared of fractions, which may be done by multiplying by 12, the least common multiple of the denominators 3, 2, and 4, we shall then have 8x2. 18x+1149x2-24x+87.

18x + 24x87—114.

27; and, by changing the signs of

Comp. square, x2-6x+9=27+9= 36.

Ext. root, x-3=√√36±6.

Transp. a6+3=9 or- 3.

Ex. 6. Given 3x2 + 2x = 161, to find the values of r.

example is, the half of which is, and, therefore, its square

In this example, both values of x are positive. This will always be the case when the coefficient of x is greater than x itself, and is, at the same time, negative.

93. All equations containing two different powers of x, may be solved in the same manner, provided the index of x in the one term be the double of that in the other.