ART. 208.—Every equation of the second degree, may be reduced to one of the forms ax2=b, or ax2+bx=c. For, in an incomplete equation, all the terms containing 2 may be collected together, and then, if the coefficient of x2 contains more than one term, it may be assumed equal to a single quantity, as a, and the sum of the known quantities, to another quantity, b, and then the equation becomes ax2=b, or ax2-b=0.

So a complete equation may be similarly reduced; for all the terms containing x2, may be reduced to one term, as ax2; and those containing x, to one, as bx; and the known terms to one, as c; then the equation is ax2+bx=c, or ax2+bx-c=0.

Hence, we infer: That every equation of the second degree, may be reduced to an incomplete equation involving two terms, or to a complete equation involving three terms.

Frequent illustrations of these principles will occur hereafter.

INCOMPLETE EQUATIONS OF THE SECOND DEGREE.

ART. 209.-1. Let it be required to find the value of x in the equation Transposing,

-16=0.

Extracting the square root of both members,

Verification.

x=4, that is, x=+4, or 4. (+4)2—16=16-16=0.

or, (-4)2-16=16-16=0.

2. Find the value of x in the equation 5x2+4=49.

Transposing,

Dividing,

5x45

4. Given ax2+b=cx2+d, to find the value of x.

ax2-cx2-d-b

or, (a—c)x2=d-b

d

a-c

From the preceding examples, we derive the

RULE,

FOR THE SOLUTION OF AN INCOMPLETE EQUATION OF THE SECOND DEGREE.

Reduce the equation to the form ax2=b. Divide both sides by the coefficient of x2, and then extract the square root of both members ART. 210.-If we take the equation ax2=b

By separating into factors, (x+m)(x—m)=0.

Now, this equation can be satisfied in two ways, and in two only; that is, by making either of the factors equal to 0.

By making the second factor equal to 0, we have

x-m=0, or x=+m.

By making the first factor equal to 0, we have x+m=0, or x=—m.

Since the equation (x+m)(x—m)=0, can be satisfied only in these two ways, it follows, that the values of x obtained from these conditions, are the only values of the unknown quantity.

Hence we conclude

1st. That every incomplete equation of the second degree, has two roots, and only two.

2d. That these roots are equal, but have contrary signs.

Find the roots of the equation, or the values of x, in each of the following examples.

REVIEW.-208. To what two forms may every equation of the second degree be reduced? Why? 209. What is the rule for the solution of an incomplete equation of the second degree? 210. Show that every incom plete equation of the second degree, has two roots, and only two; and that those roots are equal, but have contrary signs.

QUESTIONS PRODUCING INCOMPLETE EQUATIONS OF THE SECOND DEGREE.

ART. 211.-In the solution of a problem producing an equation containing the second power of the unknown quantity, the equation is found on the same principle, as in questions producing equations of the first degree. See Art. 156. 1. Find a number, whose

to 60.

Let x the number; then

multiplied by its, will be equal

2. What number is that, of which the product of its third and fourth parts is equal to 108?

Ans. 36.

3. What number is that, whose square diminished by 16, is equal to half its square increased by 16?

4. What number is that, whose square diminished equal to the square of its half, increased by 54?

Ans. 8.

by 54, is

Ans. 12.

5. What number is that, which being divided by 9, gives the

same quotient, as 16 divided by the number?

Ans. 12.

6. What two numbers are to each other as 3 to 5, and the difference of whose squares is 64?

Let 3x the less number; then 5x the greater.

And

Or

From which

(5x)-(3x)=64

25x2-9x2-16x2-64.

x=2; hence, 3x=6 and 5x=10, are

the numbers. See general directions, page 127.

REVIEW.-211. In the solution of a problem producing an equation containing the second power of the unknown quantity, upon what principle is the equation found?

7. What two numbers are those which are to each other as 3 to 4, and the difference of whose squares is 63?

8. What two numbers are those, which are to to 4, and the sum of whose squares is 100?

Ans. 9 and 12.

each other as 3

Ans. 6 and 8.

9. What number is that, to which if 3 be added, and from which if 3 be subtracted, the product of the sum and difference is 40? Ans. 7.

10. The breadth of a lot of ground is to its length, as 5 to 9, and it contains 1620 square feet; required tne breadth and length. Ans. Breadth 30, length 54 feet.

11. A man purchased a farm, giving as many dollars per acre, as there were acres in the farm; the cost of the farm was 1000 dollars; required the number of acres and the price per Ans. 100 acres, $10 per acre. 12. What two numbers are those, whose sum is to the greater, as 10 to 7, and whose sum, multiplied by the less, produces 270? Ans. 21 and 9.

acre.

Let 10x their sum; then 7x= the greater, and 3x= the less number.

13. What two numbers are those, whose difference is to the greater as 2 to 9, and the difference of whose squares is 128? Ans. 18 and 14. 14. C bought a number of oranges for 48 cents, and the price of an orange was to the number bought, as 1 to 3; how many did he buy, and how much a piece did he pay?

Ans. 12 oranges, at 4 cents a piece.

15. A person bought a piece of muslin for 3 dollars and 24 cents, and the number of cents which he paid for a yard, was to the number of yards, as 4 to 9; how many yards did he buy, and what was the price per yard? Ans. 27 yds., at 12 cents per yd 16. Find two numbers, in the ratio of to, the sum of whose squares is 225. Ans. 9 and 12.

By reducing and to a common denominator, we find they are to each other as 3 to 4. Then let 3x and 4x represent the numbers.

17. Find three numbers, in the proportion of,, and †, the sum of whose squares is 724. Ans. 12, 16, and 18.

18. A merchant sold a piece of muslin at such a rate, that the price of a yard was to the number of yards, as 4 to 5; but, if he had received 45 cents more for the same piece, the price of a yard would have been to the number of yards as 5 to 4; how many yards were there in the piece, and what was the price per yard? Ans. 10 yards, at 8 cents per yard.

COMPLETE EQUATIONS OF THE SECOND DEGREE.

1. Let it be required to find the values of x, in the equation x2-4x+4=1.

It is evident, from Article 197, that the first member of this equation is a perfect square. By extracting the square root of both members, we have x-2=±1

Whence

x=2±1=2+1=3, or 2—1=).

Verification. (3)2-4X3+4=1, that is, 9-12+4=1

also,

(1)2-4X1+4=1, that is, 1- 4+4=1.

Hence, x has two values, +3 and +1, either of which verifies the equation.

2. Let it be required to find the value of x, in the equation

x2+6x=16.

If the left member of this equation were a perfect square, we might find the value of x, by extracting the square root, as in the preceding example. To ascertain what is necessary to be added, to render the first member a perfect square, let us compare it with the square of x+a, which is x2+2ax+a2.

Hence, by adding 9, which is the square of half the coefficient of the first power of x, to each member, the equation becomes

x2+6x+9=25

Extracting the square root,

Whence

x+3=15

x=- −3±5=+2, or -8.

Either of which values of x will verify the equation.

ART. 212. We will now show the different forms to which every complete equation of the second degree may be reduced, and illustrate further, the principle of completing the square.

Since every complete equation of the second degree may be reduced to the form ax2+bx=c, if we divide both sides by a, we have

For the sake of simplicity, let=2p, and=q. The equation

then becomes

b

a

a

x2+2px=q (1.)

If is negative, and positive, the equation becomes

α

a

x2-2px=q (2.)