Ex. 17. Find two numbers which are to each other as 3 to 5, and whose product is 240.

Ans. 12 and 20.

Ex. 18. Find two numbers whose sum is to their difference as 5 to 2, and the difference of whose squares is 160?

Ans. 14 and 6.

Ex. 19. Find two numbers whose sum is to the less as 11 to 5, and whose sum multiplied by the less produces 220.

Ans. 12 and 10.

Ex. 20. A mercer bought a piece of silk for 324 shillings; and the number of shillings 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 of a yard?

Ans. 27 yards at 12 shillings per yard

COMPLETE EQUATIONS OF THE SECOND DEGREE.

(165.) A complete equation of the second degree is one which contains both the first and second powers of the unknown quantity, together with known terms.

Every complete equation of the second degree can be reduced to an equation containing but three terms. Thus, take the equation

5x2+18=3x2-4x+48. By transposing and reducing, we have

2x2+4x=30;

and, dividing by 2, x2+2x=15,

an equation containing but three terms.

QUEST.-What is a complete equation of the second degree? To what form may every complete equation be reduced?

Clearing the equation of fractions, we obtain

7x2+36+3x=4x2+12x+48.

Transposing terms, we obtain

7x2-4x2+3x-12x=48-36.

Uniting similar terms, we have

or, dividing by 3,

3x2-9x=12,

x2-3x=4.

(166.) Hence every complete equation of the second degree can be reduced to an equation of the form

x2+ax=b;

and, for this reason, complete equations are sometimes called equations of three terms. It is to be understood, however, that the signs of the quantities a and b may be either positive or negative.

Suppose we have the equation

x2-6x+9=1,

in which it is required to find the value of x.

Since each member of this equation is a complete square, if we extract the square root, we shall obtain a new equation involving only the first power of x, which may be easily solved.

In order to verify these values, substitute each of

QUEST.-Why is a complete equation called an equation of three

terms?

them in place of x in the given equation. Taking the first value, we shall have

that is,

4-6x4+9=1;

16−24+9=1,

an identical equation.

Taking the second value of x, we obtain

that is,

2'-6x2+9=1;

4-12+9=1,

an identical equation.

Hence we see that a complete equation is readily solved, provided each member of the equation is a perfect square. But equations seldom occur under this form. Take, for example,

x2-6x=-8.

The preceding method seems to be inapplicable, because the first member is not a complete square. We may, however, render it a complete square by the addition of 9, which must also be added to the second member to preserve the equality.

The equation thus becomes

x2-6x+9=9-8=1,.

which is the equation before proposed.

(167.) The peculiar difficulty, then, in resolving complete equations of the second degree, consists in rendering the first member an exact square.

In order to discover a general method of solution, let us take the equation

x2+ax=b,

which is the form to which every complete equation of the second degree can be reduced.

QUEST.-In what consists the difficulty of solving a complete equa tion? Explain the method of solving the general equation.

In order to avoid fractions, we will represent the coefficient of x by 2p, and will write the equation in the form

x2+2px=q.

We have seen that if we can by any transformation render the first member of this equation the perfect square of a binomial, we can reduce the equation to one of the first degree by extracting the square root.

Now we know that the square of a binomial, x+p, is x2+2px+p2; that is, the square is composed of the square of the first term, plus twice the product of the first term by the second, plus the square of the second term.

Hence, considering x2+2px as the first two terms of the square of a binomial, we see that the third term of the square must be p2, which is the square of half the coefficient of the first power of x,

If, then, we add p' to the first member of the proposed equation, it will become a complete square; and, in order that the equality may not be destroyed, we must add the same term to the second member of the equation, which thus becomes

x2+2px+p2=q+p3.

Extracting the square root of each member of the equation, we have

x+p=±√q+p2;

whence, by transposing p, we have

x=−p±√q+p2.

We prefix the double sign ±, because the squares both of +√q+p and also of - √q+p' is +q+p2, and therefore every equation of the second degree must have two roots.

(168.) Hence, for the solution of every complete equation of the second degree, we have the following

RULE.

1 Clear the equation of fractions. Transpose all the known quantities to one side of the equation, and all the terms containing the unknown quantity to the other side, and reduce the equation to the form x2+2px=q.

2. Take half the coefficient of the first power of x, square it, and add the result to each member of the equation.

3. Extract the square root of both members of the equation, and transpose the known term contained in the first member to the other side of the equation. Ex. 1. What are the values of x in the equation x2-10x=-16?

Completing the square by adding to each member the square of half the coefficient of the secerm, we have

them, substitute in the original equation, and we shall

have

also,

8-10x8=-16,

44-80=-16;

2'-10x2=-16, i. e., 4-20=-16;

both of which are identical equations.

QUEST.-Give the rule for solving a complete equation of the second

degree.