Show that each of these values will satisfy the equation.

Note. Clearing of fractions and transposing, we find in each member of this equation a binomial factor, which being canceled, the equation is easily solved.

PROBLEMS.

Prob. 1. What two numbers are those whose sum is to the greater as 10 to 7, and whose sum, multiplied by the less, pro

Prob. 2. What two numbers are those whose sum is to the greater as m to n, and whose sum, multiplied by the less, is equal to a?

Prob. 3. What number is that, the third part of whose square being subtracted from 20, leaves a remainder equal to 8?

Prob. 4. What number is that, the mth part of whose square being subtracted from a, leaves a remainder equal to b?

Ans. ±√m(a—b).

Prob. 5. Find three numbers in the ratio of

sum of whose squares is 724.

Prob. 6. Find three numbers in the ratio of m, n, and p, the

sum of whose squares is equal to a.

m2+n2+p2; and ±

ap2 m2+n2+p2"

Prob. 7. Divide the number 49 into two such parts that the quotient of the greater divided by the less may be to the quotient of the less divided by the greater as to .

Ans. 21 and 28. Note. In solving this Problem, it is necessary to assume a principle employed in Arithmetic, viz., If four quantities are proportional, the product of the extremes is equal to the product of the means.

Prob. 8. Divide the number a into two such parts that the quotient of the greater divided by the less may be to the quotient of the less divided by the greater as m to n.

Prob. 9. There are two square grass-plats, a side of one of

which is 10 yards longer than a side of the other, and their areas are as 25 to 9. What are the lengths of the sides?

Prob. 10. There are two squares whose areas are as m to n, and a side of one exceeds a side of the other by a. What are the lengths of the sides?

Prob. 11. Two travelers, A and B, set out to meet each other, A leaving Hartford at the same time that B left New York. On meeting, it appeared that A had traveled 18 miles more. than B, and that A could have gone B's journey in 15 hours, but B would have been 28 hours in performing A's journey. What was the distance between Hartford and New York?

Ans. 126 miles.

Prob. 12. From two places at an unknown distance, two bodies, A and B, move toward each other, A going a miles more than B. A would have described B's distance in n hours, and B would have described A's distance in m hours. What was the distance of the two places from each other?

Prob. 13. A vintner draws a certain quantity of wine out of a full vessel that holds 256 gallons, and then, filling the vessel with water, draws off the same quantity of liquor as before, and so on for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time?

Hence

256(x-1)+

=81.

Prob. 14. A number a is diminished by the nth part of itself, this remainder is diminished by the nth part of itself, and so on to the fourth remainder, which is equal to b. Required the value of n.

Prob. 15. Two workmen, A and B, were engaged to work for a certain number of days at different rates. At the end of the time, A, who had played 4 of those days, received 75 shillings, but B, who had played 7 of those days, received only 48 shillings. Now had B only played 4 days and A played 7 days, they would have received the same sum. For how many days were they engaged? Ans. 19 days.

Prob. 16. A person employed two laborers, allowing them different wages. At the end of a certain number of days, the first, who had played a days, received m shillings, and the second, who had played b days, received n shillings. Now if the second had played a days, and the other b days, they would both have received the same sum. For how many days were they engaged?

Complete Equations of the Second Degree.

256. In order to solve a complete equation of the second degree, let the equation be reduced to the form

x2+px=q.

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 its square

'root.

Now we know that the square of a binomial, x+a, or x2+ 2ax+a2, 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+px as the first two terms of the

square of a binomial, and consequently px as being twice the product of the first term of the binomial by the second, it is evident that the second term of this binomial must be 2

257. In order, therefore, that the expression x2+px may be rendered a perfect square, we must add to it the square of this second term 2; and in order that the equality of the two members may not be destroyed, we must add the same quantity to the second member of the equation. We shall then have

Taking the square root of each member, we have

Thus the equation has two roots: one corresponding to the plus sign of the radical, and the other to the minus sign. These two roots are

258. Hence, for solving a complete equation of the second degree, we have the following

RULE.

1st. Reduce the given equation to the form of x2+px=q.

2d. Add to each member of the equation the square of half the coefficient of the first power of x.

3d. Extract the square root of both members, and the equation will be reduced to one of the first degree, which may be solved in the usual manner.

259. When the equation has been reduced to the form x2+ px=q, its two roots will be equal to half the coefficient of the sec