Ans. x= a 3+ Show that each of these values will satisfy the equation. 2. Oca - 17=130-2x2. Ans. x=+7. 4x2+5 3. =45. 9 + Vab 4. 2+ab=5x 2 2a2 5. X+ vas + x2 = Ans. x=+ va toca V3 6. axé — 5c=b2_3c+d. 02 522 7 299 7. 3 1224 24 8. 12ab+=4a2 +962. X-25 9. 1122 Xcia 4 10. + x2–17= Ans. x=+41 Væc? - 17 #ta, x-a 2(a2 +1) 11. Ans. x=+1. x-axta (1+a)(1-a) 22 + 3x-7 12. =1. Ans. x=+3. 18 x +2+= 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. x+25 =3 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, produces 270? Let 10x=their sum. Then 7x=the greater number, and 3x=the less. Whence 30x2=270, and x=9; therefore x=3, and the numbers are +21 and +9. m 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? an? Ans. V am-n) and + EV m(m-n) 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. Evma-6). 1 2 3 Prob. 5. Find three numbers in the ratio of and the 2 3 4 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. Ans. am2 an2 ар? and + ma+na+pzi ma+na+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. Thus, if a“: 6::c:d, then ad=bc. 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. avm avn Ans. and Vmtva Vm+m Prob. 9. There are two square grass-plats, a side of one of V ma+na+pai 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? ανη ava Ans. and Vmvn vm - Vn 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 159 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? Vmt in ✓m-vn 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? Ans. 64, 48, 36, and 27 gallons. 1 Note. Suppose - part is drawn each time. , . 256 256(x-1) Then 256 remains after the first draught. 256(x - 1) Similarly, remains after the second draught, and x? Ans. ax So on. Hence 256(x-1) = 81. Prob. 14. A number a is diminished by the nth part of it. self, 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. va Ans. Va-Vi 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 shil. lings, 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 bave received the same sum. For how many days were they engaged? OVm-ava days. Vm-vn 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 22 +px=4. 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 squaro root. Now we know that the square of a binomial, 6c+a, or x2 + 2ax+a , 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 22 +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 P evident that the second term of this binomial must be p2 257. In order, therefore, that the expression 2c2+px may be rendered a perfect square, we must add to it the square of this second term P ; and in order that the equality of the two mem 2 bers may not be destroyed, we must add the same quantity to the second member of the equation. We shall then bave 202 -=9+ 4 4' Taking the square root of each member, we have р X+ 2 i 4 whence, by transposition, x= pa - 9+ 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 p2 р and + 4 258. Hence, for solving a complete equation of the second degree, we have the following RULE. 1st. Reduce the given equation to the form of a2 +px=9. 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 ? + px=9, its two roots will be equal to half the coefficient of the sec |