by 240; and the sum of the 2d and 3d exceeded the first by 360. What was the share of each? Ans. The 1st 200, the 2d 300, the 3d 260. 15. What two numbers are those, which, being in the ratio of 3 to 4, their product is equal to 12 times their sum? Ans. 21 and 28. 16. A certain company at a tavern, when they came to settle their reckoning, found that had there been 4 more in company, they might have paid a shilling a-piece less than they did; but that if there had been 3 fewer in company, they must have paid a shilling a-piece more than they did. What then was the number of persons in company, what each paid, and what was the whole reckoning? Ans. 24 persons, each paid 7s. and the whole reckoning 8 guineas. 17. A jocky has two horses and also two saddles, the one valued at 18 the other at 31. Now when he sets the better saddle on the ist horse, and the worst on the 2d, it makes the first horse worth double the 2d: but when he places the better saddle on the 2d horse, and the worse on the first, it makes the 2d horse worth three times the 1st. What then were the values of the two horses? Ans. The 1st 61. and the 2d 91. 18. What two numbers are as 2 to 3, to each of which if 6 be added, the sums will be as 4 to 5? Ans. 6 and 9.. 19. What are those two numbers, of which the greater is to the less as their sum is to 20, and as their difference is to 10? Ans. 15 and 45. 20. What two numbers are those, whose difference, sum, and product, are to each other, as the three numbers 2, 3, 5? Ans. 2 and 10. 21. To find three numbers in arithmetical progression, of which the first is to the third as 5 to 9, and the sum of all three is 63? Ans. 15, 21, 27. 22. It is required to divide the number 24 into two such parts, that the quotient of the greater part divided by the less, may be to the quotient of the less part divided by the greater, Ans. 16 and 8. as 4 to 1. 23. A gentleman being asked the age of his two sons, answered, that if to the sum of their ages 18 be added, the result will be double the age of the elder; but if 6 be taken taken from the difference of their ages, the remainder will be equal to the age of the younger. What then were their ages? Ans. 30 and 12. 24. To find four numbers such, that the sum of the 1st, 2d, and 3d, shall be 13; the sum of the 1st, 2d, and 4th, 15; the sum of the 1st, 3d, and 4th, 18; and lastly the sum of the 2d, 3d, and 4th, 20. Ans. 2, 4, 7, 9. 25. To divide 48 into 4 such parts, that the 1st increased by 3, the second diminished by 3, the third multiplied by 3, and the 4th divided by 3, may be all equal to each other. Ans. 6, 12, 3, 27. QUADRATIC EQUATIONS. QUADRATIC Equations are either simple or compound. A simple quadratic equation, is that which involves the square of the unknown quantity only. As ax2=6. And the solution of such quadratics has been already given in simple equations. A compound quadratic equation, is that which contains the square of the unknown quantity in one term, and the first power in another term. As ax2 + bxc. All compound quadratic equations, after being properly reduced, fall under the three following forms, to which they must always be reduced by preparing them for solution. The general method of solving quadratic equations, is by what is called completing the square, which is as follows : 1. REDUCE the proposed equation to a proper simple form, as usual, such as the forms above; namely, by transposing all the terms which contain the unknown quantity to one side of the equation, and the known terms to the other; placing the square term first, and the single power second; dividing dividing the equation by the co-efficient of the square or first term, if it has one, and changing the signs of all the terms, when that term happens to be negative, as that term must always be made positive before the solution. Then the proper solution is by completing the square as follows, viz. 2. Complete the unknown side to a square, in this manner, viz. Take half the co-efficient of the second term, and square it; which square add to both sides of the equation, then that side which contains the unknown quantity will be a complete square. * 3. Then extract the square root on both sides of the equation, and the value of the unknown quantity will be determined, - * As the square root of any quantity may be either + or, therefore all quadratic equations admit of two solutions. Thus, the square root of n2 is ei thern orn; for + n×+ n and — n × — n are each equal to +n2. But the square root of — n2, or ✔✅ —n2, is imaginary or impossible, as neither +n nor n, when squared, gives —--n3. So, in the first form, x3+axb, where x+a is found ✔b+4a2, the root may be either +✔b+4a2, or — -✅b+4a2, since either of them being multiplied by itself produces ba2. And this ambiguity is expressed by writing the uncertain or double sign ± before ✔b+‡a2; thus x= — }a. In this form, where x = ±√ba -a, the first value of x, viz. x=+ ✔b+ta2 — Ja, is always affirmative; for since tab is greater than—ja2, the greater square must necessarily have the greater root; therefore ✔b+÷a3 will always be greater thana, or its equal a; and consequently +✔b+}a2 --a will always be affirmative. 2 The second value, viz. x=−√b+}a3 —ja will always be negative, because it is composed of two negative terms. Therefore when a ax=b, we shall have ƒ= +✔b+1 a2 — ja for the affirmative value of x, and + ✔ba2 —ļa for the negative value of x. In the second form, where == ±√b+ ¦ a2 + la the first value, viz. æ➡ +✔b+1 a2+ža is always affirmative, since it is composed of two affirmative, terms. But the second value, viz.`x=-- ·√b + ¦ a2 + {a, will always be ne gative; for since b+a2 is greater than a3, therefore greater than a2, or its equal a; and consequently is always a negative quantity. b+a2 will be ✔b+{a2 + 1a a Therefore, determined, making the root of the known side either + or -, which will give two roots of the equation, or two values of the unknown quantity. Note, 1. The root of the first side of the equation, is always equal to the root of the first term, with half the coefficient of the second term joined to it; with its sign, whetheror. 2. All equations, in which there are two terms including the unknown quantity, and which have the index of the one just double that of the other, are resolved like quadratics, by completing the square, as above. Thus, x1+ax2=b, or x2o+ax”—b, or x+ax2 x3-b, are the same as quadratics, and the value of the unknown quantity may be determined accordingly. Therefore, when x2-ax-b, we shall have x=+✔b+ża2 +‡a for the affirmative value of x; and a=— ✔b+4a2+za for the negative value of x; so that in both the first and second forms, the unknown quantity has always two values, one of which is positive, and the other negative. But in the third form, where x= -±√}a2−b+}a, both the values of x will be positive when a2 is greater than b. For the first value, viz. x= = +√ia2+ will then be affirmative, being composed of two affirmative terms. The second value, viz. x 2 a2-b; and ✔aba is affirmative also; for since a is greater than ab, therefore ✔a ora is greater than consequently ✔a2-b+ga will always be an affirmative quantity. So that, when x2 -Ɑx= b, we shall have x≈ + ✔ fa2 — b + a, and also x=—✔a2 —ba, for the values of x, both positive. But in this third form, if b be greater than a2, the solution of the proposed question will be impossible. For since the square of any quantity (whether that quantity be affirmative or negative) is always affirmative, the square root of a negative quantity is impossible, and cannot be assigned. But when b is greater thana, then a—b is a negative quantity; and therefore its root ✔a2--b is impossible, or imaginary; consequently, in that case, xa±✔¦a2 —b, or the two roots or values of x, are both impossible, or imaginary quantities. EXAMPLES. EXAMPLES. 1. Given x+4x=60; to find x. First, by completing the square x2+4x+4=64 ; Then, transpos. 2, gives, x=6 or 10, the two roots. 2. Given x2-6x+10=65; to find x. First trans. 10 gives x2 6x=55; Then by complet. the sq. it is x2 −6x+9=64 ; Then trans. 3, gives x=11 or - -5. 3. Given 2x2+8x-30—60; to find x. First by transpos. 20, it is 2x2+8x=90 ; 4. Given 3x2 -3x+9=81; to find x. -x= And compl. the sq. gives x2-x+1=36; 5. Given 1x3-x+301=52%, to find x. 6. Given ax2 bxc; to find x. First by div. by a, it is x2. Then compl. the sq. gives x2. a b b α b 2a 4a3 4a2 2a ; 7. Given x1-2ax2=b; to find x. First by compl. the sq. gives x2 - 2ax2+aa—a3 +b; And |