= = greater is 5 times the less; consequently, we must have 5a2 =845, or 169 139, which gives x=13. Hence, supposing the numbers to be positive, we shall have 13 and 65 for the numbers. 2. To find two numbers, such that their sum shall be 20 and their product 75. If 10x and 10. x stand for the numbers, then one of the conditions of the question is satisfied, since their sum is 20; and to satisfy the other condition, we must have (10 + x). (10 − x) = 100 — x2 = 75, or a2=25; consequently, by extracting the arithmetical square root of 25, we get x = 5, so that 10 +x and 10 x become 15 and 5. 3. Let 2a and p stand for the sum and product of two numbers or quantities; then it is required to find them. The first condition of the question is satisfied by representing the numbers or quantities by a + x and a x, whose sum is 2a; and to satisfy the second condition we must have (a+x) (a — x) = a2 — x2 = p, or x2 = a2-p, whose arithmetical root gives x = Va2 - p. Hence, a +x and a + Vap and a - Va-p, which satisfy the conditions of the question. x become a 4. To find two numbers whose difference is 8 and product 240. Represent the numbers by + 4 and 4, and their dif x x ference will be 8, agreeably to the conditions of the question. To satisfy the other condition of the question, we must have (x + 4) (x — 4) = x2-16 240 or a2 = 256; whose arithmetical square root gives a = = = : 16. Hence, +4 and x 4 become 20 and 12, which are the sought numbers. 5. To find two numbers or quantities whose difference is 2d and whose product is p. Representing the numbers or quantities by + d and xd, we have x2- dp or x = √p + d'; consequently, we shall have the numbers or quantities represented by √p + d2 + d and Vp + ď2 – d; which satisfy the conditions of the question. 6. If 2a and 2 represent the sum, and sum of the squares of two numbers or quantities, then it is required to find them. If a + and a represent the numbers or quantities, then one of the conditions of the question is satisfied; and to satisfy the other condition we must have the equation (a + x)2 + (a− x)2 = 2a2 + 2x2 = b2 or a2 = b2 - 2a2 ; whose arith 7. Given, 2d and 6, the difference and sum of the squares of two numbers or quantities, to find them. Representing the numbers or quantities by x + d and -d, by proceeding as in the last question, we shall have b2-2d +d and b-2d2 2 2 d for the sought numbers or quantities; which will be real if 2ď2 is not greater than 62, and positive if V b2 - 2d2 is not less than d. 8. To find a number or quantity, such that if it is subtracted from 2a, and the remainder multiplied by the number or quantity, the product shall equal b. x, If we represent it by a +x, the remainder will be a and vice versa; consequently, we shall have (a + x) (a − x) = a2 — x2 = b or a2= a2-b, which gives a Va2 - b. Hence, the sought number or quantity is a since [2a (a Va2 — b]. (a ± √ a2 — b) = (a = √ a2 — b). (a ± √ a2 — b) = a2 — (a2 —— b) = b. 9. Given, 2a and b, the sum and the sum of the cubes of two numbers or quantities, to find them. Representing them by a +x and ax, the first condition is satisfied; and to satisfy the second condition, we have (a + x)3 + ( a − x)3 = 2α3 + 6ax2 = b or x2 be real if 2a3 is not greater than b, and positive if is not greater than a. 6a 10. Given, 2d and b, the difference and difference of the cubes of two numbers or quantities, to find them. x Representing them by + d and xd, we have (x + d) =6x2d + 2d3 = Hence b - (x — d)3 Ve - 2d3 6d 6d 2d3 + d and -d, are the sought numbers or quantities; which will be real if 2d is not greater than b, and positive if d is 11. To divide the number or quantity 2a into two parts, such that their product shall be to the sum of their squares, in the ratio of m to n. Since the parts may be expressed by a +x and a — x, their product and the sum of their squares will be expressed by a2a and 2a + 2x2. Hence, we have the proportion a2x2: 2a2 + 2x2:: mn, or the equivalent equation 2ma2 +2mx2= na2-na; which gives na2; which gives a = a/n. 2m n + 2m Hence, the sought numbers or quantities are expressed by positive if 2m is not greater than n; moreover, if q2 — p2 and 2(p2+92) are put for m and n (supposing a greater than p), they will become a x +P and a x 1-P; which will be 12. "A company at a tavern had £8 158. to pay for their reckoning, but before the bill was settled two of them left the room, and then those who remained had 108. apiece more to pay than before; how many were there in company?" Let +1 represent the number, then x - 1 will represent x the number after two had left; and since £8 158. x + 1 and £8 158. x 1 evidently express the money paid by an individual before and after two had left the room, it follows, from the question, - 1 Freeing the equation from fractions, we have (£8 158.) (x − 1) + 10s. × (x2 − 1) = (£8 15s.) × (x + 1); which is easily reduced to 108.£18 360s., or a 36 and a 6; con= = sequently, x+1=7 is the sought number; noticing that in getting from the equation a2 = 36 we have rejected the negative root expressed by a = 6; because it is clearly not applicable to the question. Otherwise. Representing the sought number by x, we shall have 2 for the representative of the number after two had left the room; consequently, we shall, as before, have the equation £8 158. £8 158. Freeing the equation from fractions, etc., we get the equation a3 — 2x= 35; then completing the square we have (x − 1)2 = 36, whose positive square root gives x16, and thence a= 7, as before. Remarks.-1. It is clear that the first of these solutions is preferable to the second; because it gives a simple instead of a complete quadratic, so that the unknown letter is found by extracting the square root, without the necessity of previously completing the square. 2. It is hence clear that in solving questions, we ought always (for the sake of simplicity) to avoid the use of complete quadratics, when we can so express whatever is required as to obtain the solution by simple quadratics. 3. We will add, that all the questions we have given, with the exception of the first, are ordinarily reduced to complete quadratics, or to equations containing two unknown letters; processes which are entirely unnecessary, as is clear from the solutions we have given. 13. "A person bought a quantity of cloth for £33 158., which he sold again at £2 88. per piece, and gained by the bargain as much as one piece cost him; required the number of pieces." If stands for the number of pieces, it is clear that £33 15s. x = 6758. and (£2 88.) 488. xx will equal the price of each pieee, and what was received for all the pieces; consequently, we must have the equation 48x 675 = 675 x ; noticing, that the values of 2 and -225 16x are the roots of this equation, agreeably to what has been previously shown. From (225) 16x + 4 × 4 × 225 - 225 (225 +4) 225 289 16 = 16 16 (20 + X by 16 16' extracting the square root in the arithmetical sense, we get 225 15 × 17 x + 16. 16 ; consequently, taking the half sum of this and the preceding equation, we have x= Remark. This method of solving a complete quadratic is almost as simple as the method of solving a simple quadratic; since it is suggested by the nature of the case, and does not require the formality of completing the square, etc. 14. If 2a and b equal the sum and the sum of the fourth powers of two numbers or quantities, then it is required to find them. Let a + and a stand for the numbers or quantities, then, since the first condition of the question is satisfied, we must have (a + x)* + (a − x)1 = b, or we shall have 2a + 12a2x2 + 2x1 = b; which gives x1 + 6a2x2 = b- 2a1 ing the equation, we get x= (√2 + 16a1 Solv 3a), and thence a +x and a — ≈ will become a +√(+16a |