Now since it is possible to factor a quadratic equation, the solution of the equation can at once be found by placing each factor equal to zero and solving the resulting equations; thus, in example 2, 5+4x-12x2 = (2 x + 1) (5 — 6x) = 0 a1 + a2b2+b1= aa + 2 a2b2 + b1 — a2b2 = [(a2 + b2)2 — a2b2] (a2 + b2 + ab) (a2 + b2 — ab) ===(a2 + ab + b2) (a2 — ab+b2). 5. Factor 4x1-9 x2+1. [394] 4 x1 − 9 x2 + 1 = 4 x1 −4 x2+1—5 x2 = (2 x2—1)2 — (x√/5)2 = (2 x2 − 1 + xv 5) (2 x2 — 1 — xv 5). x1 + a2 = x* + 2 ax2 + a2 − 2 ax2 = (x2 + a)2 — (xv ́2 a)2 = (x2+ a−√2a · x) (x2 + a + √ 2 ax). 2 Factor the following: 1. x2+3x+1. 3. 3x2+7x6. 5. 6x19x+15. 7. x + x2 + 1. EXERCISE LXXII 2. 4x2+13x + 3. 4. 6+5 6x2. 6. x 1. 11. Solve the equation 16 2+16x+8x+16-8x 0. (x2 + 4)o — (2 x √2) = (x2+4+2x√2) (x2+4−2 x √2) == (x2+2√2x+4) (x2 − 2 √ 2 x + 4). Either factor, x2+2v 2x+4 or x2 - 2√2x+4, equated to zero will reduce x+16 to zero. Therefore the roots of 4+16= 0 will be found by solving each of the equations found by placing each factor equal to zero. X1, X2 = √2 (1 + i). 12. Having factored examples 6, 7, 8, 9, 10, solve the equations CHAPTER V PROBLEMS INVOLVING EQUATIONS OF THE SECOND DEGREE IN ONE UNKNOWN QUANTITY 417. PROBLEM I.-Divide the number 31 into two parts such that their product shall be 228. Let x and y be the parts of 31; then Here, although two sets of values for x and y are obtained, yet there is only one way of dividing 31, so that the product of the two parts shall be 228. 418. PROBLEM II.-Given that the perimeter of a rectangle is 2p, and its area equal to that of a square of which the side is a. Calculate the sides of the rectangle. Let x and y be the sides of the rectangle; then x2 and Y2i it would be necessary only to interchange the terms length and breadth of the rectangle. In order that the problem may be possible it is necessary and sufficient that the sides found are real and positive. In order that the roots of equation (3) may be real, it is necessary that since x x1 will both be positive, is the sum of two positive quantities, p and V p2 - 4 a2, and Y1 is the difference between p and V p2 — 4 a2, which is + and less than p. Therefore, the inequality (4) is necessary and sufficient to make the problem possible. It follows from (4) that the problem is always possible, if and impossible when a is greater than 22 (8412, 3); when a2 = 2 the 4 4 radical p2 - 4 a2 is equal to zero, and the two values X1 and Y1 are From this discussion the two following theorems may be deduced: 1. Of all rectangles which have the same perimeter that which has the greatest area is a square. 2. Of all rectangles which have a given area that which has the least perimeter is a square. GEOMETRIC CONSTRUCTION.-Since the two sides of the rectangle are given by the formulae x1 it is easy to construct geometrically these sides x, and y, by means of the given dimensions, a and p. In any straight line lay off a length AB, equal to P, and on AB as a diame 2 ter construct a semicircle; then about A as a center, with a as a radius, describe an are intersecting the circumference on AB in C. From a known theorem of Geometry, it follows that About B as a center construct a circumference with BC as a radius, and let and O' be the points in which the line AB is intersected by this circumference. Hence, Therefore, AO and AO' are the sides required. This construction is impossible if a> For in this case the circumference with the P center A will not intersect the circumference on AB. This result corresponds to the algebraic fact that if a>, the radical a2 4 is imaginary, If a=2, BC= 0, then 4040' and the 2 rectangle will be a square. = = Thus we are led to the same results as those which were deduced from the algebraic discussion. PROBLEMS 3. Find two numbers such that their sum is 39 and the sum of their cubes 17199. 4. Find the number such that the sum of the number and its reciprocal is m. 5. The product of two numbers is 750, and the quotient when one is divided by the other is 31; find the numbers. 6. Find the number such that when it is (1) divided by n, (2) subtracted from n, the result is the same? 7. A number which consists of two digits has this property: when it is divided by the product of its digits the quotient is 3, and when it is increased by 10, its digits appear in the reverse order. Find the number. 8. Divide the number 53 into two parts whose product is 612. 9. Divide the quantity a2 + b2 into two parts whose product is } (a + a2b2 + b1). 10. Find two factors of 2268 whose sum is 99. 13. The sum of the squares of two numbers, one of which is 12 greater than the other, is 1130. What are the numbers? 14. The members of a society each contribute the same amount to a fund of $336. If there were three members less, each member would have to contribute $2 more. Find the number of members. 15. A person purchased a certain number of sheep for $175; after losing two of them he sold the rest at $2 a head more than he gave for them, and by so doing gained $5 by the transaction. Find the number of sheep purchased. 16. A cask contains 360 gallons of wine; a certain quantity is drawn and an equal quantity of water is put in; from this mixture the same quantity as before is drawn, and 84 gallons in addition; on replacing the drawn liquid with water it is found that the barrel contains equal quantities of wine and water. How many gallons were drawn the first time? 17. A number of men pass a certain time in a hotel and on leaving they have a bill of $12 to pay. Had there been 4 more in the party and had each spent 25 cents less, their bill would have been $15. What was the number of men? 18. A merchant paid a certain sum for a horse, later he sold the horse for $144 and thereby gained as much per cent as the horse cost him dollars originally. What did he pay for the horse? 19. What is the quotient, whose dividend is n times smaller than its divisor, and the sum of the quotient and its reciprocal is n? 20. A manufacturer had agreed to pay a capitalist $8,800 after 7 months and $5,940 at the end of 1 year. After how many months can the manufacturer pay back the capitalist the total amount, $14,740, if interest at 5% per annum is charged for the money which he paid later than it was due and if a rebate of 5% per annum is allowed for the money paid before it was due? 21. A capitalist lent k dollars at a certain rate per cent and withdrew each year b dollars; at the end of 2 years there remained k' dollars invested. At what per cent was the money lent? PROBLEMS CONNECTED WITH THE THEOREM OF PYTHAGORAS 419. In this section only those problems will be discussed which are connected with the Theorem of Pythagoras for rightangled triangles; first because it is not desired to make the discussion too extended, and secondly because the Theorem of Pythagoras for acute- and obtuse-angled triangles has more of a trigonometric interest. " |