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PROBLEMS PRODUCING PURE EQUATIONS. (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 ?

Ans. +21 and 9 (2) There are two numbers in the proportion of 4 to 5, and the difference of whose squares is 81. What are the numbers ?

Ans. +12 and 15. (3) A detachment from an army was marching in regular column, with 5 men more in depth than in front; but upon the enemy coming in sight, the front was increased by 845 men, and by this movement the detachment was drawn up in five lines. Required the number of men ?

Ans. 4550. (4) 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 been idle 4 of those days, had 75 shillings to receive; but B, who had been idle 7 of those days, received only 48 shillings. Now, had B been idle only 4 days and A 7, they would have received exactly alike. For how many days were they engaged, how many did each work, and what had each per day?

Ans. A worked 15 and B 12 days.

A received 5 and B 4 shillings per day. (5) 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 liquid 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.

PROBLEMS WHICH PRODUCE ADFECTED OR COMPLETE QUADRATIC

EQUATIONS.

PROBLEM 1. 190. To find a number such that twice its square, augmented by three times the number, is equal to 65. Let x be the number required, we have for the equation of the problem,

2x2 + 3x=65.

Solving the equation,
Hence

x=5; r=-5 The first of these two values satisfies the conditions of the problem, as stated in the enunciation ; for, in fact,

2(5): +3X5=2 x 25+15

=65. In order to interpret the meaning of the second value, let us observe, that if we substitute —x for +x in the equation 2x + 3x=65, the coefficient of 3.r alone will change its sign, for (-x)=(+x)=xo. Hence the value of 2 will no longer be

2

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but will become

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13 Hence

x=2; x=-5, where the values of r differ from those already found in sign alone.

Hence we may conclude that the negative solution -5, considered without reference to its sign, is the solution of the following problem:

To find a number such that twice its square, diminished by three times the number, is equal to 65. In fact, we have

(13)? 13 169 39

PROELEM 2. A tailor bought a certain number of yards of cloth for 12 pounds. If he had paid the same sum for 3 yards less of the same cloth, then the cloth would have cost 4 shillings a yard more. Required the number of yards purchased. Let x be the number of yards purchased.

240 Then is the price of one yard, expressed in shillings. If he had paid the same sum for 3 yards less, in that case the price of each

240 would be represented by r_5

But by the conditions of the problem, this last price is greater than the former by 4 shillings; hence the equation of the problem will be

'240 240

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..r=15; r=-12. The value of x=15 satisfies the conditions of the problem, for

240 240

15=16; 12 =20, the price of each yard in the first case being 16 shillings, and in the last case 20, which exceeds the former by 4 shillings.

With regard to the second solution, we can form a new enunciation to which it will correspond. Resuming the original equation, and changing r into -7, it becomes

240 240

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an equation which may be considered as the algebraic representation of the following problem :

A tailor bought a certain number of yards of cloth for 12 pounds. If he had paid the same sum for 3 yards more, then the cloth would have cost 4 shillings a yard less. Required the number of yards purchased. The above equation when reduced becomes

x+3x=180, instead of ro—3x=180, as in the former case; solving the above, we find

x=12; x= -15. The two preceding problems illustrate the principle explained with regard to problems of the first degree.

PROBLEM 3. A merchant purchased two bills; one for $8776, payable in 9 months, the other for $7488, payable in 8 months. For the first he paid $1200 more than for the second. Required the rate of interest allowed.

Let x represent the interest of $100 for 1 month.

Then 12.x, 9r, 8x severally represent the interest of $100 for 1 year, 9 months, 8 months.

And 100+9x, 100 + 8x represent what a capital of $100 will become at the end of 9 and of 8 months respectively.

Hence, in order to deterinine the actual value of the two bills, we have the following proportions :

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100+ 9.r

7488 x 100 100+8x:100:: 7488: The fourth terms of the above proportions express the sum paid by the merchant for each of the bills. Hence, by the conditions of the problem,

877600 748800 100+ 9.

x 100-18=1200, or, dividing each member by 400,

2194 1872

100+9x100+8x=%. Clearing of fractions and reducing,

216x2 + 4396.r=2200. Whence

2198. 2200 +(2199) x=-216 +V 216 +(216) –2198+ V5306404

216
.:. 127-_-2198+ V 5306404

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-2198 +2303.5........

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... 12x=5.86........; and 12r=-250.08........ The positive solution, 12x=5.86......, represents the required rate of interest per cent. per annum.

With regard to the negative solution, it can only be considered as connected with the other by the same equation of the second degree. If we resume the original equation, and substitute – x for +r, we shall find great difficulty in reconciling this new equation with an enunciation analogous to that of the proposed problem.

PROBLEM 4. A man purchased a horse, which he afterward sold to disadvantage for 24 pounds. His loss per cent. by this bargain, upon the original price of the horse, is expressed by the number of pounds which he paid for the horse. Required the original price. Let x be the number of pounds which he paid for the horse.

Then r–24 will represent his loss; But, by the conditions of the problem, his loss per cent. is represented by the number of units in x;

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