Problems giving rise to Equations of the Second Degree involving but one unknown quantity.

1. Find a number such that three times the number added to twice its square will be equal to 65.

Let x denote the number. Then from the conditions,

Both of these roots verify the equation: for,

2 × (5)2 + 3 × 5 = 2 × 25+ 15 = 65;

The first root satisfies the conditions of the problem as enun

ciated.

The second root will also

satisfy the conditions, if we regard

its algebraic sign. Had we denoted the unknown quantity by

We see that the

13

x=

2

and x= 5.

roots of this equation differ from those of equation (1) only in their signs, a result which was to have been expected, since we can change equation (1) into equation (2) by simply changing the sign of x, and the reverse.

2. A person purchased a number of yards of cloth for 240 cents. If he had received three yards less, for the same sum, it would have cost him 4 cents more per yard. How many yards did he purchase?

Let a denote the number of yards purchased.

Then will

240

x

denote the number of cents paid per yard.

Had he received three yards less,

-3, would have denoted the number of yards purchased, and

would have denoted the number of cents he paid per yard.

The value x = 15 satisfies the conditions of the problem, understood in their arithmetical sense; for, 15 yards for 240

or 16 cents for the price of one yard, and

12 yards for 240 cents, gives 20 cents for the price of one yard, which exceeds 16 by 4.

conditions of the following problem:

A person sold a number of yards of cloth for 240 cents: if he had received the same sum for 3 yards more, it would have brought him 4 cents less per yard. How many yards did

he sell?

If we denote the number of yards sold by x, the statement of this last problem, and the given one, both give rise to the same equation,

hence, the solution of this equation ought to give the answers to both problems, as we see that it does.

Generally, when the solution of the equation of a problem gives two roots, if the problem does not admit of two solutions there is always another problem whose statement gives rise to the same equation as the given one, and in this case the two roots form answers to both problems.

3. A man bought a horse, which he sold for 24 dollars. At the sale, he lost as much per cent. on the price of his purchase, as the horse cost him. What did he pay for the horse? Let x denote the number of dollars that he paid for the horse : then, x 24 will denote the number of dollars that he lost.

But as he lost x per cent. by the sale, he must have lost x upon each dollar, and upon x dollars he lost a number

100

Both of these values satisfy the conditions of the problem.

For, in the first

the horse and sold

place, suppose the man gave 60 dollars for him for 24, he then loses 36 dollars. But,

from the enunciation, he should lose 60 per cent. of 60, that is,

therefore, 60 satisfies the problem.

If he pays 40 dollars for the horse, he loses 16 by the sale; for, he should lose 40 per cent. of 40, or

therefore, 40 satisfies the conditions of the problem.

4. A grazier bought as many sheep as cost him £60, and after reserving 15 out of the number, he sold the remainder for £54, and gained 2s. a head on those he sold: how many did he buy? Ans. 75.

5. A merchant bought cloth for which he paid £33 15s., which he sold again at £2 8s. per piece, and gained by the bargain as much as one piece cost him: how many pieces did he buy? Ans. 15.

6. What number is that, which, being divided by the product of its digits, the quotient will be 3; and if 18 be added to it, the order of its digits will be reversed? Ans. 24.

7. Find a number such that if you subtract it from 10, and multiply the remainder by the number itself, the product will be 21. Ans. 7 or 3.

8. Two persons, A and B, departed from different places at the same time, and traveled towards each other. On meeting, it appeared that A had traveled 18 miles more than B; and that A could have performed B's journey in 15 days, but B would have been 28 days in performing A's journey. How far did each travel? A 72 miles.

Ans.

{

B 54 miles.

9. A company at a tavern had £8 15s. to pay for their reckoning; but before the bill was settled, two of them left

the room, and then those who remained had 10s. apiece more to pay than before: how many were there in the company? Ans. 7.

10. What two numbers are those whose difference is 15, and of which the cube of the lesser is equal to half their product? Ans. 3 and 18.

11. Two partners, A and B, gained $140 in trade: A's. money was 3 months in trade, and his gain was $60 less than his tock: B's money was $50 more than A's, and was in trade 5 months: what was A's stock? Ans. $100.

12. Two persons, A and B, start from two different points, and travel toward each other. When they meet, it appears that A has traveled 30 miles more than B. It also appears that it will take A 4 days to travel the road that B had come, and B 9 days to travel the road that A had come. What was their distance apart when they set out?

Ans. 150 miles.

Discussion of Equations of the Second Degree involving but one unknown quantity.

115. It has been shown that every complete equation of the second degree can be reduced to the form (Art. 113)

in which Ρ and q are numerical or algebraic, entire or frac tional, and their signs plus or minus.

If we make the first member a perfect square, by completing the square (Art. 112*), we have

x2 + 2px + p2 = q+p2,

which may be put under the form

(x + p)2 = q + p2.

Now, whatever may be the value of q+p2, its square root may be represented by m, and the equation put under the form

(x + p)2 = m2, and consequently, (+p)2 — m2 = 0.