Page images
PDF
EPUB
[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

52. x2 + bx + cx = (a+c)(a - b).

3a2x_6a2 + ab - 2b2 b2x

53. abx2+

[blocks in formation]

54. (3a2 + b2) (x − x + 1) = (3 b2 + a2) (x2+x+1).

SOLUTION OF QUADRATIC EQUATIONS BY A FORMULA

267. It was shown in Art. 264 that if ax + bx = c, then

[ocr errors][merged small][merged small]

This result may be used as a formula for the solution of quadratic equations, as follows:

1. Solve the equation 3x2 + 5x = 12.

In this case, a=3, b=5, c = 12; substituting these values in (1),

x= -5 ± √25 + 144 _ - 5 ±√169

6

=

6 -5±13 6

-3 or

4 Ans. 3'

2. Solve the equation 110m2 - 21 x= - 1.

In this case, a = 110, b = - 21, c = − 1 ; therefore,

[ocr errors][merged small][merged small][merged small][merged small][merged small]

Note. Particular attention must be paid to the signs of the coefficients in substituting.

[blocks in formation]

XXII. PROBLEMS.

INVOLVING QUADRATIC EQUATIONS.

268. 1. A man sold a watch for $21, and lost as much per cent as the watch cost him. Required the cost of the watch.

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

That is, the cost of the watch was either $70 or $30; for each of these values satisfies the given conditions.

2. A farmer bought some sheep for $72. If he had bought 6 more for the same money, they would have cost $1 apiece less. How many did he buy?

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Only the positive value of x is admissible, as the negative value does not answer to the conditions of the problem. The number of sheep, therefore, was 18.

Note 1. In solving problems which involve quadratics, there will always be two values of the unknown quantity; but only those values should be retained as answers which satisfy the conditions of the problem.

Note 2. If, in the given problem, the words "6 more" had been changed to "6 fewer," and "$1 apiece less" to "$1 apiece more," we should have found the answer 24.

In many cases where the solution of a problem gives a negative result, the wording may be changed so as to form an analogous problem to which the absolute value of the negative result is an answer.

PROBLEMS.

3. I bought a lot of flour for $175; and the number of dollars per barrel was 4 of the number of barrels. How many barrels were purchased, and at what price?

4. Separate the number 15 into two parts the sum of whose squares shall be 117.

[blocks in formation]

5. Find two numbers whose product is 126, and quotient

6. I have a rectangular field of corn containing 6250 hills. The number of hills in the length exceeds the number in the breadth by 75. How many hills are there in the length, and in the breadth?

7. Find two numbers whose difference is 9, and whose sum multiplied by the greater is 266.

8. The sum of the squares of two consecutive numbers is 113. What are the numbers?

9. A man cut two piles of wood, whose united contents were 26 cords, for $35.60. The labor on each cost as many dimes per cord as there were cords in the pile. Required the number of cords in each pile.

10. Find two numbers whose sum is 8, and the sum of whose cubes is 152.

11. Find three consecutive numbers such that twice the product of the first and third is equal to the square of the second increased by 62.

12. A grazier bought a certain number of oxen for $240. Having lost 3, he sold the remainder at $8 a head more than they cost him, and gained $59. How many did he buy?

13. A merchant bought a quantity of flour for $96. If he had bought 8 barrels more for the same money, he would have paid $2 less per barrel. How many barrels did he buy, and at what price?

14. Find two numbers, whose product is 78, such that if one be divided by the other the quotient is 2, and the remainder 1.

15. The plate of a rectangular looking-glass is 18 inches by 12. It is to be framed with a frame all parts of which are of the same width, and whose area is equal to that of the glass. Required the width of the frame.

16. A merchant sold a quantity of flour for $39, and gained as much per cent as the flour cost him. What was the cost of the flour?

17. A certain company agreed to build a vessel for $6300; but, two of their number having died, the rest had each to advance $200 more than they otherwise would have done. Of how many persons did the company consist at first?

18. Divide the number 24 into two parts, such that the sum of the fractions obtained by dividing 24 by them shall be 64 . 15

19. A detachment from an army was marching in regular column, with 6 men more in depth than in front. When the enemy came in sight, the front was increased by 870 men, and the whole was thus drawn up in 4 lines. Required the number of men.

20. A merchant sold goods for $16, and lost as much per cent as the goods cost him. What was the cost of the goods?

« PreviousContinue »