Examples of Binomial Equations. 10. What number is it, whose half multiplied by its third part, gives 864 ? Ans. 72. 11. What number is it, whose 7th and 8th parts multiplied together, and the product divided by 3, gives the quotient 298? Ans. 224. 12. Find a number such, that if we first add to it. 94, then subtract it from 94, and multiply the sum thus obtained by the difference, the product is S512. Ans. 18. 13. Find a number such, that if we first add it to a, then subtract it from a, and multiply the sum by the difference, the product is b. Ans. √ (a2 —b). 14. Find a number such, that if we first add it to a, then subtract a from it, and multiply the sum by the difference, the product is b. Ans. (a+b). 15. What two numbers are they whose product is 750, and quotient 3? Ans. 50 and 15. 16. What two numbers are they whose product is a, and quotient b? Ans. ✔✅ a b and a 17. What two numbers are they, the sum of whose squares is 13001, and the difference of whose squares is 1449 ? Ans. 85 and 76. 18. What two numbers are they, the sum of whose squares is a, and the difference of whose squares is b? Ans. √(a+b) and √ 1 (a—b). 19. What two numbers are to one another as 3 to 4, the sum of whose squares is 324900? Ans. 342 and 456. 20. What two numbers are as m to n, the sum of whose squares is a? Ans. m√ a and n√ a ✓ (m2 + n2)° Examples of Binomial Equations. 21. What two numbers are as m to n, the difference of 22. A certain capital is let at 4 per cent.; if we multiply the number of dollars in the capital, by the number of dollars in the interest for 5 months, we obtain 117041}. What is the capital? Ans. $2650. 23. A person has three kinds of goods, which together cost $5525. The pound of each article costs as many dollars as there are pounds of that article; but he has one third more of the second kind than he has of the first, and 3 times as much of the third as he has of the second. How many pounds has he of each? Ans. 15 pounds of the first, 20 of the second, and 70 of the third. 24. Find three numbers such, that the product of the first and second is 6, that of the first and third is 10, and the sum of the squares of the second and third is 34. Ans. 2, 3, 5. 25. Find three numbers such, that the product of the first and second is a, that of the first and third is b, and that of the second and third is c. 26. What number is it, whose third part, multiplied by its square, gives 1944 ? Ans. 18. 27. What number is it, whose half, third, and fourth, multiplied together, and the product increased by 32, gives 4640? Ans. 48. 28. What number is that, whose fourth power divided by 4th of it, and 167 subtracted from the quotient, gives the. remainder 12000? Ans. 11. Cases of imaginary Solution. 29. Some merchants engage in business; each contributes a thousand times as many dollars as there are partners. They gain in this business $2560; and it is found that this gain is exactly half their own number per cent. How many merchants are there? Ans. 8. 30. Find three numbers such, that the square of the first multiplied by the second is 112; the square of the second multiplied by the third is 588; and the square of the third multiplied by the first is 576. Ans. 4, 7, 12. 227. Corollary. When the solution of a problem gives for either of its unknown quantities only imaginary values, the problem must be impossible. 228. EXAMPLE. In what case would the value of the unknown quantity in example 13 of art. 226 be imaginary? and why should the problem in this case be impossible? that is, when the product of the sum and difference is required to be greater than the square of a. required number is x, this product is Now if the (a + x) (a — x) = a2 — x2; and, therefore, less than a2. Equations of the Second Degree. CHAPTER VI. EQUATIONS OF THE SECOND DEGREE. 229. It may easily be shown, as in art. 120, that any equation of the second degree with one unknown quantity, may be reduced to the form in which Ar2 denotes the aggregate of all the terms multiplied by the second power of the unknown quantity, Br denotes all the terms multiplied by the unknown quantity itself, and M denotes all the terms which do not contain the unknown quantity. 230. Problem. To solve an equation of the second degree with one unknown quantity. Solution. Having reduced the given equation to the we could easily reduce it to an equation of the first degree, by extracting its square root, if the first member were a perfect square. But this cannot be the case, unless the first term is a perfect square; the equation can, however, always be brought to a form in which its first term is a perfect square, by multiplying it by some quantity which will render the coefficient of the first term a perfect square, multiplying by this coefficient itself, for instance; thus the given equation multiplied by A becomes A2x2+ A B+ A M = 0. |