Examples of Binomial Equations.

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 1170413. 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 Ans. 48.

4640?

28. What number is that, th of whose fourth power divided by it, and 167 subtracted from the quotient, gives the remainder 12000? Ans. 114.

29. Some merchants engage in business; each contributes a thousand times as many dollars as there are partners.

Cases of imaginary Solutions.

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.

152. Corollary. When the solution of a problem gives for either of its unknown quantities only imaginary values, the problem must be impossible.

EXAMPLE.

In what case would the value of the unknown quantity in example 13 of art. 151 be imaginary? and why should the problem in this case be impossible?

that is, when the product of the sum and differ-
ence is required to be greater than the square of a.
Now if the required number is x, this product is

Equations of the Second Degree.

CHAPTER V.

Equations of the Second Degree.

153. It may easily be shown, as in art. 89, that any equation of the second degree with one unknown quantity, may be reduced to the form

Ax2 + B x + M = 0,

in which Ax denotes the aggregate of all the terms multiplied by the second power of the unknown quantity, Bx denotes all the terms multiplied by the unknown quantity itself, and M denotes all the terms which do not contain the unknown quantity.

154. Problem. To solve an equation of the second degree with one unknown quantity.

Solution. Having reduced the given equation to the form

A x2 + B x + M = 0,

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

A2 x2 + A B x + A M = 0.

Equations of the Second Degree.

Now that the equation is in this form, we can readily ascertain whether its first member is a perfect square, by attempting to extract its root, as follows :

A2 x2 + A B x + A MA x + B. Root.
A2 x2

so that the first member is a perfect square only when the

remainder is zero, that is,

AM - B2 = 0;

and, in every other case,

Ax+B

is the root of the square which differs from it by this remainder, that is,

A2 x2+ABx+AM=(Ax++ B)2 + AM-1 B2 = 0;

or, transposing A M - + B2, we have

B2

(A x + + B)2 = + B2 — А М.

Now the square root of this last equation is

B2

A x + + B = ± √ (+ B2 — АМ),

which, solved as an equation of the first degree, gives

in which either of the two signs + or -, may be used in the double sign ±, and we thus have the two roots of the given equation

AM,

which is the same as

A2 x2 + A B x + + B2 = + B2 B2

is obtained immediately from the equation

A2 x2 + A B x + A M = 0,

by transposing AM to the second member, and adding + B2 to both members. Hence

To solve an equation of the second degree with one unknown quantity.

Reduce it as in arts. 85 and 88, transposing all the terms which contain the unknown quantity to the first member, and the other terms to the second member.

Multiply the equation by any quantity, (the least is to be preferred,) which will render the coefficient of the second power of the unknown quantity an exact

square.

Add to this equation the square of the quotient, arising from the division of the coefficient of the first power of its unknown quantity, by twice the square root of the coefficient of the second power of its unknown quantity.

Extract the square root of the equation thus augmented, and the result is an equation of the first degree, to be solved as in art. 90.

154. Corollary. When we have

B2-4A M