3. A and B lay out money on speculation; the amount of A's stock and gain is $27, and he gains as much per cent. on his stock as B lays out. B's gain is $32; and it appears that A gains twice as much per cent. as B. Required the capital of each.

Ans. A's capital, $15; B's, $80.

4. There are two square fields, the larger of which contains 25,600 square rods more than the other, and the ratio of their sides is as 5 to 3. Required the contents of each.

Ans. The larger, 40,000 square rods; the smaller, 14,400

square rods.

5. Bought sugar at such a rate, that the price of a pound was to the number of pounds as 4 to 5. If the cost of the whole had been 45 cents more, the number of pounds would have been to the price of a pound as 4 to 5. How many pounds were bought, and what was the price per pound?

Extracting the square root, x = 10, the no. of pounds. Substituting the value of x in (1), y = 8, the price per pound.

6. A and B engage in a speculation. A disposes of his share for $11, and gains as many per cent. as B

invested dollars. B's gain was $36, and the gain upon A's investment was 4 times as many per cent. as upon B's. How much did each invest?

Ans. A invested $5; B, $120.

7. What two numbers are those whose difference multiplied by the less is 42, or by their sum is 133?

Ans. 13 and ± 6.

8. Two workmen, A and B, engage for a certain number of weeks, at different rates. At the end of the time, A, who had been absent 4 weeks, received $75; but B, who had been absent 7 weeks, received only $48. Now, if B had been absent only 4 weeks, and A 7 weeks, they would have received exactly alike. How many weeks were they engaged for, how many did each work, and what had each per week?

Ans. Engaged for 19 weeks; A worked 15 weeks, B 12 weeks; A received $5, B $4, per week.

Here, the conditions lead to the equation 25 (x — 7)2 = 16 (x — 4)2, which, by taking the square root of each member, gives 5(x-7)=4(x-4).

X

AFFECTED QUADRATIC EQUATIONS.

273. An AFFECTED QUADRATIC EQUATION is one which contains both the second and first powers of the unknown quantity; as,

x2 + ax = b, and ax2 + b x

c = b x2

ax + d.

Equations of this kind, containing every power of the unknown quantity from the first to the highest given, are sometimes called complete equations.

274. Any affected quadratic equation can always, by reduction, be exhibited under the general form

x2 + px = 9,

where p and q are understood to-represent any quantities, positive or negative, integral or fractional.

For we can reduce it to this form, by bringing all the terms containing the unknown quantity to the first member of the equation, and the known quantities to the other, and then dividing by the coefficient of x2.

Equations of this kind, therefore, have sometimes been denominated trinomial, or those of three terms.

275. To solve an affected quadratic equation, it is obviously required to take the square root of its members. But the member containing the unknown quantities alone cannot be a perfect square (Art. 224), nor can it often be so rendered by transposition. It is therefore necessary to have recourse to methods of completing its square.

FIRST METHOD OF COMPLETING THE SQUARE.

276. Suppose we have given 'any quadratic equation, x2 + px = 9,

in which it is required to find the values of x.

Adding (?)2

2

(ハ)

to both members (Art. 90), we have

x2 + px + 12 = q + R 2,

where the first member is a complete square. Extracting the square root, we obtain

Here, since px must be twice the product of the two terms of the root of the completed square (Art. 90),

Р

and x being one of those terms, must be the other.

2

We therefore made x2+px a perfect square by adding to

it

(2)2, or ", which had also to be added to the second

4

member to preserve the equality.

The double sign is used in the result, since the square

Every quadratic equation may have two roots.

1. Given 2x2 + 12 x+ 36 = 356, to find the values

Here, taking 13, the positive root of 169, we find

x = 10; but taking

- 13, the negative root, we find

x =

16.

2. Given x2

By transposition,

12x+30= 3, to find the values of x.

Completing the square,

Extracting the square root,

Transposing,

Whence,

12x+30

х2

12x=

3

12x + 36 = 9

where both values of x are positive.

27

63

x=6±3

X = 9, or 3;

The results obtained in each of these operations may be readily verified. Hence the following

GENERAL RULE.

Reduce the given equation to the form x2+ px = q. Complete the square by adding the square of half the coefficient of x to both members. Extract the square root of both members, and solve the simple equation thus produced.

Sometimes the first member of the equation reduces to a perfect square, and the second to 0. Then the root may be found directly; but x will really have but one value.

Thus,

It is, however, convenient in this case to consider the

quadratic equation as having two equal roots.