These values of x are imaginary, as the factor-1 is so; but they both satisfy the equation, which becomes by putting b√1 for x,

so that the equation is verified. If√-1 1 were sub

a

stituted for x, it would also be found to fulfil the equation. Since 2 is positive, whether the value of x be positive or negative, the term ax2 is always positive for a real value of a, and 2 is positive, but the sum of two positive quantities cannot be = 0, and from this circumstance alone it appears that the equation is impossible for real values of x. When the root of such an equation therefore is imaginary, it proves the equation to be impossible for real values of the unknown quantity.

2. Given 3x2 1583+x2, to find x
3. Given a2x2 - 2 — 0, to find x
4. Given ax2 +b=0, to find x
5. Given d2 + bx2 = a2, to find a
6. Given a2x2+d2 = c2x2 + b2

QUESTIONS PRODUCING PURE QUADRATIC EQUATIONS.

1. What number is that whose square being added to 50, the sum is 99?

The number is therefore +7 or -7, the square of which is 49, and 49 +50 = 99; so that the number satisfies the given condition.

2. There is a number whose 4th power is 4 times its square: required the number.

Now, 24(2)=16, and 4a2=4(+2)2=4x4=16, so that +2 or 2 is the required number.

3. Find a number such that 10 times its 5th be equal to 250 times its cube.

power shall

4. Find a number such that (m2) times its mth power shall be equal to m times its (m +2)th power.

Let a the number,

(m—2) xm = mxm+2
тат+2 =

then

or

(m-2)xm

2m, the value of a will be a fraction. If

2

m2, the numerator is imaginary, and the question impossible.

5. A number is equal to 9 times its reciprocal: required the number.

6. If the height through which a heavy body falls in different periods of time be proportional to the squares of these times, in how many seconds will a body fall through 400 feet, the space it falls through in one second being 16·1 feet?

Let the number of seconds taken in falling through 400 feet,

then

hence

16.1:400= =12:22

16.1 x2400

x=√24·9 ± 4.9, or nearly 5 seconds.

=

The negative value - 4.9 implies that a heavy body thrown upwards with such a velocity as would make it ascend to 400 feet, would take 4.9 seconds of time to reach that height. The velocity cannot be taken in an opposite sense, but the direction of the motion can be so, and by this change a proper interpretation of the negative value is determined.

7. If the force of the earth's attraction at any point not within its surface diminish inversely as the square of its distance from the centre, at how many semi-diameters distant from the earth's centre will this force be 16 times less than at the surface of the earth?

Let the force at the earth's surface, that is, at the distance of a radius or semi-diameter from the centre, be called 1, and let a be the required distance, that is, the number of

semi-diameters from the centre at which the force is 16

times less, then

therefore

a2 = 16, and x =√16= ± 4 semi-diameters.

EXERCISES.

1. Find a number such that the square of its half added to 6 shall give 10. Ans. 4. 2. Find a number such that if 3 times its cube be diminished by 5 times the number itself, the remainder shall be equal to 7 times the number.

Ans. 2. 3. Find a number such that its fourth power shall be equal to 9 times its square.

Ans. 3.

4. Find a number such that 10 times its sixth power shall be equal to 40 times its fourth power.

Ans. 2.

5. Find a number such that (m2) times its (n+2)th power shall be equal to m times its nth power.

6. In what time will a body fall through a height of 1000 feet? (See 6th example.) Ans. 7.9 seconds.

7. If the attraction of a magnet, at the distance of 10' inches from its pole, can support a small weight of 20 grains, at what distance will it support only 4 grains, its attractive power nearly varying inversely as the distance from its pole? Ans. 22.4 inches.

1. Adfected quadratics containing only one unknown quantity.

(274.) Every adfected quadratic, as before observed (272), may be reduced to the form

and there will be the four following cases, according as 6 or c is positive or negative, namely:

the signs being taken in any order.

(275.) When it happens in the first form that 4 ar= (+6)=2, the equation is a complete square (180), and is of the form

m2x2 + 2 mnx+n2 = 0

and its square root is (190) ... mx ± n = 0

Thus, if m2 and 3, the equation is

(276.) When the equation is a complete square, the value of x is thus easily found, but this seldom happens; and when it is not so, it may be solved by these two methods :I. METHOD.

(277.) Arrange the equation, if necessary, so that the first member shall consist of two terms, the first term containing the square of the unknown quantity, and the second that quantity itself, and that the second member may contain only known quantities. Free the square of the unknown quantity of its coefficient, if it have any, by dividing the equation by it.

To make the first member a complete square, add to both sides the square of half the coefficient of the second term. Then extract the square root of both sides, and the result is a simple equation, from which the value of the unknown quantity may easily be found by former rules."