PROBLEMS REQUIRING THE SOLUTION OF QUADRATIC EQUATIONS OF ONE UNKNOWN QUANTITY.

122. In forming the equation from the given question, the Rules given in page 75 must be observed. In the solution of the equations peculiarities often occur. Some of these will be discussed in the examples worked out below. 1. The sum of two numbers is 20, and the sum of their cubes is 3500; find the numbers.

In this case, therefore, the two solutions lead to the same result. This question may also be solved by two simultaneous equations. These will be

= 20

x + y 23+ y3

the solution of which is more symmetrical than that just given.

2. I have to walk a distance of 144 miles, and I find that if I increase my speed 13 miles per hour, I can do the journey in 16 hours less than if I walk at my usual rate: find my usual rate of walking. the number of miles per hour I usually walk. the increased rate.

Let
And x + 3/2

x =

=

The value 3 is the only one that applies to this problem; 3 miles per hour is therefore the rate of walking.

The negative value of x may be explained by a reference to the following problem ;

I have to walk a distance of 144 miles, and I find that if I diminish my speed 12 miles per hour, I can do the journey in 16 hours more than if I walk at my usual rate: find my usual rate of walking.

If we construct an equation from this question, we shall find it to be

which are the same numerically as the roots of the former, but are of opposite signs. That is, the negative root of the first equation, with its sign changed, is the solution of the second question; that is, 4 miles per hour.

The pupil should carefully compare the statements in the two problems. We shall return to the subject presently.

3. The length of a room is 6 yards more than its breadth, and the cost of a carpet to cover it, at 58. per square yard, is £20 11s. 3d.: find the size of the room.

The negative value may be explained by a reference to the following question.

The breadth of a room is 6 yards less than its length, and the cost of a carpet, &c. (the same as before).

In this case we see that the negative root of the first question, with its sign changed, is the solution of the second question.

4. A farmer bought one flock of sheep for £40, and another flock of four sheep less at 5s. a-head more; the whole cost of the larger flock was £4 more than the cost of the smaller find the number of sheep in each flock.

Let

x = the number of sheep in the larger flock. x-4=

lesser

And 20 The value

32 bears reference to the following question:

A farmer bought one flock of sheep for £40, and another flock of four sheep more, at 58. a-head less; the whole cost of the larger flock was £4 less than the cost of the smaller find the number of sheep in each flock.

Let

x = the number of sheep in the smaller flock. X- 4 =

Hence we get the equation

larger

The following principle will serve as a guide to the interpretation of the negative roots of quadratic equations.

State a second question, in which all the positive quantities of the original question are made negative, and all the negative positive. Construct the equation corresponding to this new question, and solve it; then the positive root of this new equation will be equal numerically to the negative root of the original equation, and will be the answer to the second question.

In practice, the following rules for writing down the question to which the negative value refers, will be found useful.

Wherever the original question implies

the new must imply

addition

subtraction

according to the explanations in p. 7.

5. Divide a line a feet long into two parts, so that the rectangle contained by the whole line and one of the parts shall be equal to the square of the other part.