By J. Parry Lewis (auth.)
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Extra info for An Introduction to Mathematics: For Students of Economics
We commence our solution as before, squaring both sides. , which has solutions x = 6 and x = 118. Once again, only one of these is correct. We must now consider why only one of the solutions obtained by this method is correct. Let us consider the second example, Jx+3 +2x=15 for which we obtained the solutions x = 6 and x x = 6 in the left-hand side we obtain = a47 • If we substitute J9+12 which gives us 15, which is equal to the right-hand side. If, however, we substitute the solution x =a; we have for the left-hand side At first sight this appears to be i+a27 which comes to 22, which is not equal to the right-hand side.
10) 1-l 4(1-rtf24) =8 X l igH =7iU It is often necessary to approximate slightly by evaluating r" to an adequate number of places with logarithms. Example: I am entitled to an annual income of £1,000 now and in each of the next nine years. Because the promise of money in the future is less useful to me than having that sum of money now, I would be prepared to exchange the promise of £1,000 next year for £900 now. I would also be prepared to exchange the promise of £1,000 in the following year for £810 now.
If we square - 2 we again obtain +4. In fact, as far as elementary algebra and arithmetic go, we cannot find the square root of a negative quantity. All the numbers we have met have been either positive or negative (or zero). The square of any positive number is bound to be positive ; and so is the square of any negative number. For reasons which we shall see later, we call a root of this kind, involving the square root of a negative quantity, a complex root. We shall have a great deal more to say about them later on.