is required to find how much he is to receive. Here the calculation may be made as follows:

So that he may claim in ready money £432 19s. 1d.

Generally, then, the present value of the rent (A), to continue n years at 5 per cent., would be equal to the sum of the

Now if the annual rent be A, commencing at present, and lasting n years, and the rate of interest be represented generally by R, it will be thus actually worth

which is a geometrical progression, and the whole is reduced

Ex. 1. What will 3757. amount to in five years at 6 per cent. compound interest?

ANS. £503.

Ex. 2. What will be the amount of 1000l. in 10 years at

5 per cent. compound interest?

ANS. 1628. 16s.

Ex. 3. In what time will 3651. amount to 400l. at 4 per

cent. compound interest?

ANS. 2 years and 145 days.

Ex. 4. At what rate per cent. will 400l. amount to 4907. 11s. 7d. in 3 years

ANS. 6 per cent.

Ex. 5. Required the present worth of 7007., payable three years hence, allowing interest at 5 per cent.?

ANS. £604. 13s. 8d.

Ex. 6. What will 3452l. amount to in 64 years, at 6 per cent. compound interest?

ANS. 143763.

Ex. 7. What will be the amount of 17., placed out to compound interest at 5 per cent. for the space of 500 years?

ANS. £39323200000.

Ex. 8. What will 650l. amount to in five years at 5 per cent. compound interest?

ANS. £829. 11s. 74d.

Ex. 9. What sum must be paid down to receive 600%. in nine months, at 5 per cent.?

ANS. £578. 6s. 34d.

Ex. 10. What will 50l. amount to in five years at 5 per cent. compound interest?

ANS. £63. 16s. 34d.

Ex. 11. What will the same sum at the same time amount to, supposing the same rate of interest to be payable halfyearly?

ANS. £64. Os. 1d.

Ex. 12. What is the present value of an annuity of 70l., to commence six years hence, and then to continue for 21 years, allowing interest at 5 per cent.?

ANS. £669. 14s. Od.

CHAP. LI.

OF THE NATURE OF EQUATIONS OF THE SECOnd degree.

423. It has been already shown (Art. 210), that all equations of the second degree admit of two solutions; and this property ought to be attentively considered in every point of view, because the nature of equations of a higher degree will be considerably illustrated by such an examination. It will be necessary, therefore, to examine with more attention the reasons why every equation of the second degree admits of a double solution.

424. It is true that we have already seen that this double solution arises from the circumstance, that the square root of any number may be taken either positively or negatively; but, as this principle will not easily apply to equations of a higher degree, it may be proper to illustrate this property in another and a more distinct manner. Assuming, therefore, for example, the quadratic equation,

x2=12x-35,

we shall give a new reason why this equation is resolvible in two ways, by admitting for x the values 5 and 7, both of which will satisfy the terms of the equation.

This equation, then, becomes by transposition,

x2-12x+35=0;

and

and it is now required to find a number such, that if we substitute it for x, the quantity x2-12x+35 may be really equal to nothing, and it will remain to show how this may be done in two different ways.

425. Now the whole of this consists in showing clearly, that a quantity of the form of x2-12x+35 may be considered as the product of two factors. In reality, the quantity before us is composed of the two factors (x−5) × (x−7); and since the above quantity must become 0, we must have the product (x −5). (x −7)=0; but a product, of whatever number of factors it be composed, becomes equal to 0, only when one of those factors is reduced to 0. This is a fundamental principle, to which particular attention must be paid, more especially in the consideration of equations of higher degrees.

426. Now, it will be easily understood, that the product (x−5) × (x−7) may become 0 in two ways: first, when the factor x-5=0; and, secondly, when the factor x-7=0. In the first case x=5; in the second x=7. The reason, therefore, is very evident, why such an equation as x2-12x +350 admits of two solutions: that is, why we can assign two values of x, both of which equally satisfy the terms of the equation for it depends upon this fundamental principle, that the quantity x2-12x+35 may be represented by the product of two factors.

427. The same circumstances are found in all equations of the second degree, for in any equation, as x2-ax+b=0, the quantity 2-ax+b may be always considered as the product of two factors, which we shall represent by (x-p) × (x-g). Now as this product must be = 0, it is clear that this may happen in two different ways, first when x=p, and secondly when xq, and these are the two values of a which satisfy the terms of the equation, and are termed the roots of the equation.

428. It

as

428. It will here be necessary to consider the nature of these two factors, in order that their product may exactly produce x-ax+b. By actual multiplication, we obtain x2 - (p+q)x+pq, which quantity must be the same x2-ax+b, and therefore we have p+q=a and pq=b. Hence is deduced this very remarkable property, that in every equation of the form of x-ax+b=0, the two values of x are such that their sum is equal to a and their product to b; from whence it follows, that if we know one of the values, the other is easily found.

429. We have as yet however only considered the case in which the two values of x are positive, and which requires the second term of the equation to have the sign — and the third to have the sign+. Let us also consider the cases, in which either one or both the values of x become negative. The first occurs, when the two factors give a product of the form (x−p) × (x+q), for then the two values of x are x=p and xq, and the equation itself becomes

a2+(q-p)x-pq=0;

the second term having the sign+, when q is greater than p, and when p is greater than q; and the third term being always negative.

The second case occurs, when the two factors are (x+p) × (x+q), for in this case x=-p and x-q, and the equation becomes

x+(p+q)x+pq=0.

430. The signs, therefore, of the second and third terms, show us the nature of the roots of any equation of the second degree. For if the second and third terms be both positive, the roots are both negative; if the second term be negative, and the third positive, both the roots are positive; lastly, if the third term be negative, one of the roots is positive. But, in every case, the second term contains the sum of the roots, and the third their product.

431. It