837. When the middle term which contains the first power of x is wanting, the equation assumes the form axx+c=0, or axx = c, as the sign of c is positive or negative. Such an equation is called a pure equation of the second degree, since its resolution is without difficulty, for we have only to divide by a, which gives xx= =2; and, taking the square root of both sides, we have x= ✔, which resolves the equation.

a

838. Three cases are however to be considered; I. when is a square number (whereof, therefore, the root can be assigned) we obtain for the value of x a rational number, either integer or fractional. Thus, xx=144 gives x=12; and xx= gives . II. When

is not a square, in which case the sign ✔ must be used. If, for example, xx=12x=√12, the value whereof may be determined by approximation, as heretofore shown. III. When becomes a negative number, then the value of x is altogether impossible and imaginary; this result, indeed, proves the question, that such an equation is in itself impossible.

a

839. It must be here observed, that whenever the extraction of the square root is required the root has two values, the one negative, the other positive. Take the equation xx= 49, the value of x is as well -7 as +7, which is expressed by ±7. Hence all these questions admit of a double answer; but it will easily be perceived that there are many cases in which a negative value cannot exist.

840. In such equations as axx=br, where the known quantity c is wanting, two values of r may exist, though dividing by r we only find one. Thus, in the equation xx= 3x, wherein a value of x is required, such that xx may become = 3x. This is done by supposing x=3, a value found by dividing the equation by x. Besides this value, however, there is another equally satisfactory, namely, x=0, for then xx=0, and 3x=0. Thus equations of the second degree admit of two solutions, whilst simple equations admit only of one. 841. We here submit three examples for the illustration of pure equations of the second degree.

I. Find a number the half whereof multiplied by the third produces 24.

Let the number = x. fr multiplied by jr must produce 24. Therefore rx=24. Multiplying by 6 we have xx=144, and the root extracted gives x= ± 12. We put +12; for if x = +12, we have fr=6 and fr=4, and the product of these quantities is 24. If x=-12, we have r = −6, and jr = −4; the product of which is likewise 24. II. Find a number such that by adding 5 to it and subtracting 5 from it the product of the sum by the difference would be 96.

Let the number be x. Then +5 multiplied by x-5 must give 96. Therefore

xx-25=96.

Adding 25 we have rr=121, and extracting the root we have x=11.

Thus x+5=16, and x-5=6, and lastly, 6 x 16=96.

III. Find a number such that by adding it to 10 and subtracting it from 10 the sum multiplied by the remainder or difference will give 51.

Let the number = x. Then 10+ multiplied by 10-x must make 51. So that

100-xx-51.

Adding xx, and subtracting 51, we have xx=4 -49; the square whereof gives x=7.

RESOLUTION OF MIXED EQUATIONS OF THE SECOND DEGREE.

842. If three kinds of terms are found in an equation of the second degree, namely, — I. That which contains the square of the unknown quantity, as arr; II. That in which the unknown quantity is only known in the first power, as be; III. The kind of terms composed of known quantities only, such an equation is said to be mixed or complete ; since two or more terms of the same kind may be brought into one, and all the terms may be brought to one side of the sign The general form of a mixed equation of the second degree will be xxxbx c=0.

=.

And we now propose to show how the value of r may be derived from such an equation; which may be done in two ways.

843. Such an equation may be by division reduced to such a form that the first term may contain only the square xr of the unknown quantity x. Leaving the second term on the same side with x, we will transpose the known term to the other side of the sign =. Thus the equation assumes the form xx± px = ±q, in which p and q represent any known numbers, positive or negative; and all we have to do is to find the true value of x. Now if xx+pr were a real square, no difficulty would attend the solution; because it would only be required to take the square root on both sides. It is, however, evident that xx + px cannot be a square; for we have already seen that if a root consists of two terms, for example x + n, its square must contain three terms, namely, twice the product of the two parts, besides the square of each part; that is, the square of x+n is xx + 2nx + nn. Having,

then, already on one side rx + px, we may consider xr as the square of the first part of the root, and in this case pr must represent twice the product of the first part of the root by the second part, whence the second part must be p; and indeed the square of + p is found to be xx + px + pp. Now this last being a real square, which has for its root x + p, if we resume the equation xx+px=q, we have only to add pp to both sides, and we obtain zx+px + \pp=q+App, the first side being a square and the other containing only known quantities. Taking, therefore, the square root of both sides, we find a + p= √(pp + q) ; and subtracting p, we obtain a = = − { p + √ } pp +9; and as every square root is positive or negative, we shall have for a two values, thus expressed

This formula contains the rule whereby all quadratic equations may be resolved, and it should be well remembered, so that it may not be necessary to repeat it. The equation may be always arranged in such a manner that the pure square rr may be found on one side, and the above equation have the form xx=-px+q, where it is evident that x=-} + √ipp + q.

844. The general rule deduced, therefore, to resolve the equation xx=-px + q, depends upon the consideration that the unknown quantity a is equal to half the coefficient or multiplier of x on the other side of the equation, plus or minus the square root of the square of this number, and the known quantity which forms the third term of the equation. 845. Thus, having the equation xx=6x+7, we should immediately say that x=3± √9+7=3±4, when we have for values of x, I. x=7; II. x = −1. So, also, the equa

tion zx=10x-9 would give a=5± √√25-9=5±4; that is, the two values of x are 9 and 1. The rule will be better understood by the arrangement under the following cases: I. Let p be an even number, and the equation such_that_xx= 2px+q, we shall, in that case, have r=p± √pp+q.

II. Let p be an odd number, and the equation xa=px + q ; we shall here have x=p± √\pp+q; and since ipp + q = we may extract the square root of the

=

pp+1q

4 "

III. Lastly, if p be a fraction, the equation may be resolved in the following manner.

The other method of resolving mixed quadratic equations is to transform them into pure equations, which is effected by substitution; thus, in the equation xx=pr+q, we write another unknown quantity y, instead of the unknown quantity x, such that x=y+p, by which, when y is found, the value of x may be readily determined.

846. Making this substitution of y+p instead of x, we have xx=yy+py+\pp, and pr=py+pp; hence our equation becomes yy +py + {pp=py + App+q, which is first reduced by subtracting py to yy + pp=pp+q; and then by subtracting {pp to yy=\pp+q. This is a pure quadratic equation, which directly gives y= ± √}pp+q. And since z=y+p, we have x={p± √pp + q as before. We shall now illustrate the rule by some examples.

I. What are those two numbers, one whereof exceeds the other by 6, and whose product is 91?

If the less ber, the other is x + 6, and their product xx+6x=91.
Subtracting 62, the remainder is ax=91-6x, and the rule gives

=-13.

x=−3± √9+ 91 = −3 ± 10; so that x=7, and x=

The question admits of two solutions: by one, the less number x=7, and the greater x+6=13; by the other, the less number r=-13, and the greater x+6=-7.

II. Find a number from whose square if 9 be taken, the remainder is a number as many units greater than 100 as the number sought is less than 23.

Let the number sought. We know that xx-9 exceeds 100 by xx-109, and since z is less than 23 by 23-x, we have the equation xx-109=23—x.

Wherefore xx= −x + 132, and by the rule = − − ↓ ± √ } + 132 − ¦± √/529 = − } ± 23. So that x=11, and x = -12.

If, then, a positive number be required, the number is 11, the square of which minus 9 is 112, and, consequently, greater than 100 by 12, in the same manner as 11 is less than 23 by 12.

III. To find two numbers in a double ratio to each other, such that if we add their sum to their product we may obtain 90. U

=

Let one of the numbers = x. Then the other will be 2x, their product also= 2xx.
If we add to this 3r, or their sum, the new sum should make 90.
So that 2xx+3x=90, 2xx=90−3x, and xx – –

-+45; whence we obtain

IV. To find such a number that if its half be multiplied by its third, and to the product half the number required be added, the result will be 30.

Let the number = x. Its half multiplied by its third will make frx, so that дxx+ x = 30.

Multiply by 6, and we obtain xx + 3x=180, or xx = −3x+180, which gives

x = − } ± √} +180 = − ¦ ± 77 ;

consequently x is either =12, or -15.

PURE EQUATIONS OF THE THIRD DEGREE.

847. An equation of the third degree is pure when the cube of the unknown quantity is equal to a known quantity, and neither the square of the unknown quantity nor the unknown quantity itself is found in the equation.

x3 = 125; or, more generally, x3=ɑ, x3 =
= are such equations.

The method of deducing the value of x from such an equation is obvious, for we have
only to extract the cube root on both sides. Thus the equation x3=125 gives x=5; the
Va
equation 3=a gives x= Va; and the equation = gives x= v2, , or x=

a

To re

solve such equations it is only necessary to know how to extract the cube root of a given number. In this way, however, we obtain only one value for x; but as every equation of the second degree has two values, we have reason to suspect that an equation of the third degree has also more than one value. To investigate this is the object of what follows.

848. Let us, then, take the example x3-8 to find from it the numbers whose cube = 8. As = 2 is such a number, x3-8=0, and must be divisible by x-2. The division is as follows:

Hence it follows that the equation r3-8=0 may be represented by these factors, –

(x − 2) + (xx + 2x + 4)=0.

The question then is, what number is to be substituted for x in order that x3=8, or that 3-8=0; and it is manifest that the condition is answered by supposing the product just found to be equal to 0. This, however, occurs not only when the first factor x-2=0, whence we have r=2; but also when the second factor xx + 2x+4=0. We will there

fore make + 2x+4=0, and we shall have xx= -2x-4; and thence x=-1+ − S. From which we learn that besides the case in which r=2, which corresponds to the equation x3=8, there are also two other values of a whose cubes are also 8. These are I. x = −1 + √−3; and, II. x= − 1 − √ −3; which will be evident from cubing them Thus

These values, it is true, are imaginary or impossible, yet they must not be neglected; and, indeed, what has been said applies to every cubic equation, such as xa; namely, that besides the value x= Va we shall always find two other values. To abridge the calcu

lation, let us suppose √a=c, so that a=c3; the equation will then assume this form, x-3=0, which will be divisible by rc, as under.

Consequently the equation in question may be represented by the product (x-c) × (xx + cx + cc)=0, which, in fact, is =0, not only when x-c=0, or x=c, but also when rx + cx + cc=0. This expression contains two values of x, inasmuch as it gives xx= -cz-ce, and r=Scc, that is, x=

сс

-cc, or x=

2

2

x c.

Now, c having been substituted for Va, we conclude that every equation of the third degree of the form x3 = a furnishes three values of x, expressed thus:

a.

III. x=

x ya.

2

This shows that every cube root has three different values, but that one only is real, and the other two impossible: and this is the more remarkable, since every square root has two values; and if we were to pursue the subject, (which is not our intention, since the subject is unnecessary for the architect, but if he wishes he must refer to books on the subject,) we should find that every biquadratic root has four different values, and so on with fifth roots, &c. In ordinary calculations only the first of those values is employed, the other two being imaginary. We subjoin some examples:

I. To find a number whose square multiplied by its fourth part shall produce 432. Let the number =; then the product of xx multiplied by a must = 432: that is, 3432, and 3=1728.

By extracting the cube root we have r=12.

The number sought, then, is 12 for its

square multiplied by its fourth part, or by 3=432.

II. To find a number whose fourth power divided by its half, with 14 added to the product, is 100.

Let the number = r; its fourth power will be xa.

Dividing by the half, or i̟x, we have 2æ3 ; and adding to that 144, the sum must be 100. We have, therefore, 2x3 + 14}=100; subtracting 14 the remainder is 2r3:

343
89

Dividing by 2 we have r3= and extracting the cube root we find x =

343

RESOLUTION OF COMPLETE EQUATIONS OF THE THIRD DEGREE.

849. An equation of the third degree is said to be complete when besides the cube of the unknown quantity it contains the unknown quantity itself and its square; so that the general formula for these equations, bringing all the terms to one side, is,

850. We here propose to show the method of deriving from such equations the values of I, which are also called the roots of the equation. There is no doubt, as we have seen in the last section, that such an equation has three roots, as in respect of pure equations of the same degree.

X

851. First, then, considering the equation x3-6xx ÷ 11x−6=0; since an equation of the second degree may be considered as the product of two factors, an equation of the third degree may be represented by the product of three factors, which in the present instance are (x-1)x(x-2) x (x-3)=0; for by actually multiplying them we obtain the given equation. Thus, (x-1) × (x-2) gives xx-3x+2, and this multiplied by (x-3) gives x3-6xx+11x-6, which are given quantities, and =0. This happens when the product (x−1) × (x−2) × (x−3) becomes nothing; and as it is sufficient for this purpose that one of the factors =0, three different cases may give this result; namely, when x-1=0, or x=1; secondly, when x-2=0, or, x=2; and lastly, when x-3=0, or, x=3. If any number whatever besides one of the above three were substituted for a none of the factors would become = 0, and consequently the product would no longer = 0, which proves that the equation can have no other root than those three. If in every other case three factors of such an equation could be assigned in the same manner, we should immediately have three roots. Let us, then, consider more generally these three factors, x-p, x-q, x-r. Seeking their product, the first multiplied by the second gives xx-(p+q)x+pq, and this

product multiplied by x-r makes x3-(p+q+r) xx + ( pq + pr + qr) x − pqr. If this formula must become =0, it may happen in three cases: first, when x-p=0 or x=p; second, when r−q=0, or x=q; third, when x-r=0, or x=r. Let us then represent the quantity found by the equation x3-axx + bx − c = 0. That its three roots may be,

I. x=p; II. x=q; III. x=r, it is evident we must have, 1st. a=p+q+r; 2d. b=pq+ pr+qr; and 3d. c=pqr; from which we find that the second term contains the sum of the three roots; that the third term contains the sum of the products of the roots taken two by two; and lastly, that the fourth term consists of the product of all the three roots multiplied together. From this last property is deduced the truth, that an equation of the third degree can have no other rational roots than the divisors of the last term, for that term being the product of the three roots must be divisible by each of them. root by trial, we immediately see what numbers we are to choose.

Hence, to find a

As this

852. Let us, for instance, consider the equation x3 = x + 6, or x3—r—6=0. equation can have no other rational roots but numbers which are factors of the last term 6, we have only the numbers 1, 2, 3, 6 to try with, the result whereof will be as follows:

From which we see that x=2 is one of the roots of the given equation; and it will now be easy to find the other two; for a = 2 being one of the roots, -2 is a factor of the equation, and the other factor is to be sought by means of division, as follows:

853. Since, then, the formula is represented by the product (x-2) × (xx + 2x + 3), it will become 0 as well when x-2=0 as when xx + 2x+3=0. This last factor gives rr = -2x -3; and consequently r=-1 ± √ −2. These are the other two roots of the equation, and they are evidently impossible or imaginary.

854. The operation explained is, however, only applicable when the first term 3 is multiplied by 1, the other terms of the equation having integer coefficients. When this is not the case, a mode must be adopted by which the equation is transformed into another having the condition required, after which the trial mentioned may be made.

y

8

4

8

855. Let us, for instance, take the equation x3- - 3xx + 4x-3=0. As there are four 13 Зуу lly parts in it, let us make x=2; we shall then have + -3=0, and multiplying by 8 we obtain the equation y3-6yy+11y-6=0; the roots of which are, as we have already seen, y=1, y=2, y=3; whence in the given equation we have, I. x={; II. x=1; III. x=}.

Di

Let us take an equation in which the coefficient of the first term is a whole number, different from 1, and whose last term is 1: for instance, 6x3-11xx + 6x − 1 =0. viding by 6 we have x-xx+x−}=0. The equation may be cleared of fractions by

y

y3

216

216

+-1=0; and

the method just shown. First, supposing a=%, we shall have 6' multiplying by 216 the equation becomes y3-11yy +36y-36=0. It would be tedious to try all the divisors of the number 36, and as the last term of the original equation is 1, it is better, in this equation, to suppose x= for we shall then have 3-+-1=0. Transposing the terms 23-6zz + 11z-6=0, the roots are here z=1, z=2, z=3, whence in our equation a=1, x={}, x=}.

1

6

6

856. It has been heretofore shown that to have all the roots in positive numbers the signs plus and minus must succeed each other alternately; by this means, the equation takes the form a3-axx + bx-c=0; the signs changing as many times as there are positive roots. Had the three roots been negative, and the three factors r+p, x+q, x+r had been multiplied together, all the terms, would have had the sign plus, and the form of the equation would have been x3 + arx + bx + c=0, wherein the same signs follow each other three times, that is, the number of the negative roots.

857. From this we may learn that as often as the signs change the equation has positive roots, and when the same signs follow each other the equation has negative roots; and this teaches us whether the divisors of the last term are to be taken affirmatively or negatively, when we wish to make the trial that has been mentioned.