356 x and y have each two values; in employing these for any purpose it is necessary to take together the roots which have the two upper signs and also the roots which have the two lower signs. The two solutions which are thus obtained differ from each other only by the change of x into y and y into x, a result which was to be anticipated, since this change makes none in the equations x+y=a, x2+y2=b2. The same observation applies to the following example, in which x2+y2=a2, and xy=b2, are given to find the values of x and y. b2 Since xy=b2, y=, and y2= x Substituting this value of y2 in the equation x+y=a2, 64 .*.x++b+=a2x2, and x-a2x2+b+=0. Considering x as the unknown quantity, x2=a2±√a^—4ba, •'•x=±√{a2±{√/aa—4b*. This expression gives four roots of x, substituting these values of x in the b£ equation y the four corresponding values of y are, y= To simplify this expression, let both terms of the fractional value of y be multiplied by +}a°F{√a^—4ba, then y= ±√¿a2±{{√a^—4b*×±√ {a2+√a*—4ba The denominator of this value of y becomes ..y= +b2√ {a2 + {{√a^—4ba ̧ a*—} (a*—4b*)=+V. = = ± √ { a2 + {√a^—4ba. The equations x2+y2=a2; xy=b2, may be resolved more simply as follows: To the first equation x2+y2=a2, then x+2xy + y2 or (x+y)2=a2+262, •••x+y=±√a2 +26o. Subtracting twice the second equation from the first, x2-2xy+y, or (x-y)2=a2-2b2; •*.x-y=+√a2—2b2, then since x+y=+√/a2+2b2, and x-y=√a2-2b2, by addition 2x=±√a2+2b2+√/a2-2b2, and by subtraction 2y=+a+2b2+/a2-2b2, 9 and y=3(±√a2+2b2 +√/a2—2b2). 126. The four solutions obtained in this manner differ from the four obtained by the preceding method. It can, however, be shown that the solutions are identical. For this purpose it is necessary to investigate a method for the extraction of the square root of a quantity which is partly commensurable and partly incommensurable. If a, b are rational quantities; √a, √ their square roots; and if the square of √a+√b is formed, the result, which is a+b+2√√ab, may be represented by the general expression A+B, in which A, B are rational quantities. From this it is concluded, conversely, that the square root of an expression of the form ✔A+√B may, in certain cases, be reduced to the form √a+√b. It is required to effect this transformation when it is possible. It is necessary, therefore, to consider A, B as rational quantities, B as an irrational quantity, and to determine for a, b certain rational values such that, (√a+√b)2=A+√B, or a+b+√4ab=A+√B. In the first member of this equation a+b is by hypothesis a rational number and only the part √4ab irrational. A being rational and ✅B irrational, it is to be proved that a+b=A and that √√4ab√✅B or 4ab=B 1, 2. For suppose that these equalities do not subsist, and transpose the terms of the equation, a+b+√4ab=A+√/B. So that ab A—a−b+√B, where a+b is by supposition not equal to A, or A-a-b is not equal to zero. To abridge this expression make A-a-b-k; then, 4ab=k+B. Squaring both members, 4ab=k2+B+2k√✅B, and transposing, 4ab-k2— B=2k/B, an absurd equality; for a rational cannot be equal to an irrational quantity. The absurdity can be removed by alone making k=0, or a+b=a. Then the last equality is reduced to 4ab-B=0, or 4ab=B. Since a+b=A, (a+b)2=A2, or a2+2ab+b2=A2. therefore a A+√A2—B, The question requires that the values of a and b shall be rational. Now in order that this condition may be fulfilled, it is necessary that A2-B shall be a square. Let A2-B-C2. Then a= ; b= A=C; A+C A+C .. 2 A 2 Since B has been tacitly supposed positive, the radicals of the second member ought both to be taken with the sign + or both with the sign otherwise, if this member is raised to the second power, -✔B and not + ✔B is reproduced. It is assumed that a is greater than b, and for this reason the radical A'-B is taken with the sign +. This explanation of itself proves that if the expression is ✔✅A—√B, it is necessary to take the two radicals with contrary signs; and consequently that √A-√B=√ A+C As a particular case, let A+B=/12+√140; then A=12, B=140, A2-B=144-140=4=C2. A-C 5. As a second particular case, let VA-√B=√1-2x√√1-x2. In this case, A=1, B=4r2—4x4. A2-B=1—4x2+4x+=(1—2x2)2=C2, and... C=1—2x2. Whence 1-2x √/1 · − x2= √√ 2 2 When A2-B is not a square, the values of a, b are irrational; yet by substituting them in a+b, the result is still an expression equivalent to √A+√B. When B is imaginary, if B is changed into -B2, A+B becomes A+B 1. It deserves notice that the square root of A+B-1 is reducible to an expression of the form a+ẞ√√—1, in which a, ß are real quantities which, as well as A, B, need not be rational. This proposition is rendered evident by the substitution of -B2 for B in the investigation contained in this article. By this change C becomes equal to √A2+B2 instead of √A-B; thereA+ A+B2 ; b= 2 2 The radical A+B being fore a= Denoting these values by a2 and—ß2, √a=±a, √b=±ß√−1; and consequently A+B√~1=+(a+ẞ√√−1), . . . . 6 ; in which expression a, B are both taken with the sign+, or both with the sign; because the product 2aß ought to be equal to B which is assumed to be positive. If the sign- precedes B-1, it is evident the expression must be, If the expression ✔✔-1 is compared with A+B√−1, = A=0, B=1, a=' 2 2' If formulæ 6 and 7 are added together (taking only the upper sign where occurs) the imaginary parts destroy each other, and √A+B√−1+√A−B√−1=2a. A+√A+B2 But a&a= .. 2a=√4a=√2A+2√A2+B2; 2 ..√A+B√—1+A-B√-1=√2A+2√A+B2. √+√−1+√~√=1=√/2√T=√2. 127. As a third example, let it be required to find two numbers, such that the sum of their products by the numbers a, b respectively may be equal to 2s, and the products of the numbers to p. Let x, y denote the two numbers. x(28-ax) Substituting this value of y in the 2d equation b =p. Whence ax2-2sx=-bp, 2 Since s is greater than /s2-abp x has two values affected with the same sign; for the same reason y has two values also affected with the same sign. When s2<abp these values are imaginary. If a=b=1, x=s+√/s2—p‚ y=s+√/s2—p. In this case the first value of x is equal to the second value of y, and the second value of x to the first value of y. The equations from which these values are deduced are x+y=2s, xy=p. If x+y or 2s is considered the sum, and ry or p the product of the roots of a quadratic equation (Art. 121), two corresponding values of x and y are given by the equation x2-28x+p=0; and these values are, as above, 4th problem. Given 2e, the sum of the extreme terms of four proportional numbers, 2m, the sum of the mean terms, and 4s2, the sum of the squares, to find the four proportional numbers. Let v, x, y, z represent the four proportionals. Then, by the question, v+z=2e, x+y=2m, v2+x2+y2+z2=4s2; and by Article 332, Part I, vz=xy. To simplify the solution, let vz or xy=u (u being a quantity which is unknown for the present), then the equations v+2=2e} give, as in the last problem, And in like manner the equations x+y=2m } give { x = m + √/m2 -u u ·y=m—√/m2- -U. v=e+ √e2-u -U. Substituting these values of v, x, y, z in the equation v2+x+y2+z2=482, (e+√e2-u)2+(e−√√ e2―u)2+(m+ √ m2 —u)2 + (m—√m2—n)2=48°. Squaring the terms, and reducing 4e2+4m2-4u=4s2, ... u=e2+m2—§2. Introducing this value of u into the expressions of the values of v, x, y, z v=e+√/s2—m2, x=m+√/s2-e2, 9 z=e— √ s2 m2, y=m-√s2-e2. Verification. vz=(e+√/s2—m2) × (e—√/s2 —m2)=e2—s2+m2, The solution of this problem is much facilitated by the employment of the unknown auxiliary u. 128. The most general equation of the second degree containing one unknown quantity has been reduced to the form ax2+bx+c=0. In like manner the most general equation of the second degree containing two unknown quantities is reducible to the form ax2+byx+cy2+dx+ey+f=0, in which a, b, c, d, e, f, are known quantities, numerical or algebraic. Since two equations are required for the determination of two unknown quantities, let two proposed equations of the second degree in x, y be ax2+byx+cy2+dx+ey+f=0 and from these equations let it be required to determine the values of x, y. Arranging the terms of equations 1, 2 according to the powers of x, ax2+(by+d)x+cy2+ey+f=0 3, 4. Next, multiplying the terms of equation 3 by a', and the terms of equation 4 by a, and subtracting the second product from the first, au'x2+(by+d)a'x+ca'y2+ea'y+fa'=0, aa'x2+(b'y+d')ax+ac'y2+ae'y+af'=0; ((ba'-ab')y+(da'—ad'))x+(ca'-ac)y2+(ea-ae)y+fa'—aƒ'=0; (ac-ca') y2+(ae'—ea) y+af' —fa' ..x=- (ba'-ab')y+da-ad If this expression of a is substituted in either of the equations 1, 2, the result is an equation involving only the unknown quantity y; but it is evident, without making the substitution, that since the numerator of the value of x is of the form my2+ny+p, the expression of a must be of the fourth degree in y. Now this expression of x forms a part of the result of this substitution. Whence in general the resolution of two equations of the second degree involving two unknown quantities depends on the resolution of one equation of the fourth degree containing one unknown quantity. 129. The analysis of the second degree in those cases in which the number of unknown quantities exceeds the number of equations is not sufficiently elementary to be considered in this treatise. The subject is discussed in the "Théorie des nombres" of Legendre. 130. To resolve an equation of the second degree. Rule. Reduce the equation to the form x2+px=q; add to both members the square of half the coefficient of x, ; extract the square root of both members, prefixing to the second the sign ±, and deduce the value of x from the new equation. Additional equations and problems of the second degree: |