This problem shows how much the introduction of an unknown auxiliary facilitates the determination of the principal unknown quantities. There are other problems of the same kind, which lead to equations of a degree superior to the second, and yet they may be resolved by the aid of equations of the first and second degrees, by introducing unknown auxiliaries. 152. We will now consider the case in which a problem leads to two equations of the second degree, involving two unknown quantities. An equation involving two unknown quantities is said to be of the second degree, when it contains a term in which the sum of the exponents of the two unknown quantities is equal to 2. Thus, 3x2-4x+y3—xy-5y+6=0, 7xy-4x+y=0, are equations of the second degree. Hence, every general equation of the second degree, involving two unknown quantities, is of the form ay2+bxy+cx2+dy+fx+g=0, a, b, c,... representing known quantities, either numerical or algebraic. Take the two equations a y2+b xy+c x2+dy+fx+g=0, a'y2+b'xy+c'x2+d'y+f'x+g'=0. Arranging them with reference to x, they become cx2+(by+f)x+a y2+d y+g=0, Now, if the co-efficients of x in the two equations were the same, we could, by subtracting one equation from the other, obtain an equation of the first degree in x, which could be substituted for one of the proposed equations; from this equation, the value of x could be found in terms of y, and by substituting this value in one of the proposed equations, we would obtain an equation involving only the unknown quantity y. By multiplying the first equation by c', and the second by c, they become cc'x2+(by+ƒ)c'x+(ay2+dy+g)c'=0, cc'x2+(b'y+f')c x+(a'y2+d'y+g')c=0, and these equations, in which the co-efficients of x2 are the same, may take the place of the preceding. Subtracting one from the other, we have [(bc'—cb')y+fc' — cf']x+(ac' —ca')y2+(dc'—cd')y+gc'—cg'=0, which gives (ca'—ac')y2+(cd'—dc')y+cg'-gc' This expression for x, substituted in one of the proposed equations, will give a final equation, involving y. Y But without effecting this substitution, which would lead to a very complicated result, it is easy to perceive that the equation involving will be of the fourth degree; for the numerator of the expression for a being of the form my2+ny+p, its square, or the expression for x2, is of the fourth degree. Now this square forms one of the parts of the result of the substitution. Therefore, in general, the resolution of two equations of the second degree, involving two unknown quantities, depends upon that of an equation of the fourth degree, involving one unknown quantity. 153. There is a class of equations of the fourth degree, that can be resolved in the same way as equations of the second degree; these are equations of the form x2+px2+q=0. They are called trinomial equations, because they contain but three kinds of terms; viz. terms involving a1, those involving x2, and terms entirely known. In order to resolve the equation +p+q=0, suppose 2=y, we have We perceive that the unknown quantity has four values, since which affect the first radical, can be each of the signs + and combined successively with each of the signs which affect the second; but these values taken two and two are equal, and have contra ry signs. Take for example the equation by supposing 2=y, it becomes whence x4-25x2-144; y2-25y=-144; Substituting these values in the equation ay there will result 1st. x2=16, whence x=4; 2d. x2=9, whence x=±3. Therefore the four values are +4, 4, +3 and -3. Again, take the equation x-722-8. Supposing 2=y, the equation becomes y2-7y=8; whence y=8, y=—1. Therefore, 1st. 2-8, whence x=2√2; 2d. x=-1; whence x=: =√1; the two last values of x are imaginary. Let there be the algebraic equation taking 2=y, the equation becomes (2bc+4a2)x2=—b2c2 ; y2 − (2bc+4a2)y =—b2c2 ; bc + 2a2±2a√bc + a2. And consequently x= bc + 2a2±2a √bc + a2. 154. Every equation of the form y2"+py”+q=0, in which the exponent of the unknown quantity in one term is double that of the other, may be solved by the rules for equations of the second degree. For, put y"=x, then y2=x2, and y2”+py”+q=x2+px+q=0. Extraction of the Square Root of Binomials of the form a ± √ō. 155. The resolution of trinomial equations of the fourth degree, gives rise to a new species of algebraic operation: viz. the extrac tion of the square root of a quantity of the form a± √√b, a and b being numerical or algebraic quantities. By squaring the expression 3± √5, we have (3± √5)2=9±6 √5+5=14±6 √5: hence, reciprocally 14±6 √5=3±√5. Whence we see that an expression of the form Va±√b, may sometimes be reduced to the form a'±√b' or √a'±√b'; and when this transformation is possible, it is advantageous to effect it, since in this case we have only to extract two simple square roots, whereas the expression Va±√b requires the extraction of the square root of the square root. 156. If we let p and q denote two indeterminate quantities, we can always attribute to them such values as to satisfy the equations √a+ √b=p+q... (1). √a— √b=p—q · · (2). These equations, being multiplied together, give √a2-b=p2-q2. . . . . . (3). Now, if p and q are irrational monomials involving only single radicals of the second degree, or if one is rational and the other irrational, it follows that p2 and q3 will be rational; in which case, p-q3, or its value, Va2-b, is necessarily a rational quantity, or aa—b is a perfect square. When this is the case, the transformation can always be effected. For, take a2-b, a perfect square, and suppose Va-b-c; the equation (3) becomes p2-q2=c. Moreover, the equations (1) and (2) being squared, give p2+q2+2pq=a+√b, p2+q-2pq=a— √b; whence, by adding member to member, Hence, by adding these last equations, and subtracting the second from the first, we obtain These two formulas can be verified; for by squaring both members of the first, it becomes a+ √b=- +. +21 a+c a-c a2-c2 =a+ √ a2 — c2 ; 4 but the relation √a2-b=c, gives c2=a2-b. |