a ral one. ta a a a 2 a known quantities can be substituted, alternately, for each other, without producing a new equation, are always capable of being reduced to others of lower dimensions. In order to such a reduction, let the equation, if it be of an even dimension, be first divided by the equal powers of its two quantities in the middle term: then assume a new equation, by putting some quantity (or letter) equal to the sum of the two quotients that arise by dividing those quantities one by the other, alternately ; by means of which equation let the said quantities be exterminated; whence a numeral equation will emerge, of half the dimensions with the given lite But, if the equation propounded be of an odd dimension, let it be first divided by the sum of its two quantities, so will it become of an even dimension, and its resolution will therefore depend upon the preceding rule. Let there be given the equation 2*-4ax:+50x?—4ata=0. za 4x 4a a Here, dividing by aʻzé, we have +5 = 0, (or a ade, a? + 4X-++5= 0, by joining .the corresponding terms ;) zona and by making z = ent, and by squaring both sides, we have + , zca a? x2 a? also z=+2+ or z1-27 Therefore, by substituting 22° these values, our equation becomes 2–2—42+5, or 2–4z=3; whence z=3. But + being - = 2, we have xizax = =-a; and consequently x={za+N (1az2—23) = jax{zEN (2-4)} =fax{3EN (5)}, in the present case. 2. Let there be given 25 740x4—12a*x3—12aRx?+4aʻxtar In this case, we must first divide by zta, and the quotient will come out 24 +3ax—15a*x +3aRxta -0: whence, by proceeding za Q3 as in the former example, we have it to ++3x+ 15 = 0, a or 2-2 +3z-15—0, and from thence z=1(73)-3. Given 7x0–26ax'—26aRx+7=0, which, divided by a®X®, be23 a a? Gt , 2013 z? a? e di Deco we likewise have have 23 -22 = 十一十 a a a a =0. : comes 7x + –26 x+y=0. Now making, as before, tzt; and therefore 2–37= 3 a a a with a zas a 2003 a® + : which values be. 2003 a zao ing substituted above, our equation becomes 7 X (28—32)—26X (7_2=0, or 723–262—212 + 52—. Where, trying the divisors of the last term, which are 1, 2, 4, 13, &c. the third is found to answer; z, consequently, being = 4. 4. Wherein let there be given 2x?—13a%2—13aRx+2a=0. Here dividing first by x+a, the quotient will be 2x8–2ax11a*x* +11a%23—11a*z?_2a2+2a=0; which, divided again by 2013, as 22 a? dozo, gives 2x + 3)-2xã+ a3 -11X6)+11=0, that is, 2X (28_3z)—2X(2-2)—112+11=0, or 278_2z+_17z+15 =0; whence 2=3. 107. A literal equation may be made to .correspond. numeral one, by substituting an unit in the room of the given quantity (or letter :) and equations that do not seem, at first, to belong to the preceding class, may sometimes be reduced to such by a proper substitution, that is, by putting the quotient of the first term divided by the last, equal to some new unknowù quantity (or letter) raised to the power expressing the dimension of the equation. Thus, if the equation given be 2x4 + 2428 2x+ 3152 +216x+162—0; by putting = y*, we have a = 162 whence, after substitution, the given equation becomes 162y + 6487–2835y2 +648y+162=0; which now answers to the rule, and may be reduced down to 2y +8y_354*+8y+2=0. 4. Given 7x6—2625_26x+7=0, to find the roots of the equation. Ans. The only real roots are 2+13. 5. Given 2x”—13a%25_13az +2a'=0, to find the roots of the equation. Ans. The only reals roots are may alj + İN5), and {(2+~14) EN (2+4w14)}. 108. Of Binomial Equations, or such as are of the form "ta"=0. Equations of this kind, which are a peculiar species of reciprócal equations, may be reduced to a more simple form by putting day, and then dividing the result by a"; in which case we shall have a"y+"=0, or yo+1=1, where the several values of y are the roots of -1, or +1; and consequently those of u are the same roots multiplied by a. Whence, as the first of these forms, 2+1=1, or "=1, (using a instead of y,) is a reciprocal equation, wanting all the intermediate terms, its solution may be obtained from the rules bě 3y; fore given for this purpose, by making the coefficients, p, q, r, s, each equal to 0, and finding the result accordingly. It therefore only remains to determine the several roots of the equation 2"_=0, or 2"=1, which, for the first ten orders, may be done as follows. Case I. When the index of the first term is an even number. RULE I. If the equation be of the fourth power, as x-1=0, where two of the roots are, evidently, +1 and -1, find the two value of x in the equation x+1=0, and they will be the other two roots of the proposed equation. 2. If it be of the sixth power, as 26—1=, where two of the roots are 1 and -1, as before ;' find the two values of x in each of the equations 2*+*+1=1, and m-x+1=1, and they will be the other four roots of the proposed equation. 3. If it be of the eigth power, as 28-1=1, where two of the roots are also 1 and -1, find the two values of z in the equation 2–21=0, or =21, and let them be denoted by r and r'; then the roots of the 3 quadratics g+1=0, x-rx+1=1, and —r'x +1=0, will be the other six roots of the proposed equation. 4. And if it be of the tenth power, as 270—1=1, where two of the roots are 1 and 1, as they are in all even powers, find the four roots of the equation 4-3z*+1=1, and let them be denoted by r, go! mondo, ""; then the roots of the four quadratics r3+1=1, 2-p2+1=0,2—4x+1=Ò, & inq""x+1=, will be the other eight roots of the proposed equation. CASE II. When the index of the first term is an odd number. Rule I. If the equation be of the third power, as 28—1=0, where one of the roots is evidently 1 ; find the two values of x in the equation z++1= 0, and they will be the other two roots of the proposed equation. 2. If it be of the fifth power, as 2 - 1 : 0, where one of its roots is 1, as before ; find the two values of z in the equation -1=0, and let them be denoted by r and r'; then the roots of the two quadratics 2-rx+1=1, and 2mpx+1=1, will be the other four roots of the proposed equation. 3. If it be of the seventh power, as x?—1=0, where one of its roots is also 1 ; find the three values of z in the equation 28+z 22—1=1, and let them be denoted by r, r', q'; then the roots of the 3 quadratics, a?-rx+1=1,2—r'x+1=1, & m_p"2+1=1, will be the other six roots of the proposed equation. 4. And if it be of the ninth power, as 24—1=0, where one of its roots is likewe 1, as in all the odd powers, find the four values of z in the equation z4 + 28—37—22+1=0, and let them þe denoted by r, gir gold pow; then the roots of the four quadratics +7 -2 2m -6 X 2m 2m rx+=1, 2—r'x+1=1,2—7"x+1=1, & *_p"2+1=1, will be the other eight roots of the proposed equation. 109 Since two of the values of x, in every case of even powers, are +1 and -1, if the first member of an equation of the form 22m – 1=0, be divided by 2?–1, the quotient will be 22m+m+m+...+x+x+=0, 1 And because one of the values of x, in every case of the odd powers, is 1, if the first member of an equation of the form quam+=0,be divided by :-1, the quotient will be ccm1 quam 1-2m+220 +220-3+...+?+x+1=0, both of which are reciprocal equations, having all the coefficients of their several terms=1; and, 'consequently, each of them may be resolved by the method employed in the last article, without recurring to any separate considerations. 1. Required to find the four roots of the equations x-1=0. Here, the index of x being an even number, two of the roots are +1 and — 1. by Case I, we have x+1=1, or x =EN-1; whence the four roots are 1, -1, +w-1, and -»_1. 2. It is required to find the three roots of the equation x-1=0. Here, the index of x being an odd number, one of the root is 1. And, by Case II, we shall have m+x+1=0, or +2=). Whence, by quadratics, x=-*N(1-1)=-IN-3. ' ( - I+IN_3 . 1,-1+İN_3, and -1-N-3, are the roots sought. And if it be required to find the three roots of the equation 28+1=1, we shall have, in this case, one of the roots -1, And, by dividing 2 Tol by 2+1, there will arise 2-3+1=0,0 _=- . -1 Consequently,x=l+N (1-1)={#IN_3. that is -1, 4+1-3, and -N-3, are the roots sought. 3. It is required to find the five roots of the equation 25—1=0. Here 5 being an odd power, 1 root of the given equation is = 1 Therefore, by the rule, Case II, +22=0, orz+z=1; z=-*twi+1)=-+*5=r, and z=-115=r', Whence, also, 2-rx+1=0, or r-63+15) 1. And, 2=(-1+5)EN{16(-1+x5)-1}=(-1+»5) (-10—2x5=1-1+5)+(10+2x5)-1. In like manner, 2—r'x+1=0, or ztli+5)=-1, where x=-11+5)ŁN 16(1+x5)-15-1(1+5) xw 4(10—2N5)n-1. Therefore the five roots are 1, (-1+5)+(10+2x5)-1, and -1(1+5)+İN 10-215N-1. 1. It is required to find the four roots of the equaton 2+1=0. See Index. Ans. x=2+N--2, and -12+1N-2. 3N— 2. It is required to find the five roots of the equation 2+1=0. Ans. -1, 1+5)+1(10-25) N-1, . a and *(1-5)+1(10+25)-1. 3. It is required to find the six roots of the equation 24a=0. Ans. a, a, al-3+N-3), and a(3+N-3). 4. It is required to find the eight roots of the equation 28-1=0. Ans. 1,-1,-1,-1, (W2E-2), (32EIN-2). 5. Find the roots of the equations 10_a"=0, and 226_1=0. Of Equations that have equal Roots. 110. Besides the classes of equations before treated of, there are others of a different kind, that are equally susceptible of being reduced to those of lower dimensions ; the most useful of which are such as have two or mose equal roots, with the same or contrary signs; in which case the method of resolving them, as far as the 4th order, inclusively, are as follows:* RULE I. If a quadratic equation of the general form af taxt b=0, has two equal roots, with the same sign, they will be each =ja, or +ła, according as the coefficient of the second term is positive or negative. And if the equal roots have contrary signs, the equation must be of the form 2 –B=0, in which case 2 + Ņb, or=+b, and —b. 2. If a cubic equation of the general form, 2+azi+bx+c=0, has two equal roots, with the same sign, each of them will be a root of the equation 3x+2ax+b=1, and the remaining root will be = twice one of the equal roots. And if the equal roots have contrary signs, we shall have in that case, ab = -C; and the roots sought will be and — and the remaining ♡ -wroot will be =a. 3. If a biquadratic equation of the form , 38 210 538 289,"; } has two equal roots, with the same sign, each of them will be a root of the equation 22416 25068 (34-86)x+2(ab-6c)x+ac-160=0; el 50 2652 2652 22416 25058 1868 2089 By Case II 2652 2652 221 221 1868 934 2089 872356. 1334025 934 za +(221)=1 + 221 221 48841 46841 221 * Other methods of resolving equations that have equal roots, have been given by Maclaurin, and several of the best foreign writers on Algebra, which, though not so simple as those here laid down, are more commodious for equations of the higher orders, which, if treated in the manner employed in the text, would tedious. 23* с or z? 1 become very |