2. It is required to find the five roots of the equation x-1=0. Here 5 being an odd power, one root of the given equation is = 1. Therefore, by the rule, x2+x-1=0, or z2 + z=1, where x = −1 + √(+1)= −1+√5=r, and x2-rx+1=0, or x2 - (+√5)x = −1. x=4( − 1 +√5) ± √{ 1 - 1 + √5)2 — 1 } = 16 = (−1 + √5) ± √-10-2√5(- 1 + √5) ± = (√10+2√5)√−1. In like manner, x2-r'x+1=0, or x2 + (1⁄2 + 1⁄2√5)x = −1. Where x = − (1 + √/5) ± √{ // (1 + √5)2 — 1 } = − · 4 — (1 + √5) + (√10— 2√5)√ − 1. 1, ( − 1 +√5) ±(√10+2√5)√−1, and − (1 +√5) ±(√10—2√5)√ — 1. 3. It is required to find the four roots of the equation x-1=0. --1 Ans. +1,-1, +-1, and -4. It is required to find the five roots of the equation x3 + 1 = 0. Ans. 1, (1 +√5) ±(√10−2√/5)√ — 1, and 5. It is required to find the six roots of the equation x-ao=0. Ans. a, a, a-±√-3), and a(1±√3) 6. It is required to find the eight roots of the equation x3 — a3=0. Ans. a, -a, a√ −1, − a√ — 1, a(± √2±± √ ·− 2), and a(√2±1⁄2✓ −2) 7. It is required to find the ten roots of the equation x1-a1o=0. 10 OF EQUATIONS THAT HAVE EQUAL ROOTS. (U) BESIDES the several 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 them being such as have two or more equal roots, with the same or contrary signs; in which case> the methods of resolving them, as far as to the 4th order, inclusively, are as follows: RULE. 1. If a quadratic equation, of the general form x2+ax+b=0, have two equal roots, with the same sign, they will be each -a. And if the equal roots have contrary signs, the equation must be of the form x2-b=0; in which case, x=√b2, or + b and -b. 2. If the cubic equation x3 + ax3 + bx+c=0, has two equal roots with the same sign, each of them will be a root of the equation 3x2 + 2ax+b=0, and the remaining root will be a- twice one of the equal roots. And, if they have contrary signs, we shall have, in that case, abc, and the roots sought C will be ✓ and will be=a. and the remaining root 3. If a biquadratic equation x2 + ax3 + bx2 + cx+d=0, has two equal roots, with the same sign, each of them will be a root of the equation (3a-8b)x+2(ab-6c)x+ae-16d=0; and the other two roots will be found from the equation x2 + (a+2r)x+b+2ar +3r2 = 0, C > being one of the equal roots before found. And if the equal roots have contrary signs, we shall have a(bc — ad) = co, and x= ±√——, +- as before; the other two roots being found from the equation x2 + ax + b + r2 =0. 4. Also, if the biquadratic x2 + ax3 + bx2 + cx + d=0, has three equal roots, with the same sign, each of them will be a root of the equation 6x2 + 3ax = −b. And if two of the three equal roots have contrary signs, we shall have a(bc-ad) = c2, and x = ±√-as in the last case (g). (g) The mode of solution, here given, might have been easily extended to equations of the 5th, 6th, &c. power; but in those above the 4th degree, the expressions for the equal roots, when they have the same sign, are too complicated to be of any practical use. Rules might also have been introduced for determining a priori, by means of the relations of the coefficients alone, whether or not any given equation has equal roots; but, except in the case where they have contrary signs, this may, in general, EXAMPLES. 1. Supposing that two of the roots of the equation x + x2-33x+63=0, are equal to each other, and have the same sign, it is required to determine them. Here a being =1, and b= −33, we shall have, by the rule, *2 3x2 + 2ax+b=3x2 + 2x−33=0; or a2 +3x=11. * The former of which (3) being substituted for x, in the original equation, is found to succeed. Whence, two of the three roots are, each =3; and the remaining root = -7. 2. Supposing that two of the roots of the equation are equal to each other, and have the same sign it is required to determine them. be more readily done, by finding the root of the quadratic equation to which it is reducible, and then seeing whether the root, thus obtained, be a root of the original equation. In all cases, of this kind, however, it may be of use to know, that an equation cannot have equal roots when the last term and the coefficient of the last but one, are prime to each other; but the reverse of this is not always true. 51 (3a2-8b)x2 + 2(ab-6c)x+ac - 16d=x2 2 188 and 459 And, by trial, the first of these, will be found 3" to be a root of the original equation proposed; whence the two roots required are each = And, since r=3, we shall have, by the rule, x2 + (a + 2r)x+b+2ar + 3r2 = x2 + 213 3. It is required to determine whether the equation x2+11x2 + 19x2-99x-252=0, has equal roots, and, if so, what they are. Here, since a=11, b=19, c=-99, and d= − 252, a(bc-ad) = 11( − 19 × 99 + 11 × 252) = 11(-1881 +2772) = 9801 = 992 = c* ; whence the equation has two equal roots, with contrary signs. 99 11 And consequently x= ±√===±√==±√9=3 |