And consequently +2.601676, -2.261806, and ―.339870 are the three roots required. 5. Given +9x=30, to find the root of the equation, or the value of x. 6. Given x3-2x=5, to equation, or the value of x. 2.180849 Ans. x find the root of the Ans. x=2.0945514 7. Given x3-23x=-16, to find the root of the .equation, or the value of x. Ans. x5.472136 8. Given x3-27x=36, to find the three roots or values of x. Ans. 5.765722, -4.320684, and -1.445038 OF BIQUADRATIC EQUATIONS. (R) A biquadratic equation, as before observed, is one that rises to the fourth power, or that is of the general form x2 + ax3 + bx2 + cx+d=0, the root of which may be determined by means of the following formula; substituting the numbers of the given equation, with their proper signs, in the places of the literal coefficients a, b, c, d. RULE I. Find the value of x in the cubic equation x' + ( {ac — — b' — d) x = 12 b 10, b3 + (c2+da3) — 2/24 (ac +8d) by one of the former rules; and let the root, thus determined, be denoted by r. Then find the two values of x in each of the following quadratic equations. x2 + ( a + √ { { a2 + 2 (r− }{b}) } )x = − (r + }b) + 1 x2 + ( a − √ { { a2 + 2 (r — } b) } ) x = − (r + b) — 4 ~{ (r+ b) * -d} and they will be the four roots of the biquadratic equation required. Or, if the equation be of the more commodious form x2 + bx2 + cx+d=0, to which it can be always reduced, by taking away its second term, the rule will be as follows: Find the value of x in the cubic equation and let the root thus determined be denoted by r. Then find the two values of x, in each of the following quadratic equations, x2 + (√ {2(r—b)}) x = − (r + b) + ~ { (r + b)° — d} x2 + (√ { 2 (r — b) } ) x = − (r + } b) — ~ { (r + 'b)' — d} --- and they will be the four roots of the biquadratic equation required. Or the four roots of the given equation, in this last case, will be as follows: x= −{v{2(r− }b)} + √{ − 2 T 2 r b + √[(r + {b) • — d] } 3 b 3 b 2 3 1. Given the equation -10x+35x-50x+ 24=0, to find its roots. Here a= -10, b=35, c= −50, and d=24; Whence, by substituting these numbers in the cubic equation x2 + ({ ac— — b2 - d) x = b2 + (c2 + da2) — — (ac + 8d), which equation, being solved according to the rule before given for that purpose, gives (b) Biquadratic equations, in certain particular cases, are reducible to quadratics, as may be readily shown from the demonstration of the general rule, given in Vol. II of the present work; but as such equations seldom occur in practice, and are of little importance as an object of investigation, any formal enumeration of the methods that might be given for resolving them would be unnecessary; particularly as it can be easily discovered, from the relation of the coefficients, whether any given numeral equa. tion is of this kind or not. ≈= {√(35+18√ − 3)+ '/(35 — 18√ − 3) } 2= But, by the rule for binomial surds, given in the Note to Art. I. Case II, And if this number be substituted for r, -10 for a, 35 for b, and 24 for d, in the two quadratic equations x2+ ['a + √ ¦ (a2 + 2(r− + b))] x = − (r + b) + x2 + [ a ~ ~ { a2 + 2 (r − b ) } ] x = − (r + b) — 3 1 3 1 From the first of which a=+=+=2 or 1, 2. Given x+12x-17=0, to find the four roots of the equation. Here a=0, b=0, c=12, and d= −17; Whence, by substituting these numbers in the cubic equation Where it is evident, by inspection, that x=1. And if this number be substituted for r, o for b, and 17 ford in the four expressions for x, given above, they will become x = + 1⁄2 √2 + √ − 1 −√18=+2√2+√ − 1 −3√2 x = + √2 −√ − 1 −√18=+1√2−√-1-3√2 Which are the four roots of the proposed equation; the two first being real, and the two latter imaginary. RULE II. The roots of any biquadratic equation of the form x*+ ax2 + bx + c=0, may also be determined by the following general formulæ first given by Euler; which are remarkable for their elegance and simplicity. Find the three roots of the cubic equation ≈3 +2ax2 + (ao − 4c) x = b2 by either of the former rules, before given, for this purpose; and let them be denoted by r', r", and r'”. Then, we shall have |