(2) What is the cube root of 27054036008 ? Ans. 3002. (3) What is the cube root of 332300979637? Ans. 3,2151+. Obs. The solution of these questions will be found in the Key, after the Dutch Method of solving Cubic equations. Cardan's Method of Solving Cubic Equations. EXAMPLES. (1) GIVEN 3+3x2+2x=24 to find the value of x. To exterminate the second term, put x=-1. Then will a3=23—322+32—I 3x2+32-62+3 Multiplying by v3, v©++=24v3 By transposition, v6-24v-2 3v Complg. the sq. v6-24v3+12=144-1-3881-143,9629 To find the other roots. Bring all the terms of the original equation to one side, thus; x3+3x+2x-24-0, and x-2=0. Dividing by this second expression, we have x−2) x3+3x2+ 2x−24 (x2+5x+12=0 x3-2.x2 Solving the quadratic eq. x2+5x=-12, gives two roots or (2) Given a3+9x2+32x=-42, to find the values of x. Destroy the second term as before, and the equation becomes z3+5=0, or, z2=-5 .. 2=±√-5 And x=-3=±√(-5)-3. In this example the solution has brought out the impossible roots, but it is evident that the equation must also have one root possible. To find which, let the original equation be divided by the product of the two simple equations x+5+3=0, and x−√−5+3=0, which product is x2+6x+14=0; x2+6x+14) x3+9x2+32x+42 (x+3=0 x+6x2+14x 3+18x+42 Hence x=-3 the possible root of the equation. (3) Given x3+x=2, to find the values of x. Obs. In this equation there is no second term, therefore we can immediately proceed with the solution. Multg. by v3 and trans.; v6—2v3=27 Extg. the roots; v=3/(1+3√21) To find the other two roots, we proceed as before, and find Hence the solution of cubic equations having two impossible roots is a matter of great simplicity, for having reduced the original equation to ax= +b, we have only to put v3F. the form = 2763 +b and proceed a as in quadratics, where it must be observed that x=v=37 EXAMPLES FOR PRACTICE. (1) Given x3-9x2+29x=30, to find the values of x. (2) Given y3+11y2+48y=-60, to find the values of y. Ans. -2, and -9-39 2 Obs. We have given the preceding examples, such, that in each, two of the roots will be impossible; for, if the equation should have all its roots possible, this method will not ascertain them; it will therefore be of use to give some rule by which we may know when the above simple process will apply. For this purpose observe, that after destroying the second term of the equation, if it be of the form x-ax±b, and if 4a3 be greater than 2762, the solution cannot be effected by this rule. EXAMPLE.-Given x3-9x=3. 3 Here by proceeding as before, and putting v+- for x, we by reason of -11 being imaginary, or rather inconceivable. Cardan's Rule. We shall now show another method of raising the theorem, by which all cubic equations having two impossible roots are solved, commonly called CARDAN's Rule. Let the equation be x3+bx=c, and for x substitute v+w, then the eq. will become (vw)3+b(v+w)=c, or by expanding the first term v3+3v2w+3vw2+w3+b(v+w)=c ; but Svw+3vw2=3 vw (v+w). Now, 3vw(v+w)+b(v+w)=(3vw+b)(v+w); 3 Assume 3vw+b=0, and the equation becomes v3+w3=c. By trans. 3vw=−b; .. vw=—— Cube each side, and we have v3w3: b 3 b3 27 (v3+w3)2=y6+2v3w3+w¤=c2, from this, let four times the last equation be taken, and the remainder is b3 Extg. the roots; v= √ √{ ¿c + √ (12 + 27 ) }, And we shall find 4 Hence b3 Obs. This formula will solve all cubic equations, except in the eq. as_bx+c; when 63 is greater than c2. EXAMPLE.-Given x3-6x=12, to find the value of x. Here b-6, and c=12; then we have, by the formula, x=3/{2+√(144—216)}+$/{12 — √ (144—216)} Or x=3/6+√(36—8) +$6−√√(36—8) x=3/11,2915+/,7085=2,2436+,8915-3,1351. Ans. EXAMPLES FOR PRACTICE. (1) Given x3-6x=9, to find the value of x. (2) Given x3-3x=2, to find the value of x. Ans. 3. Ans. 2. (3) Given x3-3x2-6x=8, to find the value of x. Ans 2. (4) Given 2y3-12y2+36y=44, o find y. Ans. 2,32748. |