560. We have seen that if the equation ƒ(x) = 0 has r roots equal to a, the equation f'(x) = 0 has r- 1 roots equal to a. But f(x) is the first derived function of f'(x); hence the equation f(x)=0 must have r− 2 roots equal to a; similarly the equation f(x) = 0 must have r-3 roots equal to a; and so on. These considerations will sometimes enable us to discover the equal roots of ƒ (x) = 0 with less trouble than the method of Art. 559. 561. If a, b, c,...k are the roots of the equation f(x) = 0, to prove that We have f(x) = (x − a) (x − b) (x −c)... (x−k); == writing x + h in the place of x, f(x + h) = (x − a + h) (x − b + h) (x − But ƒ (x+h) = f (x) + hƒ′ (x) + |2 ƒ'' (x) + :.. ; hence f'(x) is equal to the coefficient of h in the right-hand member of (1); therefore, as in Art. 163, ƒ'(x) = (x − b ) (x − c) ... (x − k) + (x − a) (x − c) ... (x − k) + ... ; that is, 562. The result of the preceding article enables us very easily to find the sum of an assigned power of the roots of an equation. Example. If S denote the sum of the kth powers of the roots of the equation x5+px2+qx2+t=0, Now = x2 + (a+p) x2 + (a2 + ap) x2 + (a3 + a2p + q) x + a1+a3p+aq; x-a and similar expressions hold for f(x) f(x) f(x) f(x) x-b' x-c' x-d' x-e Hence by addition, 5x+4рx3+2qx=5x2+(§1+5p) x2+(S2+pS1) x2 +(S3+pS2+5q)x+(S ̧+pS3+qS1). By equating coefficients, S1+5p=4p, whence S1=-p; S2+pS1=0, whence S2=p2; S3+pS2+5q=2q, whence S.=-p3 – 3q ; S1+pS2+qS1=0, whence S4=p1+4pq. To find the value of S, for other values of k, we proceed as follows. Multiplying the given equation by x-5, x2 +рx2-1+qx-3+ tx2-5=0. Substituting for x in succession the values a, b, c, d, e and adding the results, we obtain Sk+pSx-1+qSx-3+tSx-5=0. 563. When the coefficients are numerical we may also pro ceed as in the following example. Example. Find the sum of the fourth powers of the roots of + 1 hence S, is equal to the coefficient of in the quotient of f'(x) by f(x), 25 which is very conveniently obtained by the method of synthetic division as follows: 1. EXAMPLES. XXXV. C. If ƒ (x) = x2+10x3+39x2+76x+65, find the value of ƒ (x − 4). If f(x) = x2 - 12x3+17x2 - 9x+7, find the value of ƒ (x+3). 3. If f(x)=2x4 - 13x2+10x-19, find the value of ƒ (x+1). 2. 4. If f(x)=x+16x3 + 72x2+64x-129, find the value of ƒ (x – 4). 5. If f(x)=ax3+bx5 + cx+d, find the value of ƒ (x + h) − f (x − h). 6. Shew that the equation 10x3 - 17x2+x+6=0 has a root between 0 and -1. 7. Shew that the equation x1 − 5x3+3x2+35x-70=0 has a root between 2 and 3 and one between 2 and -3. 8. Shew that the equation x-12x2+12x-3=0 has a root between 3 and 4 and another between 2 and 3. 9. Shew that x5 + 5x4 - 20x2-19x-2=0 has a root between 2 and 3, and a root between 4 and -5. Solve the following equations which have equal roots: 10. 12. x4-9x2+4x+12=0. 11. x-6x3+12x2 - 10x+3=0. x5-13x2+67x3- 171x2+216x-108=0. 13. x5-x3+4x2 - 3x+2=0. 14. 8x+4x3- 18x2+11x-2=0. 15. x-3x+6x3-3x2 - 3x+2=0. 16. x6-2x5 – 4x4 + 12x3-3x2-18x+18=0. 17. x-(a+b) x3-a (a - b) x2+a2 (a+b) x-a3b=0. Find the solutions of the following equations which have common roots: 18. 2x1-2x3+x2+3x-6=0, 4x1 – 2x3+3x-9=0, 19. 4x4+12x3 – x2 – 15x=0, 6xa+13x3 – 4x2 – 15x=0. 20. Find the condition that x" - px2+r=0 may have equal roots, 21. Shew that x2+qx2+8=0 cannot have three equal roots. 22. Find the ratio of b to a in order that the equations ax2+bx+a=0 and x3- 2x2+2x-1=0 may have (1) one, (2) two roots in common, 23. Shew that the equation 24. If the equation x5 - 10a3x2+b1x+c2=0 has three equal roots, shew that ab1 — 9a5+c5=0. 25. If the equation xa+ax3+bx2+cx+d=0 has three equal roots, 6c - ab 3a2-86' shew that each of them is equal to 26. If x5+qx3+rx2+t=0_has two equal roots, prove that one of them will be a root of the quadratic 15rx2-6q2x+25t — 4qr=0. 27. In the equation x3-x-1=0, find the value of S. 28. In the equation x-x3-7x2+x+6=0, find the values of S4 and S.. TRANSFORMATION OF EQUATIONS. 564. The discussion of an equation is sometimes simplified by transforming it into another equation whose roots bear some assigned relation to those of the one proposed. Such transformations are especially useful in the solution of cubic equations. 565. To transform an equation into another whose roots are those of the proposed equation with contrary signs. Let f(x)=0 be the proposed equation. Put y for x; then the equation ƒ (− y) = 0 is satisfied by every root of f(x)=0 with its sign changed; thus the required equation is f(- y) = 0. If the proposed equation is n-2 ...... +Pn-x+PR = 0, then it is evident that the required equation will be which is obtained from the original equation by changing the sign of every alternate term beginning with the second. 566. To transform an equation into another whose roots are equal to those of the proposed equation multiplied by a given quantity. Let f(x)=0 be the proposed equation, and let q denote the y given quantity. Put y=qx, so that x= then the required " q equation is f(2)=0. The chief use of this transformation is to clear an equation of fractional coefficients. By putting q=4 all the terms become integral, and on dividing by 2, we obtain 567. y3 - 3y2 −y+6=0. To transform an equation into another whose roots are the reciprocals of the roots of the proposed equation. Let f(x)=0 be the proposed equation; put y= 1 so that One of the chief uses of this transformation is to obtain the values of expressions which involve symmetrical functions of negative powers of the roots. |