TRACT 11.

THE ROOTS OF ECtUATIONS.

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negative terms are found to exceed the positive terms by 5,thus showing that 1 If is very near, but something above, themiddle root, and that therefore the roots are not in arith-metical progression. It is therefore probable that 11 may bethe true value of the root, and on trial it is found to succeed.Then dividing .r 3 — 39-r* -+ 479.r — 1881 by x — 11, thequotient is x z — 28x -+ 171 =0, the roots of which quadra-tic equation are 9 and 19, the two other roots of the pro-posed equation.

8. If the equation be .r 3 — 6a' 1 -+ 9.r — 2 = 0 ;

Ave shall have p — 6, q — 9, and r — 2 ; then x =

1 _jo 3 —27r_ 1 6 3 — 27 x 2 _ 2 3 — 2 6

~9 X p z - 3 q ~~ T X 6 Z - 3x9 "" 12 -9 — T ~ 2 '

This value of x being substituted for it in the proposed equa-tion, causes all the terms to vanish, as it ought, thus showingthat 2 is the middle root, and that the roots are in arithmeti-cal progression. Accordingly, dividing the gi\ T en quantityx 3 — (ix z + 9x — 2 by x — 2, the quotient is x z — 4.r-+ 1 =0,a quadratic equation, Avhose roots are 2 + ^/2 and 2— V2,the two other roots of the equation proposed.

9. If the equation be x 3 — 5x z -+ 5x — 1 = 0 ;

Ave shall have = 5, q = 5, and r = 1 ; then x —

1 _ 5 3 —27 X 1 1 _ 125-27 1 98 49

¥ X 5-~ 3x5 — ¥ X ' 25—15 — ¥ X It) “ 45 “ ^From Avhich one might guess the root ought to be l,andAvhich being tried, is found to succeed. But Avithout suchtrial, we might make use of this value l-jf T , or l x ’ r nearly,and approximate with it in the common way.

Having found the middle root to be 1, divide the givenquantity x 3 — 5x z -f 5x — 1 by x — 1, and the quotient is-x 1 — 4x + 1=0, the roots of which are 2 -+ a/2, and2— V2, the tAvo other roots, as in the last article.

10. If the equation be x 3 — lx z -+ IS.r — 18 = 0;Ave shall have p =r 7, q = 18, and r = IS; then x —1 7 3 —27 X 18 1 343-486 143

' T~— 3x18 9 X

. - = 3 5 s ? or 3 nearly.45 5

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