All the Roots of an Equation diminished by the same Quantity. SECTION II. Transformation of Equations. 169. Problem. To transform a given equation into another in which the roots are all diminished by the same quantity. Solution. Let the given equation be x2 + a xn−1 + bxn−2+ &c. += 0; the roots of which are x', x", x'", &c., and let e be the excess of these roots above those of the required equation, which must consequently be x' e, x" •e, xl e, &c.; or if u is the unknown quantity of this new equation, we have x = e + u; and this value of x, being substituted in the given equation, produces the required equation, or (e+un+a (e+u)n−1+b (e+u)n−2+ &c. = 0; which, being arranged according to powers of u, is 1 170. Corollary. Since e is now entirely arbitrary, it may be taken to satisfy any proposed condition, such for instance as that the coefficient of un may be equal to zero, in which case the second term vanishes, and the equation becomes of the form un + b'un - 2 + cun- 3 + &c... = 0. All the Roots of an Equation diminished by the same Quantity. This condition is represented by the equation ne + a = 0, which gives for the value of e, a e n 171. Corollary. Since the roots of the equation just attained are e, xill e, &c., they become, when e is taken, equal to one of the roots x', x", x", &c. of the proposed equation, such as x for in and, therefore, one of the roots of the equation thus obtain 1 ed must be zero. Now this equation is which must satisfy it, its first member is reduced to the last term, and we have x'n + ax'n-1 + b x' n − 2 + &c. = 0 ; which equation is evidently correct, since it only differs from the given equation by the substitution for x, of one of its roots x'. Hence the above equation, divided by u, is reduced to un−1 + n x'un⋅ 2 + &c... + n x' n—1 + a = = 0. + (n-1) ax'n→ 2 + &c. Derived Polynomial. 172. Corollary. If two or more of the roots of the given equation are equal to each other, such as 0, x' -x"= = 0, x' — x1, x' — xˇ, &c. and it is, therefore, satisfied by the value of u, which reduces it to u = 0, -3 nxn−1 + (n-1) a x'n−2 + (n − 2) b x'n−3+ &c. = 0, or since x' is either of the equal roots the accent may be omitted, and we have 3 nxn−1 + (n − 1) a xn−2 + (n − 2) b x2¬3 + &c. = 0, which must be satisfied by a value of x equal to either of the equal roots of the given equation, and these equal roots can therefore be obtained by means of the process of elimination of art. 116. But it is evident that two different equations with one unknown quantity cannot be satisfied by the same value of this unknown quantity, unless their first members have a common divisor, which is reduced to zero by this value of the unknown quantity. The first member of the equation last obtained is' called the derived polynomial of the given equation, and is obtained from it by multiplying each term by the exponent of the unknown quantity in that term, and diminishing this exponent by unity. The equal roots of an equation are, therefore, obtained by finding the greatest common divisor of its first member and its derived polynomial, and solving the equation obtained from putting this common divisor equal to zero. 1. Find all the roots of the equation x3 - 7 x2 + 16x 12= which has equal roots. = 0 the greatest common divisor of which and the given first member is Now since the given equation has two roots equal to 2, it must be divisible by and we have (x — 2)2 = x2 — 4 x + 4, x37x216x-12= (x-2)2 (x-3)=0; whence x=3 is the other root of the given equation. 2. Find all the roots of the equation x2-9x5+6x+15x3-12x2-7x+6=0 which has equal roots. Solution. The derived polynomial of this equation is 7x6-45x24x3 +45x2-24x-7, the greatest common divisor of which and the given equation gives x3 x2 x+1= = 0, Examples of finding equal Roots. which is an equation of the third degree, and we may consider it as a new equation, the equal roots of which are to be found, if it has any. and the common divisor of this derived polynomial and the first member gives x3-x2-x+1= (x−1)2 (x + 1) = 0. The equal roots of the given equation are, therefore, x= and = -- and its first member is divisible by 1; (x − 1)3 (x + 1)2, and is found by division to be (x − 1)3 (x + 1)2 (x2 + x − 5). · The remaining roots are, therefore, found from solving the quadratic equation |