Question

The cubic equation x^3 + Rx^2 + Sx + M = 0 has three different non zero roots R, S, and M. compute the value of the expression 1/S + 1/M

Answer 1

Hope that helps

Answer 2

Thank you for the medal!

Answer 3

i want get an exact value. @agent0smith any idea sir ?

Answer 4

from the Vietas formula, get the equations :
R + M + S = -R ... (1)
RM + RS + MS = S ... (2)
RMS = -S ... (3) or RM = -1

Answer 5

so, the 2nd equation can be RS + MS = S + 1

Answer 6

this is similar like x^3 + ax^2 + bx + c = 0 with a, b, and c are roots (non zero)
a + b + c = -a or
b + c = -2a ... (1)

ab + ac + bc = b
a(b + c) + bc = b ... (2)

abc = -c
ab = -1
a = -1/b ... (3)

substitution (3) into (1), get
b +c = -2(-1/b) = 2/b
c = 2/b - b = (2 - b^2)/b

and the 2nd equation :
-1/b (b + 2/b - b) + b(2/b - b) = b
-1/b (2/b) + 2 - b^2 = b
-2/b^2 + 2 - b^2 = b
times b^2 to both sides, get
-2 + 2b^2 - b^4 = b^3
b^4 + b^3 - 2b^2 + 2 = 0

Answer 7

by using horner diagram, get
b^4 + b^3 - 2b^2 + 2 = 0
(b + 1)(b^3 - 2b + 2) = 0
b = -1 or the roots of b^3 - 2b + 2 = 0

Answer 8

if b = -1 is satisfying this case. i get
a = 1 and c = - 1.
of couse it is not allowed , bcz the roots must be diferent

Answer 9

and finally i think we must see the last solution from b^3 - 2b + 2 = 0

Answer 10

is this allowed ?

b + c = -2a = -2(-1/b) = 2/b ---> or c = (2/b - b)

a(b + c) + bc = b
bc = b - a(b + c)
bc = b - (-1/b)(b + c)
bc = b + (b+c)/b = b + (b + 2/b - b)/b = b + 2/b^2 = (b^3 + 2)/b^2

1/b + 1/c
= (b + c)/(bc)
= (2/b) / (b^3 + 2)/b^2
= 2b/ (b^3 + 2)
= 2b /2b
= 1

Answer 11

but i cant give a proving that a , b, and c are not same :v