Let f(x)=ax^3+bx^2+cx+d, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)

Jefferson Booth 2022-11-21 Answered
Let f ( x ) = a x 3 + b x 2 + c x + d, be a polynomial function, find relation between a,b,c,d such that it's roots are in an arithmetic/geometric progression. (separate relations)
So for the arithmetic progression I took let α = x 2 and r be the ratio of the arithmetic progression.
We have:
x 1 = α 2 r , x 2 = α , x 3 = α + 2 r
Therefore:
x 1 + x 2 + x 3 = b a = 3 α
x 1 2 + x 2 2 + x 3 2 = 9 α 2 2 c a 4 r 2 = b 2 3 a c 3 a 2
x 1 x 2 x 3 = α ( α 2 4 r 2 ) = d a
and we get the final result 2 b 3 + 27 a 2 d 9 a b c = 0.
How should I take the ratio at the geometric progression for roots?
I tried something like
x 1 = α q , x 2 = α , x 3 = α q
To get x 1 x 2 x 3 = α 3 but it doesn't really work out..
Note:
I have to choose from this set of answers:
(a)   a 2 b = c 2 d (b)   a 2 b 2 = c 2 d (c)   a b 3 = c 3 d
(d)   a c 3 = b 3 d (e)   a c = b d (f)   a 3 c = b 3 d
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Answers (2)

erlentzed
Answered 2022-11-22 Author has 22 answers
Step 1
Using your notations
a x 3 + b x 2 + c x + d = a ( x α q ) ( x α ) ( x α q )
Expand the rhs to get after simplifications
a x 3 a α ( q 2 + q + 1 ) q x 2 + a α 2 ( q 2 + q + 1 ) q x a α 3
Step 2
Compare the coefficients to get
b = a α ( q 2 + q + 1 ) q
c = a α 2 ( q 2 + q + 1 ) q
d = a α 3
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Widersinnby7
Answered 2022-11-23 Author has 7 answers
Step 1
To sum up, a cubic has its roots in arithmetic progression if and only if the arithmetic mean of the roots is a root of the cubic ( so equals one of the roots).
For the geometric progression, we could use the same trick, and say that the geometric mean of x 1 , x 2 , x 3 is a root of P. Alternatively, to avoid cubic roots, one considers the equivalent statement that x 1 x 2 x 3 must equal one of the x i 3 . So we set up the equation Q = 0 with roots x 1 3 , x 2 3 , x 3 3 and impose the condition that x 1 x 2 x 3 is a root.
The equation for x i 3 can be obtained readily by eliminating x from the equalities y = x 3 , a x 3 + b x 2 + c x + d. We get a cubic equation for y
a 3 y 3 + ( 3 a 2 d 3 a b c + b 3 ) y 2 + ( 3 a d 2 3 b c d + c 3 ) y + d 3 = 0
and the condition is that d a is a root of this equation.
Step 2
In general, given a polynomial P of degree n, one can get the condition on the coefficients so that for some ordering of the roots we have an algebraic condition F ( x 1 , , x n ) = 0. We take all the possible permutations of F ( x σ ( 1 ) , , x σ ( n ) ) of F. The condition is that the product of all these permutations is 0.
If we want m conditions F 1 = = F m = 0, we set up F = t i F i , where t 1 , t m are variables, consider all the possible permutations of F. The condition is that the product of all these is 0, as a polynomial in t 1 , t m
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