question: polynomials (1 Viewer)

[Rahul]

okay, there is this q i have been tryin to do,

if ax^3 + cx + d = 0 has a double root, show that 27ad^3 + 4c^3 = 0

 

[Rahul]

i got the sum and product of the roots of the 3 roots of P(x)
then the same with P'(x).

simultaeously solvingthe roots i got one of the roots to = -3d/c.

then i sub'd it into P(x). i ended up getting:

27ad^3 + 2c^3 = 0

intead of

27ad^3 + 4c^3 = 0

so close....i think i have made a sily mistake sumwhere. i found that if the root was actually a positive then it would work out fine ( 3d/c ).

 

[Ultimate]

Originally posted by Rahul
okay, there is this q i have been tryin to do,

if ax^3 + cx + d = 0 has a double root, show that 27ad^3 + 4c^3 = 0

let P(x) = ax^3 + cx + d ........ (1)
P'(x) = 3ax^2 + c ........ (2)

For a double root, t, P(t) = P'(t) = 0

thus roots of equation are t,t, q

Sum of roots = 0

2t + q = 0

q = -2t ........ (3)

Product of roots = -d/a

thus t^2q = -2t^3 = -d/a

t = cube root of (d/2a) = double root......... (4)

sub (4) into (1), thus P(t) = 0

thus d/2 + c . cube root (d/2a) + d = 0

- 3d/2 = c . cube root (d/2a)

-3d/2c = cube root (d/2a)

d/2a = -27d^3/8c^3

1/a = -27d^2/4c^3

thus, 4c^3 = -27ad^2

27ad^2 + 4c^3 = 0

I got d^2 instead of d^3, maybe made some little error, maybe the answer is wrong.

 

[ND]

Re: Re: question: polynomials

Originally posted by Ultimate




I got d^2 instead of d^3, maybe made some little error, maybe the answer is wrong.

I got the same thing, i guess the answer is wrong.

 

[McLake]

I agree that it is d^2 ...

 

[Rahul]

yeh sorry guys.....i wrote the final answer up wrong.

u r all right....thnx for the help

 

[RIZAL]

here is another solution aside from Ultimates already very good one;
let w be the double root
p(w) = aw + cw + d = 0 (1)
p'(w) = 3aw + c = 0 (2)

w = -c/3a .. sub in (1)

aw(-c/3a) + cw + d = 0

etc etc, finding w...

w = -3d/2c .. sub in (2)

3a(9d/4c) + c = 0

27ad/4c + c = 0

27ad + 4c=0