What are the (real) values for A, B, C and D if y^5 - Dy^3 - 2Cy^2 - 4By - 8A = 0 for all y in 1.6 < y < 2?

Answer 1

There are no solutions to this problem. Consider that by the fundamental theorem of algebra, a polynomial of degree N will have exactly N (not necessarily distinct) roots in the field of complex numbers, regardless of its coefficients. This means that the polynomial will be zero at at most N points, which in this case is five. However, there are infinitely many points in the range (1.6, 2), and as such this polynomial cannot be zero at all of them, regardless of what A, B, C, and D are (even if they are allowed to be complex).

Answer 2

1

Answer 3

Pascal has the right idea but not quite the right answer. If the polynomial is zero for all y in a specified range, then it is zero everywhere (by Taylor series expansion), including at y = 0. This leaves us with:
-D -2C -4B -8A = 0, or D + 2C + 4B + 8A = 0, which can be solved for any one of A, B, C, D if the other three are given. Now suppose that y = 1. This gives us:
1 - D - 2C - 4B - 8A = 0. But this is inconsistent with the previous answer, so the bottom line is that there is in fact no solution.

Answer 4

17

Answer 5

for ALL y ??? ((such that 1.6 < y < 2 )) ?

....

hmmm,
D = y^2
C=B=A=0

satisfies the stated condition ...

C = (1/2) y^3
D=B=A=0

also satisfies

B = (1/4)y^4
D=C=A=0

does too

... pattern emerging ???

Answer 6

there is no solution for the problem ...

the final equations obtained are ...

d+c-b=5
d+b=6
2d+c=11

if u could solve the above eqution, then that is the solution for your problem ...

Answer 7

this is Yahoo! Answers, not Yahoo! Do My Homework.

do it yourself and try to learn something.

Answer 8

yes

Answer 9

oh man...why must u ppl post this kind of Q and make me feel so stupid? haha - my fault!

Answer 10

i'll give you tree fifty for da lot