-2
$\begingroup$

how to do question c. What does it mean by "deduce"?

  • 3
    deduce. verb (used with object), deduced, deducing. 1. to derive as a conclusion from something known or assumed; infer: It is just asking you to derive the result. – 2017-02-11
  • 0
    so, find some steps that link what they're giving, to what they want you to prove – 2017-02-11
  • 1
    As in S.C.B's comment, it means "show", but it also strongly suggests that the statement to be shown follows as a consequence of previously shown results -- e.g., the results of parts (a) and (b) can be used to prove (c). – 2017-02-11
  • 0
    [What does it mean to study?](https://www.google.com/search?q=studying&ie=utf-8&oe=utf-8). [What does it mean to "deduce"?](https://www.google.com/search?q=studying&ie=utf-8&oe=utf-8#q=deduce+meaning) – 2017-02-12

2 Answers 2

0

HINT If the root $\alpha \ne \pm1$ then $1/\alpha\ne\alpha.$ Since $1/\alpha$ is a root as well, this means given any root $\alpha$, there is a distinct root $1/\alpha.$ This tells you something about the possible multiplicities.

  • 0
    my teacher showed us that if alpha is a multiple root, then 1/alpha must also be a multiple root. How does that work? – 2017-02-12
  • 0
    @Infinis That's certainly not true in general... consider $f(x) = (x-\alpha)^2(x-1/\alpha).$ But here we know that the cubic coefficient equals the linear coefficient. Maybe use an approach where you write $(x-\alpha)^2(x-1/\alpha)(x-r)$ express both those coefficients in terms of the roots (vieta's formulas). That should let you solve for $r$ and obtain $r =1/\alpha.$ There might be an easier way, not sure. – 2017-02-12
  • 0
    but how do I prove the multiplicity of alpha is 2?? I am really confused on this part of the question. – 2017-02-13
  • 0
    @Infinis Well, it's assumed that it's greater than one. So you need to show it's not three or four. Recall from the previous discussion that $1/\alpha$ is a root, distinct from $\alpha,$ whose multiplicity is greater than or equal to two... – 2017-02-13
  • 0
    Can i just say that using the remainder theorem, alpha has a multiplicity of 2 and 1/alpha has a multiplicity of 2? I still can't really figure out why alpha cannot have a multiplicity of 3. is it because of the remainder theorem? – 2017-02-13
  • 0
    @Infinis If $\alpha$ has multiple roots, $1/\alpha$ must as well. – 2017-02-13
  • 0
    how does alpha having multiplicity has a direct impact on 1/alpha having multiplicity? – 2017-02-13
  • 0
    @Infinis You were the one who told me your teacher showed you that if $\alpha$ was a multiple root, so was $1/\alpha$. And then I told you one way to show it using vieta. – 2017-02-13
0

Hint -

If above both conditions followed.

Then show that $\alpha$ is multiple of 2 and $4B = 8 + A^2$.

Or you can say $\alpha$ has two multiples 2 and $4B = 8 + A^2$.