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# FP1 Complex Numbers watch

1. given that 1-2i and 1+2i are complex solutions of f(x)=0

Solve f(x)=0 and find p

what are you suppose to do?
2. multiply out (x-(1-2i))(x-(1+2i))
3. (Original post by PeeWeeDan)
multiply out (x-(1-2i))(x-(1+2i))
why? where did you get that from? why x- the roots?
4. Hmm, ok: are you aware of the relationship between the roots of a polynomial and the polynomial's coefficients?

It is: suppose we have a polynomial with zeros , then we have:
;

;

So, can you use this to find our third root, (say)?

Spoiler:
Show
We have that by one of the relations above.

Now we have our roots, we can easily compute p.
5. (Original post by emmaxoxo)
why? where did you get that from? why x- the roots?
Because the equation will factorize to form (ax-b)(x-(1-2i))(x-(1+2i))=0 where the third solution is x=b/a.

Just for reference the answer I get without writing anything down(so pinch of salt please) is p=12 and x=1/2... but I'm not 100% would need to check it.
6. (Original post by PeeWeeDan)
Because the equation will factorize to form (ax-b)(x-(1-2i))(x-(1+2i))=0 where the third solution is x=b/a.

Just for reference the answer I get without writing anything down(so pinch of salt please) is p=12 and x=1/2... but I'm not 100% would need to check it.

so far i have x^2-2x+5, now what do I do?
7. (Original post by emmaxoxo)

so far i have x^2-2x+5, now what do I do?
(ax-b)(x^2-2x+5)= whatever that first polynomial was.

Solve.
8. (Original post by PeeWeeDan)
(ax-b)(x^2-2x+5)= whatever that first polynomial was.

Solve.
Yep done that got x=1/2, but where do you get p=12, i know it's 10+2 from answers, but where have those numbers come from?
9. (Original post by emmaxoxo)
Yep done that got x=1/2, but where do you get p=12, i know it's 10+2 from answers, but where have those numbers come from?
By multiplying out the brackets and having a look at the equation.

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