Question

Condition on \(a,b\) that if \(\alpha\) is a root of \(x^2 + ax + b\), then so is \(\alpha^2 -2\). Hold on for my attempt...

Answer 1

\[\alpha^2 + a\alpha + b = 0\]So I divide \((\alpha^2 -2)^2 + a (\alpha^2 - 2) + b \) by \(\alpha^2 + a\alpha + b\) and equate the dividend with the remainder. Correct?

Answer 2

Well, long division is turning out to be really long.

Answer 3

how about using the fact that
product of roots is b
and sum is -a

Answer 4

\[c + c^2 -2 = -a\Rightarrow c^2 + c + a-2 = 0\]\[c(c^2 - 2) = b \Rightarrow c^3 - 2c - b = 0\]In the first equation, \(1 - 4(a - 2) > 0\).
In the second equation... well...

Answer 5

I find the long division method better - but am too lazy to perform it. :P

Answer 6

Or maybe I should solve the first equation and plug that into the second.

Answer 7

\[c = \dfrac{-1 \pm \sqrt{1 - 4(a-2)}}{2} = \dfrac{-1 \pm \sqrt{4a - 9}}{2}\]

Answer 8

Whoops.

Answer 9

The radical should contain \(9 - 4a.\)

Answer 10

Not that it makes the problem any simpler...

Answer 11

@ParthKohli - surely what @hartnn said is the simplest way to get the conditions on a and b?

Answer 12

is it sufficient to express both a and b in terms of \(\alpha\)?

Answer 13

I don't know. What I do know that the number of values of \(a,b\) taken together would be finite.

Answer 14

how?

Answer 15

Because that's what the question is. It asks me for the number of values of a,b taken together.

Answer 16

For that, I'd have to get a relation between a and b - I suppose.

Answer 17

Making use of what @hartnn said we get:\[a=2-\alpha-\alpha^2\]\[b=\alpha(\alpha^2-2)\]you can set \(\alpha\) to any integer between \(-\infty\) and \(\infty\) and it will generate an infinite number of a's and b's - wouldn't it?

Answer 18

unless there is some other condition like both a and b have to be positive integers, etc?

Answer 19

Nope - real.

Answer 20

then I cannot see how the number of solutions to this is finite?

Answer 21

There must be another restricting factor. :|

Answer 22

You're the smart guy in the room - you should tell me. x)

Answer 23

@ganeshie8

Answer 24

is it that ax^2+bx+c=0 has its two roots as alpha and alpha^2-2

Answer 25

Yes.

Answer 26

\[(x-\alpha)(x-(\alpha^2-2))=0 and ax^2+bx+c=0\]
represents same

Answer 27

\[a(x- \alpha)(x - \alpha^2 + 2) = ax^2 + bx + c\]But again, that is almost the same as hartnn's method.

Answer 28

by comparing co efficients