mathematics----polynomial

Let a and 2a be the roots of the equation.

我唔知你有冇學sum of roots product of roots,我用另一個方法做:
4a^2 + ka + 4 = 0 ,,,(1)
4(2a)^2 + k(2a) + 4 = 0 ...(2)
From (2)
8a^2 + ka + 2 = 0 ...(3)

(1) - (3), we have
-4a^2 + 2 = 0
a = +/- sqrt(1/2)
Sub a = sqrt(1/2), we have
4(1/2) + k*sqrt(1/2) + 4 = 0
k = -6sqrt(2)
至於 a = - sqrt(1/2)時,相應的k,我諗應冇問題

From the given, we can see that the product of roots is equal to 1.

Therefore, let the roots be α and 1/α with α > 1, then

α > 1/α

So, α = 2/α
α^2 = 2
α = ±√2

Therefore, the possible roots are √2 and 1/√2 or -√2 and -1/√2

So the sum of roots may be:

√2 + 1/√2 = 3/√2 or -√2 - 1/√2 = -3/√2

So,

-k/4 = 3/√2 or -3/√2
k = ±6√2

From the given, we can see that the product of roots is equal to 1.

Therefore, let the roots be α and 1/α with α > 1, then

α > 1/α

So, α = 2/α
α^2 = 2
α = ±√2

Therefore, the possible roots are √2 and 1/√2 or -√2 and -1/√2

So the sum of roots may be:

√2 + 1/√2 = 3/√2 or -√2 - 1/√2 = -3/√2

So,

-k/4 = 3/√2 or -3/√2
k = ±6√2