a-maths!!!!!

let α be a root of equation 2x²+kx-1=0
then 1/α is a root of equation x²+kx-6=0

so we have,
2α²+kα-1=0..........(1)

and

(1/α)²+k(1/α)-6=0
1/α²+k/α-6=0
by multipled α² both sides,
we have,
1+kα-6α²=0
6α²-kα-1=0...........(2)

3X(1)-(2),
4kα-2=0
α=1/2k.................(3)

Sub (3) into (1)
2(1/2k)²+k(1/2k)-1=0
2/4k²+1/2-1=0
k²=1
k=±1

let
@² + k@ - 6 = 0 and hence we have

2(1/@)² + k(1/@) - 1 = 0
@² - k@ - 2 = 0

combine the above deductions,

2k@ - 4 = 0
@ = 2/k

sub @ = 2/k into @² + k@ - 6 = 0,

(4/k²) + 2 - 6 = 0

k² - 1 = 0
k = 1 or k = -1