A-maths~

Let the required roots be α and 1 / α respectively.

α2 + kα– 6 = 0

kα = 6 -α2

k = 6 / α – α --- ( 1 )



2( 1 /α2 ) + k ( 1 /α) – 1 = 0

2 /α2 + k /α – 1 = 0 --- ( 2 )



Put ( 1 ) into ( 2 ).

2 /α2 + k /α – 1 = 0

2 /α2 + (6 / α – α) /α – 1 = 0

2 /α2 + 6 /α2 - 1 – 1 = 0

8 /α2 – 2 = 0

8 = 2α2

α = 2 or –2



k = 6 / 2 – 2 = 1 or k = 6 / -2 + 2 = -1

Let @ be the root of x^2 + kx - 6 = 0,then 1/@ is the root of 2x^2 + kx - 1 = 0
@^2 + k@ - 6 = 0 and [2/@^2] + [k/@] - 1 = 0
@^2 + k@ - 6 = 0 and @^2 - k@ - 2 = 0
2k@ - 4 = 0
@ = 2/k
so,
(2/k)^2 + k(2/k) - 6 = 0
(4/k^2) + 2 - 6 = 0
k^2 = 1
k = 1 or k = -1