Question

how do you solve for k? 6(2k-1/k)^2-22k+11/k=35

Answer 1

this expands to

24x^4 - 22k^3 - 59k^2 + 11k + 6 = 0

because of 24 and + 6 roots could be 2,4,6, 8 , 12 ..and the negatives of these
f(2) = 0

so one zero = 2

dividing by (x - 2) to give a cubic expression would be the next step in the process

there are another 3 roots to find

Answer 2

this is quite a laborious method but i can't figure out an easier way

Answer 3

graphing it would help

Answer 4

I see. I tried this method and thought the same thing but it seems like the only way. Anyway, thank you! :)

Answer 5

yw

Answer 6

ah ! - a substitution can be made
6(2k-1/k)^2-22k+11/k=35
rearrange:
6(2k-1/k)^2-11(2k - 1/k) - 35 = 0
let 2k - 1 = x, then
6x^2 - 11x - 35 = 0
(3x + 5 )(2x - 7 ) = 0

x - -5/3 , 7/2

thus
2k - 1/k = 7/2
and
2k - 1/k = - 5/3


solve these to find the 4 values of k

Answer 7

how didn't i see that earlier!!???

Answer 8

Thank you! :)

Answer 9

yw

Answer 10

its a lot easier this way....

Answer 11

It is. :) I'll keep it in mind when something like this come up again.

Answer 12

yep - i should have seen it earlier - the 22k + 11/k and 2k - 1/k should have jogged my brain!

Answer 13

.. but maybe yesterday was a 'dog-day'

Answer 14

.. but maybe yesterday was a 'dog-day'