I've did this exactly same question yesterday with my
classmates, these solution is provided by my teachers when we went
through.
(a)
Let the roots be α and -α, where α>0
Sum of roots, α+(-α) = -(k-1/k)
------- (1)
Product of roots, α(-α) = 2k+3/k ------- (2)
From (1), 0 =
1-k/k
Therefore, k =1
(b)
Let the root be α
Therefore, the other rooy is 1/α
Sum of roots, α +1/α = -(k-1/k) -------
(3)
Product of roots, α(1/α) = 2k+3/k ------- (4)
From (4),
1 = 2k +3/k
k = 2k+3
Therefore, k = -3
(c)
Let one root be α
Therefore the other root is 2α
Sum of roots, α +2α = -(k-1/k)
------- (5)
Product of roots, α(2α) = 2k+3/k ------- (6)
From 5
Therefore α = 1-k/3k ------- (7)
Subst (7) into (6)
2(1-k/3k)^2 = 2k + 3/k
2(1-2k+k^2/9k^2) = 2k+3/k
2k -4k^2 +2k^3 = 18k^3 +27k^2
0 = k16k^3 +31k^3
0 = k(16k^2 +13k - 2)
k = 0 (N.A) or k = 16k^2 + 31k - 2 =0
(16k-1)(k+2)
= 0
Therefore, k= 1/16 (N.A) or k = -2
Please do not blindly copy this solution, read how it's done,
it's very commonly tested in exams. Note, this question is
very important, one part can easily cost 2-3 marks.
Regards,
Nathpoop ;D