Given that equation kx² + (k-1)x +2k + 3 = 0, where k is a non-zero integer, find the value of k for which
a) one root is the negative of the other
b) one root is the reciprocal of the other
c) one root is twice the other
Note: Alpha + beta thingy is killing me. Need help pls. Thanks a lot. May have more questions to ask.
Since one root is the negative of the other, let's assume alpha = -beta, from there solve your equation, if you get k = 0, then it means that beta = -alpha instead.
Do these for b and c, whereby you take one of each respectively and you might get the answer.
I have completely forgot about alpha and beta so I can't help you solve everything out, but as far as I know, this will be the method I would use if posed with such question.
The other smarter forumers please provide the accurate methods, heh. :D As my thinking might be wrong, ^^"
First use the quadratic formula to get general formula of the two roots. so x1 = -(k-1)/2k + sqrt[ (k-1)^2- 4(2k+3)(k)]/2k while x2 = -(k-1)/2k - sqrt[ (k-1)^2 -4(2k+3)(k)]/2k
then for a) x1 = -x2; sub in what i wrote up there and you will get an equation in k
for b) x1= 1/x2; sub in and blahblah solve
c) x1 / x2 = 2 or x1/x2= 0.5
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
thx