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AS Maths Question

Hello, any guidance on how to complete the following question would be much appreciated:

Given that the equation 3jx - jx^2 + 1 = 0, where j is a constant, has no real roots, find the range of possible values of j.

My ideas are:
b^2 - 4ac < 0
(not sure how to treat the j for this?)

Thanks for your time :smile:
(edited 6 years ago)
Yup you are correct, use that inequality and solve.
That's correct. Use the inequality.
Yep using b^2 - 4ac > 0 will work.
b^2 - 4ac > 0 --> 3j^2 + 4j > 0
Solving that quadratic inequality give j > 0 or j < -4/3
Reply 4
Original post by TheBloodlessMan
Yep using b^2 - 4ac > 0 will work.
b^2 - 4ac > 0 --> 3j^2 + 4j > 0
Solving that quadratic inequality give j > 0 or j < -4/3


Thanks very much :smile:
Reply 5
Original post by TheBloodlessMan
Yep using b^2 - 4ac > 0 will work.
b^2 - 4ac > 0 --> 3j^2 + 4j > 0
Solving that quadratic inequality give j > 0 or j < -4/3


However, the original equation has the context of no real roots.
Reply 6
Original post by simon0
However, the original equation has the context of no real roots.


I agree, I used the same method but
b^2-4ac < 0
so getting b<0 and b<-4/27
so b<-4/27

Thanks for picking up on that

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