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    AmmarTa, thanks for the big help on the second part, I now understand the second section of the question Am I getting it completely wrong for the first part of the question? Is there a way to prove there is positive roots?
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    (Original post by BanglaBOSS)
    AmmarTa, thanks for the big help on the second part, I now understand the second section of the question Am I getting it completely wrong for the first part of the question? Is there a way to prove there is positive roots?
    My first reply proved it, that's all you need to state.
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    (Original post by BanglaBOSS)
    Thanks for the replies guys! I used b^2 - 4ac to get 5k^2 +4. As f(x) has 2 roots, 5k^2 + 4 > 0. I then subtracted 4 and divided by 5 on both sides, to get k^2 > -4/5.
    Your logic is completely wrong. Why are you assuming f(x) has two distinct roots? That's what you're trying to show.

    You're trying to show thag, no matter the value of k, the quadratic f(x) has two distinct roots.

    That means you want to show that the discriminant of the quadratic is positive for all real k. NOT(!!!) assume that it is positive ajd then solve for k (as you are doing).

    The discriminant is 5k^2 + 4 which is the sum of two terms, one of which is always positive and the other is always non-negative, so the discriminant 5k^2 + 4 is always positive. Hence the quadratic always has two distinct roots, no matter the value of k - which is what you were asked to show.
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    (Original post by AmmarTa)
    As far as I can see, to solve part 2, you'd work out the 2 roots using the quadratic formula [(-b +- root(b^2 - 4ac))/2a] using your coefficients of a, b and c:
    ...
    Please don't post full solutions - it's against the terms of the forum.


    Note that for part 2 there is no need to be solving quadratics at any stage! If you think about the relationship between the coefficients and the product of the roots of the equation then the required result drops out in one line
 
 
 
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