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    I'm stuck on this strange question:

    kx^2 - (3k+2)x + (k+3) = 0 , has roots A and B.

    i) Show that the roots are real and distinct for all numbers k,

    ii) If A= 2/B, find K

    iii) Given this value of K, without solving the equation, find the equation whose roots are -A^2 and -B^2.

    I'll be honest, i've been on the first part for 2 hours now and I have no clue where to go. I've tried to use b^2 = 4ac but I can't get anywhere. Any people willing to help?

    Thanks in advance!

    By the way, my maths skills are only up to A Level.
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    (Original post by BanglaBOSS)
    I'm stuck on this strange question:

    kx^2 - (3k+2)x + (k+3) = 0 , has roots A and B.

    i) Show that the roots are real and distinct for all numbers k,

    ii) If A= 2/B, find K

    iii) Given this value of K, without solving the equation, find the equation whose roots are -A^2 and -B^2.

    I'll be honest, i've been on the first part for 2 hours now and I have no clue where to go. I've tried to use b^2 = 4ac but I can't get anywhere. Any people willing to help?

    Thanks in advance!

    By the way, my maths skills are only up to A Level.
    b^2 - 4ac > 0
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    (Original post by BanglaBOSS)
    I'm stuck on this strange question:

    kx^2 - (3k+2)x + (k+3) = 0 , has roots A and B.

    i) Show that the roots are real and distinct for all numbers k,

    ii) If A= 2/B, find K

    iii) Given this value of K, without solving the equation, find the equation whose roots are -A^2 and -B^2.

    I'll be honest, i've been on the first part for 2 hours now and I have no clue where to go. I've tried to use b^2 = 4ac but I can't get anywhere. Any people willing to help?

    Thanks in advance!

    By the way, my maths skills are only up to A Level.
    You need to plug in the coefficients into that and then you get a different quadratic involving k.
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    You need to plug in the coefficients into that and then you get a different quadratic involving k.[/QUOTE]

    Thanks for the speedy reply. I solved b2 - 4ac for k, but what does this K mean? And what can I do with it to find out if b2 - 4ac is bigger, smaller or equal to 0?
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    Sorry, i actually incorrectly solved for K.
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    I used b^2 - 4ac, and i've got it down to 5k^2 + 4.
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    (Original post by BanglaBOSS)
    I used b^2 - 4ac, and i've got it down to 5k^2 + 4.
    and how's this expression ? can it be less than 0 ?
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    That's the problem. The question says f(x) has two distinct roots, but if i solve b^2 - 4ac for K, i get a negative number.
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    (Original post by xAikx)
    You need to plug in the coefficients into that and then you get a different quadratic involving k.
    I'm down to 5k^2 + 4. Solving for k gets me a negatice number, but the original equation is meant to have 2 distinct roots
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    (Original post by BanglaBOSS)
    That's the problem. The question says f(x) has two distinct roots, but if i solve b^2 - 4ac for K, i get a negative number.
    I don't think you do . Let's just break it down

    first term 5k^2 can this term ever be less than 0 ? for any real k
    second term 4 which is obviously positive regardless of k
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    So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?
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    (Original post by BanglaBOSS)
    So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?
    for this to happen delta(b^2 - 4ac) must be strictly greater than 0 ( real = delta is greater or equal to 0 , distinct = delta is not 0)
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    That means my delta was wrong. Would you be able to work out the delta please?
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    (Original post by BanglaBOSS)
    That means my delta was wrong. Would you be able to work out the delta please?
    your delta is right 5k^2 +4
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    (Original post by BanglaBOSS)
    So what can I do to prove kx^2 - (3k+2)x + (k+3) = 0 has 2 distinct roots?
    Solve in terms of b^2-4ac > 0 showing that it has 2 distinct real solutions
    If it comes out with (3k+2)^2 - 4k(k+3) you get (9k^2+12k+4) - (4k^2+12k) which is 5k^2 + 4
    Since k^2 is guaranteed to be positive even if k is negative (look up squares mate) then the discriminant will always be > 0 as even if k = 0 then the +4 constant means the min value will be 4 .
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    (Original post by AmmarTa)
    Solve in terms of b^2-4ac > 0 showing that it has 2 distinct real solutions
    If it comes out with (3k+2)^2 - 4k(k+3) you get (9k^2+12k+4) - (4k^2+12k) which is 5k^2 + 4
    Since k^2 is guaranteed to be positive even if k is negative (look up squares mate) then the discriminant will always be > 0 as even if k = 0 then the +4 constant means the min value will be 4 .
    Thanks for the really detailed reply , where I am getting stuck is where K^2 > -4/5

    You can't square root negatives right?
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    (Original post by BanglaBOSS)
    Thanks for the really detailed reply , where I am getting stuck is where K^2 > -4/5

    You can't square root negatives right?
    out of context : you can indeed root square negative numbers but the result will be a complex number
    but here k is real so the square is obviously greater than any negative number
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    (Original post by BanglaBOSS)
    Thanks for the really detailed reply , where I am getting stuck is where K^2 > -4/5

    You can't square root negatives right?
    As far as A Level maths goes (further covers it), no. You only really deal with "real" and "natural" numbers and square rooting a negative doesn't give you an answer that's either of those (real is just a regular number, can be any number; natural is all positive integers starting at 1, it's the numbers that would've been used for counting for maths' most original and primitive uses). I'm interested in where you got k^2 > -4/5 from?
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    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.
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    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:
    a = k, b = 3k + 2, c = k + 3
    x = [-(3k+2) + or - root(5k^2+4)]/2k
    We'll say that the positive form is A and the negative form is B:
    A = (-(3k+2) + root(5k^2+4)) / 2k
    B = (-(3k+2) - root(5k^2+4)) / 2k
    Therefore since A = 2 / B we can set this to be true
    (-(3k+2) + root(5k^2+4)) / 2k = 2 / ((-(3k+2) - root(5k^2+4)) / 2k)
    From here, there's a few ways to solve it - we could painfully multiply the LHS by the denominator on the RHS to get an equation involving the k terms on once side and simply 2 on the other side. I'm not typing that monstrosity of an equation so instead I'll just tell you k = 3 (I'm sure you can multiply and expand brackets).
 
 
 
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