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    (Original post by musicangel)
    For 2b) how did you get 36(4+k) > 0 as I did 144-36k > 0
    Also, why is k on the graph instead of x?
    With regards to where the function is positive, is that everything to the right of the y axis?

    Thank you for all of this!
    2(b)

    (-12)^2 - 4(k)(9) = 144 - 36k = 36(4-k) - you're right, that was a silly error on my part!

    Because the equation with the x's will have two solutions whenever k^2 - 64 > 0, so you forget x exists for now and k's are the only things that matter to you know. You want to find what values of k makes the graph of y = k^2 - 64 > 0 (positive).

    The graph of y against k is a quadratic with a minimum at k= 0, it's negative there and then only becomes positive at k =8 and then it stays positive. So k > 8 is one solution to your inequality.

    But the graph is also positive whenever k < some value, can you tell me what value that is?
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    (Original post by musicangel)
    Ok - that makes sense.
    With qu 4, how could I do this as I get the discriminant to be 4-4q^2 > 0 ==> q^2 >0 Or have I done something wrong
    4-4q2 > 0

    is correct

    but then
    1-q2 >0
    (1-q)(1+q) >0
    diagram
    -1<q<1

    bedtime now
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    (Original post by TeeEm)
    4-4q2 > 0

    is correct

    but then
    1-q2 >0
    (1-q)(1+q) >0
    diagram
    -1<q<1

    bedtime now
    I was wondering how you got 1-q2 > 0 and the rest of them

    I really appreciate all of your help! Thank you!
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    (Original post by Zacken)
    2(b)

    (-12)^2 - 4(k)(9) = 144 - 36k = 36(4-k) - you're right, that was a silly error on my part!

    Because the equation with the x's will have two solutions whenever k^2 - 64 > 0, so you forget x exists for now and k's are the only things that matter to you know. You want to find what values of k makes the graph of y = k^2 - 64 > 0 (positive).

    The graph of y against k is a quadratic with a minimum at k= 0, it's negative there and then only becomes positive at k =8 and then it stays positive. So k > 8 is one solution to your inequality.

    But the graph is also positive whenever k < some value, can you tell me what value that is?
    Is it when k < -8?
    This all makes much more sense now! Thank you so much for this!
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    (Original post by musicangel)
    I was wondering how you got 1-q2 > 0 and the rest of them

    I really appreciate all of your help! Thank you!
    I divided the inequality by 4
    then I factorized
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    (Original post by musicangel)
    Is it when k < -8?
    This all makes much more sense now! Thank you so much for this!
    Correct! Good job.

    So, now you can say that for the equation y = x^2 + kx + 16, the equation will have two roots if k is any number above 8 or below - 8. So y = x^2 + 9x + 16 has two roots, x^2 - 10x + 16 has two roots but x^2 + 5x + 16 doesn't.

    Cool 'innit?

    Glad I helped!
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    (Original post by musicangel)
    I was wondering how you got 1-q2 > 0 and the rest of them

    I really appreciate all of your help! Thank you!
    4 - 4q^2 = 4(1-q^2) &gt; 0 \iff 1 - q^2 &gt; \frac{0}{4} \iff 1- q^2 &gt; 0

    since 0 divided by anything is still 0.
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    (Original post by Zacken)
    Correct! Good job.

    So, now you can say that for the equation y = x^2 + kx + 16, the equation will have two roots if k is any number above 8 or below - 8. So y = x^2 + 9x + 16 has two roots, x^2 - 10x + 16 has two roots but x^2 + 5x + 16 doesn't.

    Cool 'innit?

    Glad I helped!
    Thank you!!
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    (Original post by TeeEm)
    I divided the inequality by 4
    then I factorized
    Oh - ok. Thank you!!
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    (Original post by musicangel)
    Oh - ok. Thank you!!
    all the best
 
 
 
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