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    Question is this:

    The equation kx^2 + kx + 3-k =0, where k is a constant, has no real roots.
    a) Show that 5k^2 - 12k <0 (I have done this, I understand that)
    b) Find the set of possible values of k (4 marks)

    I don't have a clue how to do part B. Please could you help me and explain it to me?
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    (Original post by blobbybill)
    Question is this:

    The equation kx^2 + kx + 3-k =0, where k is a constant, has no real roots.
    a) Show that 5k^2 - 12k <0 (I have done this, I understand that)
    b) Find the set of possible values of k (4 marks)

    I don't have a clue how to do part B. Please could you help me and explain it to me?
    Try sketching a graph
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    Factorise.
    You have k(5k - 12) < 0.
    remember to get a negative when you multiply two numbers together, the numbers must have different signs.
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    (Original post by 13 1 20 8 42)
    Factorise.
    You have k(5k - 12) < 0.
    remember to get a negative when you multiply two numbers together, the numbers must have different signs.
    So I would factorise it, getting k=0 and k=12/5. However, the answer is 0<k<2.4

    How on earth do you get that answer?
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    (Original post by blobbybill)
    So I would factorise it, getting k=0 and k=12/5. However, the answer is 0<k<2.4

    How on earth do you get that answer?
    You've solved the equality. This is an inequality. But yes the same critical values are important; btw 2.4 = 12/5.
    The point is that if you want the whole expression to be negative the two factors have to have different signs. Positive*negative = negative, negative*positive = negative while positive*positive = positive, negative*negative = negative.
    So this can happen in one of two ways: k < 0 and 5k-12 > 0 or k > 0 and 5k-12 < 0
    Can you take it from there?
 
 
 

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