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    Find the value of k which results in the equation kx2 (sqaured) + 2kx -1 = 0having
    equal roots, given that k is not equal to 0?
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    You must use the discriminant, making it equal 0 because the answer has equal roots.

    Try that and then show me your working.
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    (Original post by h8skoooooool)
    You must use the discriminant, making it equal 0 because the answer has equal roots.

    Try that and then show me your working.
    I am not sure what to do

    b2-4ac=0

    a=k b=2k c=-1

    2k2-4(k)(-1)=0
    4k2-4k=0
    4k(k+1)=0 im not sure if thats right, quite confused what to do next
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    yes now solve for k
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    (Original post by Misshussain23)
    I am not sure what to do

    b2-4ac=0

    a=k b=2k c=-1

    2k2-4(k)(-1)=0
    4k2-4k=0
    4k(k+1)=0 im not sure if thats right, quite confused what to do next
    That's correct!! Now find k like you would from any other quadratic equation.
    Here's a clue, there's two answers and one of them is zero (though looking at the original question, just ignore the answer of 0)
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    (Original post by Misshussain23)
    Find the value of k which results in the equation kx2 (sqaured) + 2kx -1 = 0having
    equal roots, given that k is not equal to 0?
    If it ever mentions roots (I'm assuming this is C1) in a C1 paper/question then use the discriminant of a quadratic equation (b^2-4ac)

    Just be careful, remember b is equal to 2k not 2. Things like that can cost you marks.
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    (Original post by h8skoooooool)
    That's correct!! Now find k like you would from any other quadratic equation.
    Here's a clue, there's two answers and one of them is zero (though looking at the original question, just ignore the answer of 0)
    okay thank you soo much for your help!
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    (Original post by AlphaArgonian)
    If it ever mentions roots (I'm assuming this is C1) in a C1 paper/question then use the discriminant of a quadratic equation (b^2-4ac)

    Just be careful, remember b is equal to 2k not 2. Things like that can cost you marks.
    alright, thank you!
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    (Original post by Misshussain23)
    alright, thank you!
    My pleasure
 
 
 
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