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    If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.


    Find the range of values for p for which the followed equations have no real roots,


    (2p + 1)x^2 - 3px + 2p = 4


    a=(2p + 1)
    b=3p
    c=2p

    b^2 -4(a)(c)

    9p^2 - 4(2p + 1)(2) = 4


    9p^2 - 16p - 8 = 4

    9p^2 - 16p - 4 = 0

    (3p - 4)^2 - 16 - 4 = 0
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    (Original post by Shakaa)
    If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.


    Find the range of values for p for which the followed equations have no real roots,


    (2p + 1)x^2 - 3px + 2p = 4


    a=(2p + 1)
    b=3p
    c=2p

    b^2 -4(a)(c)

    9p^2 - 4(2p + 1)(2) = 4


    9p^2 - 16p - 8 = 4

    9p^2 - 16p - 4 = 0

    (3p - 4)^2 - 16 - 4 = 0

    (2p + 1)x^2 - 3px + 2p = 4
    (2p + 1)x^2 - 3px + 2p - 4 = 0

    a = 2p + 1
    b = -3p (dont forget the minus sign)
    c = 2p - 4

    and no real roots!! therefore b^2 - 4ac < 0
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    Firstly, get all your stuff in the LHS so RHS = 0

    Then work out your b^2 - 4ac which is a little different because c differs.

    Then remember that b^2 - 4ac must < 0 for no real roots

    It's an inequality, not equation
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    (Original post by Shakaa)
    If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.


    Find the range of values for p for which the followed equations have no real roots,


    (2p + 1)x^2 - 3px + 2p = 4


    a=(2p + 1)
    b=3p
    c=2p

    b^2 -4(a)(c)

    9p^2 - 4(2p + 1)(2) = 4


    9p^2 - 16p - 8 = 4

    9p^2 - 16p - 4 = 0

    (3p - 4)^2 - 16 - 4 = 0
    does the bit i highlighted make sense?
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    (Original post by dexter -1)
    does the bit i highlighted make sense?
    It does dexter cheers.
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    Still having issues with this :s I have corrected all the mistakes i can find but i still cant get the correct values

    (2p + 1)x^2 - 3px + 2p = 4

    a=(2p + 1)
    b=-3p
    c=2p - 4

    b^2 -4(a)(c)

    9p^2 -4(2p + 1)(2p - 4)

    9p^2 -16p^2 + 32p - 8p + 16

    -7p^2 + 24p + 16 <0
    (Is this right and if so where do i take it from here?)
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    You have to find the range(s) of p that satisfy the inequality.

    You say this is a C1 question. Solving the inequality feels like an FP1 question, but hey.

    You could graph it.

    Or you could factorise the LHS into two linear factors (take the minus sign out first).

    By thinking about when each factor is positive/negative you can solve the inequality
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    (Original post by ian.slater)
    You have to find the range(s) of p that satisfy the inequality.

    You say this is a C1 question. Solving the inequality feels like an FP1 question, but hey.

    You could graph it.

    Or you could factorise the LHS into two linear factors (take the minus sign out first).

    By thinking about when each factor is positive/negative you can solve the inequality
    I know its somewhat frowned up, but could you give me a example of how you would proceed with this question?
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    (Original post by Shakaa)


    -7p^2 + 24p + 16 <0
    (Is this right and if so where do i take it from here?)
    -(7p^2 -24p -16) < 0

    7p^2 - 24p -16 > 0 note reversal of sign

    (7p+4)(p-4) > 0

    When p > 4 both brackets are positive so satisified
    When p < -4/7 both brackets are negative so satisfied

    When p lies between these critical values, first bracket is +ve, second is -ve, so product is -ve.
 
 
 
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