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Core 1 quadratic function and inequalities Watch

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    find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

    anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
    Just thrown off a bit by having 2 x2 k2
    values.
    Thank you😊
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    Just ignore the X and use -2(K+1)
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    (Original post by Flask)
    find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

    anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
    Just thrown off a bit by having 2 x2 k2
    values.
    Thank you😊
    Yes you use b^2-4ac > 0

    A= 1
    B= -2k-2
    C= -7

    Now just use the formula, where you will end up with a with a quadratic inequality in k, and then solve.


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    (Original post by Jimmy20002012)
    Yes you use b^2-4ac > 0

    A= 1
    B= -2k-2
    C= -7

    Now just use the formula, where you will end up with a with a quadratic inequality in k, and then solve.


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    C is 2k2-7

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    (Original post by brother aldi)
    C is 2k2-7

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    Haha misread it, my fault.


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    (Original post by Flask)
    find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

    anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
    Just thrown off a bit by having 2 x2 k2
    values.
    Thank you😊
    ax^2 + bx + c = 0

    For your question:

    a = 1
    b = -2(k+1)
    c = 2k^2 - 7
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    Worked through: - (Where '^2' = Squared)
    Using b^2-4ac:

    a=1
    b=-2k-2
    c=2k^2-7

    b^2-4ac = (-2k-2)^2 -4(2k^2-7)>0
    =(4k^2+8k+4) -4(2k^2-7)>0
    =4k^2+8k+32 -8k^2+28 >0
    =4k^2-8k-32<0
    =k^2-2k-8<0=(k-4)(k+2)<0
    = -2<k<4
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    (Original post by AndrewC19)
    Worked through: - (Where '^2' = Squared)
    Using b^2-4ac:

    a=1
    b=-2k-2
    c=2k^2-7

    b^2-4ac = (-2k-2)^2 -4(2k^2-7)>0
    =(4k^2+8k+32) -4(2k^2-7)>0
    =4k^2+8k+32 -8k^2+28 >0
    =4k^2-8k-32<0
    =k^2-2k-8<0=(k-4)(k+2)<0
    =k<-2 k>4 (I may have this last part the wrong way round).
    This is wrong. You shouldn't post full solutions, let OP work through the question themselves.
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    (Original post by lizard54142)
    This is wrong. You shouldn't post full solutions, let OP work through the question themselves.
    I will delete it, I didn't realise. New to this, sorry.
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    (Original post by AndrewC19)
    I will delete it, I didn't realise. New to this, sorry.
    No problem it's just much better for people to work through a question with only a few pointers rather than just tell them the right answer. Also not sure how you got (-2)^2 = 32 in line 1, and check the inequalities last line!
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    (Original post by lizard54142)
    No problem it's just much better for people to work through a question with only a few pointers rather than just tell them the right answer. Also not sure how you got (-2)^2 = 32 in line 1, and check the inequalities last line!
    Lol you can post a worked question if you like, don't listen to this fool
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    That was a mistake, I did this on paper first you see. I meant to put +4
    I get confused with the inequalities, I wasn't ever taught it. I just guess.

    Could you tell me how to know when to use which inequality? (If that makes sense :confused:)
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    (Original post by shamsaidk)
    Lol you can post a worked question if you like, don't listen to this fool
    http://www.thestudentroom.co.uk/showthread.php?t=403989

    Read the bit about full solutions.
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    (Original post by AndrewC19)
    That was a mistake, I did this on paper first you see. I meant to put +4
    I get confused with the inequalities, I wasn't ever taught it. I just guess.

    Could you tell me how to know when to use which inequality? (If that makes sense :confused:)
    were you taught how to draw the graphs associated with the question?
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    Briefly. Should it have been -2<k<4?
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    (Original post by AndrewC19)
    That was a mistake, I did this on paper first you see. I meant to put +4
    I get confused with the inequalities, I wasn't ever taught it. I just guess.

    Could you tell me how to know when to use which inequality? (If that makes sense :confused:)
    Consider the graph of y = (x-4)(x+2).

    For what values of x is y < 0 ?
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    (Original post by AndrewC19)
    Briefly. Should it have been -2<k<4?
    Yep
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    Ok this makes sense now. It's coming back to me!
    Thanks so much Lizard
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    (Original post by AndrewC19)
    Ok this makes sense now. It's coming back to me!
    Thanks so much Lizard
    No problem as soon as you see an inequality, think graph!
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    (Original post by lizard54142)
    No problem as soon as you see an inequality, think graph!
    I'll keep this fresh in my head ready for Wednesday. Thanks again.
 
 
 
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