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Hoy do you solve this core 1 maths question? Watch

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    It's OCR June 2012 question 7. ( I can't copy and paste it on here)
    Any help you can give would be much appreciated.
    Link to question paper:
    http://www.ocr.org.uk/Images/136127-...hematics-1.pdf
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    (Original post by BlossomSerena)
    It's OCR June 2012. ( I can't copy and paste it on here)
    Any help you can give would be much appreciated.
    You're gonna have to give us the question somehow lol
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    A link to the paper or a question number would help
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    (Original post by jamestg)
    You're gonna have to give us the question somehow lol
    (Original post by SeanFM)
    A link to the paper or a question number would help
    agreed..
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    and here I was thinking it was a thread I could possibly help with x)
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    Hoydo you solve this core 1 maths question?
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    (Original post by BlossomSerena)
    It's OCR June 2012 question 7. ( I can't copy and paste it on here)
    Any help you can give would be much appreciated.
    Link to question paper:
    http://www.ocr.org.uk/Images/136127-...hematics-1.pdf
    quite simple actually, i struggles on one of these, see how it resembles x²-6x+2?

    factorise it.

    instead of x² it's x
    and instead of 6x it's  6\sqrt x remember that  6x^{1/2} \ is\ the\ same\ as\ 6 \sqrt x
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    (Original post by BlossomSerena)
    It's OCR June 2012 question 7. ( I can't copy and paste it on here)
    Any help you can give would be much appreciated.
    Link to question paper:
    http://www.ocr.org.uk/Images/136127-...hematics-1.pdf
    Try substituting y = x1/2 into the equation, and then solve the quadratic - then substitute back to get your answers
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    Make the substitution y=x^{\frac{1}{2}} and solve for y then for x.

    It's a disguised quadratic.

    Edit: ^^ it appears others have got there first.
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    it is a "disguised quadratic"

    :yep:
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    (Original post by Kvothe the arcane)
    Make the substitution y=x^{\frac{1}{2}} and solve for y then for x.

    It's a disguised quadratic.

    Edit: ^^ it appears others have got there first.
    yay! hiya found you again
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    (Original post by Riya06)
    Or you could just square the whole equation and solve it like a normal quadratic equation.
    Uh, nopes.
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    (Original post by Riya06)
    Or you could just square the whole equation and solve it like a normal quadratic equation.
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    (Original post by Riya06)
    Oh? That's weird. You can't do that? :/
    Would you, perchance, be thinking that (x + \sqrt{x} + 2)^2 =^{?} x^2 + x + 2^2?
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    (Original post by Riya06)
    Oh damn, lol yes. Now that I look at it, it is wrong.
    Yep, pretty much - glad that's sorted. :-)
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    (Original post by Riya06)
    Thanks for correcting me ^.^
    No problem. :-)
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    OCR core 1 is so much harder than AQA and Edexcel.... wtf
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    (Original post by d1ck)
    OCR core 1 is so much harder than AQA and Edexcel.... wtf
    I can't really see a difference from looking at the paper posted. :dontknow:
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    (Original post by Zacken)
    I can't really see a difference from looking at the paper posted. :dontknow:
    I'd probably have to go back over my notes to try and figure this one out since I'm quite bad at translations Q5 part ii) - I would copy-paste the question but it won't let me :/


    also how would you q10 ii)

    I thought maybe I could put P's X coord into the circle equation to get a Y where they intersect, and P's Y coord to get an X. And then use y-y1 = m x-x1, but I don't have a gradient and tbh what I'm doing doesn't really make alot of sense to me :/
    oh wait I could get the gradient from the centre to P, but I still think my overall method is wrong.
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    Solve for  \sqrt{x} then sqare
 
 
 
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