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# Hoy do you solve this core 1 maths question? watch

1. 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
2. (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
3. A link to the paper or a question number would help
4. (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..
5. and here I was thinking it was a thread I could possibly help with x)
6. Hoydo you solve this core 1 maths question?
7. (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 remember that
8. (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
9. Make the substitution and solve for y then for x.

It's a disguised quadratic.

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

11. (Original post by Kvothe the arcane)
Make the substitution 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
12. (Original post by Riya06)
Or you could just square the whole equation and solve it like a normal quadratic equation.
Uh, nopes.
13. (Original post by Riya06)
Or you could just square the whole equation and solve it like a normal quadratic equation.
14. (Original post by Riya06)
Oh? That's weird. You can't do that? :/
Would you, perchance, be thinking that ?
15. (Original post by Riya06)
Oh damn, lol yes. Now that I look at it, it is wrong.
Yep, pretty much - glad that's sorted. :-)
16. (Original post by Riya06)
Thanks for correcting me ^.^
No problem. :-)
17. OCR core 1 is so much harder than AQA and Edexcel.... wtf
18. (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.
19. (Original post by Zacken)
I can't really see a difference from looking at the paper posted.
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.
20. Solve for then sqare

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