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Hoy do you solve this core 1 maths question?

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-question-paper-unit-4721-core-mathematics-1.pdf
(edited 8 years ago)

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Reply 1
Avatar for jamestg
jamestg
21
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
A link to the paper or a question number would help :h:
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 :h:


agreed..
Reply 4
Avatar for C-rated
C-rated
2
and here I was thinking it was a thread I could possibly help with x)
Reply 5
Avatar for M14B
M14B
18
Hoy do you solve this core 1 maths question?
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-question-paper-unit-4721-core-mathematics-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 6x 6\sqrt x remember that 6x1/2 is the same as 6x 6x^{1/2} \ is\ the\ same\ as\ 6 \sqrt x
(edited 8 years ago)
Reply 7
Avatar for Alexion
Alexion
17
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-question-paper-unit-4721-core-mathematics-1.pdf


Try substituting y = x1/2 into the equation, and then solve the quadratic - then substitute back to get your answers
Make the substitution y=x12y=x^{\frac{1}{2}} and solve for y then for x.

It's a disguised quadratic.

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

:yep:
Original post by Kvothe the arcane
Make the substitution y=x12y=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
Reply 11
Avatar for Zacken
Zacken
22
Original post by Riya06
Or you could just square the whole equation and solve it like a normal quadratic equation.


Uh, nopes.
Original post by Riya06
Or you could just square the whole equation and solve it like a normal quadratic equation.


Reply 13
Avatar for Zacken
Zacken
22
Original post by Riya06
Oh? That's weird. You can't do that? :/


Would you, perchance, be thinking that (x+x+2)2=?x2+x+22(x + \sqrt{x} + 2)^2 =^{?} x^2 + x + 2^2?
Reply 14
Avatar for Zacken
Zacken
22
Original post by Riya06
Oh damn, lol yes. Now that I look at it, it is wrong.


Yep, pretty much - glad that's sorted. :-)
Reply 15
Avatar for Zacken
Zacken
22
Original post by Riya06
Thanks for correcting me ^.^


No problem. :-)
Reply 16
Avatar for d1ck
d1ck
OCR core 1 is so much harder than AQA and Edexcel.... wtf
Reply 17
Avatar for Zacken
Zacken
22
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:
Reply 18
Avatar for d1ck
d1ck
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.
(edited 8 years ago)
Solve for x \sqrt{x} then sqare

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