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C1 exam question, quadratics?



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Hi, I've been having some problems with this question. But mainly with the last part.

I've got the first part, which is 4 +/- r5 and the second part.

But for the third part, after doing y^1/2 = 4+/-r5, and then doing y = (4+/-r5)^2. I'm not sure what to do. As what I did was (4+r5)(4-r5) and I got 11. :/
Reply 1
You got the first part right I think, I'll just have a look at the second part.
Reply 2
You need to do (4+sqrt5)^2 and (4-sqrt5)^2 - not a mixture of the two.
Reply 3
Is the y to the power of 2/3 for question (iii)?
Reply 4
Original post by Mr Gum
You need to do (4+sqrt5)^2 and (4-sqrt5)^2 - not a mixture of the two.


How'd you do that?

Original post by usycool1
Is the y to the power of 2/3 for question (iii)?


Sorry for the bad scan, but it's 1/2 :redface:
Reply 5
(4+5)2=42+45+45+(5)2(4+\sqrt 5)^2=4^2+4\sqrt5+4\sqrt5+(\sqrt 5)^2

similarly for (45)2(4-\sqrt 5)^2
Reply 6
Original post by Mr Gum
(4+5)2=42+45+45+(5)2(4+\sqrt 5)^2=4^2+4\sqrt5+4\sqrt5+(\sqrt 5)^2

similarly for (45)2(4-\sqrt 5)^2


oh, and then you add them together? :/
Reply 7
No, there are two separate solutions for y (as hinted by the word "answers" in the question). It is a quadratic equation, so having two answers is not a problem.
Reply 8
Original post by Mr Gum
No, there are two separate solutions for y (as hinted by the word "answers" in the question). It is a quadratic equation, so having two answers is not a problem.


So you're saying it's 21 + 8r5

And 21 - 8r5.

Thanks :biggrin:
Reply 9
Original post by MedicalMayhem
So you're saying it's 21 + 8r5

And 21 - 8r5.

Thanks :biggrin:


Yes that's right :smile: I'm sorry I couldn't help you too much, I was very busy :frown:
Original post by usycool1
Yes that's right :smile: I'm sorry I couldn't help you too much, I was very busy :frown:


No worries :biggrin:

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