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# Another P1 thing watch

1. Yeah anyone can use a calculator lol. The question asked for it in surd form.
2. I was just checking whether he was right. The solution is on its way...I don't see anyone else doing it at the moment.
3. (Original post by mik1a)
Yeah anyone can use a calculator lol. The question asked for it in surd form.
I haven't got a clue..

11 + 2√30 = √121 + √120

and then √ the whole lot? no idea sorry.
4. The answer is not a surd...it is 1861469/397278
5. Well what is the surd form of that?

I think the point of the question is to use algebra to find it, not a calculator.
6. Its fairly simple really,
(a+b)^2 = a^2 + 2ab + b^2

So if a and b are surds, then the only rational numbers are going to come from the squared parts. So comparing to 11 + 2sqrt(30), the 2 factors of 30 that add to 11 are 5 and 6

(sqrt5 + sqrt6)^2
7. (Original post by mik1a)
m and n are negative constants such that

(m - √3)(1 + n√3) = 5√3 - 7

Find the exact values of m and n.

I tried multiplying out, then realised that was going the wrong way and that they were identities, but I'm used to adding ones. So I put the first bracket equal to 1, and got a value for n, then realised it said "m and n are...constants", and screamed. So I gave multiplying out a go, then realised equating coefficients is again what you do when the letters are variables.

Help!
m - 3n = -7 => m = 3n - 7

mn - 1 = 5

n(3n-7) - 1 = 5
3n² -7n -6 = 0
(3n+2)(n-3) = 0

=> n = -2/3 or n=3
but n<0 => n = -2/3
=> m = -9
8. (Original post by Sahir)
I haven't got a clue..

11 + 2√30 = √121 + √120

and then √ the whole lot? no idea sorry.
Originally Posted by lgs98jonee
11+2√30 =sqrt(121+120)
=sqrt(241)
and sqrt of that =241^0.25

11+2root30=(root121+ root140)

√ (√ 121+√ 140)

dunno wot to do next though
9. (Original post by lgs98jonee)
Originally Posted by lgs98jonee
11+2√30 =sqrt(121+120)
=sqrt(241)
and sqrt of that =241^0.25

11+2root30=(root121+ root140)

√ (√ 121+√ 140)

dunno wot to do next though

Its

√ (√ 121+√ 120)
10. (Original post by chrisbphd)
Its

√ (√ 121+√ 120)
yeah :-)
11. This must be the hardest P1 question ever. I'm stumped.
12. (Original post by chrisbphd)
This must be the hardest P1 question ever. I'm stumped.
Errr...look up a few posts, I've done it
14. wot syllabus is this?
15. (Original post by JamesF)
Its fairly simple really,
(a+b)^2 = a^2 + 2ab + b^2

So if a and b are surds, then the only rational numbers are going to come from the squared parts. So comparing to 11 + 2sqrt(30), the 2 factors of 30 that add to 11 are 5 and 6

(sqrt5 + sqrt6)^2
looks so easy now..u just had to take unknown answer and square it to get 11+2√ 30
16. This is EdExcel P1, I'm doing practice questions from the Heinemann books. They are pretty challenging, more so that the past papers that I've been doing, which means Im getting better.

Try this one:

Given that x^2 + y^2 = 23
xy = 1

and by considering an appropriate identity, find the possible values of x + y. Find the positive value of x - y in surd form.
17. (Original post by mik1a)
This is EdExcel P1, I'm doing practice questions from the Heinemann books. They are pretty challenging, more so that the past papers that I've been doing, which means Im getting better.

Try this one:

Given that x^2 + y^2 = 23
xy = 1

and by considering an appropriate identity, find the possible values of x + y. Find the positive value of x - y in surd form.
(x+y)² = x²+y²+2xy = 23 + 2 = 25

=> x+y = 5 or -5

(x-y)² = x² + y² - 2xy = 23-2 = 21

=> x-y = √3√7 or -√3√7
18. ah nice one. I didn't work out the relevance of that. Sometimes you need to step back and work out what method you're going to use, because I jumped straigh into simultaneous equations there.

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