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Maths yr10 Help?!

Please help urgently, I may or may not have got the right answer but I have no clue how to do it!IMG-20200925-WA0001-compressed.jpeg
Dr Frost 🤮
Copy and paste to google. I always do that x
it didn't work nothing came up...
My first step would be to simplify root 18.
yes, 3√2, i can do that
Original post by oufoufoufouf
yes, 3√2, i can do that


Then multiply both sides by 3-√a.
Expand the right hand side.
Equate the whole number parts on each side to find a.
Equate the coefficients of the √2 parts on each side to find b.
ah I think I understand, thanks
Original post by oufoufoufouf
ah I think I understand, thanks

Once you think you've got the answer it's easily checked on a calculator.
a=8 and b=17 I think
Original post by oufoufoufouf
a=8 and b=17 I think

Yes.
Original post by MarkFromWales
Yes.


wait a minute, can you go back a step, so "equate the whole number parts", what would that be?
Apologies. What I said doesn't work because it turns out that √a is itself a multiple of √2 so, as you imply, it's not clear which are the whole number parts.
Maybe someone else has a suggestion? Your answer is correct, a = 8 and b = 17 works.
Original post by MarkFromWales
Apologies. What I said doesn't work because it turns out that √a is itself a multiple of √2 so, as you imply, it's not clear which are the whole number parts.
Maybe someone else has a suggestion? Your answer is correct, a = 8 and b = 17 works.


yes it works, but I have no clue why. anyone else?
Original post by oufoufoufouf
yes it works, but I have no clue why. anyone else?

I've sort of fixed my method in a semi-satisfactory way.
4 + 3√2 = (24 + b√2)(3 - √a)
4 + 3√2 = 72 + 3b√2 - 24√a - b√a√2

Case I : a is not a multiple of 2
(details omitted - this doesn't lead to an integer solution)

Case 2 : a is a multiple of 2
Let a = 2m
4 + 3√2 = 72 + 3b√2 - 48√m√2 - 2√mb

Equating the non-√2 terms:
4 = 72 - 2√mb
√mb = 34
b = 34/√m (Equation A)

Equating the √2 terms:
3 = 3b - 24√m√2
1 = b - 8√m
b = 8√m + 1 (Equation B)

Substituting equation B into equation A:
34 / √m = 8√m + 1
34 = 8m + √m
8m + √m - 34 = 0
(√m - 2)(8√m + 17) = 0
√m - 2 = 0
m = 4

Hence a = 2*4 = 8
and from equation B we have b = 8*2+1 = 17√mb

Equating the integer terms:
4 = 72 - 2√mb
√mb = 34
b = 34/√m (Equation A)

Equating the √2 terms:
3 = 3b - 24√m√2
1 = b - 8√m
b = 8√m + 1 (Equation B)

Substituting equation B into equation A:
34 / √m = 8√m + 1
34 = 8m + √m
8m + √m - 34 = 0
(√m - 2)(8√m + 17) = 0
√m - 2 = 0
m = 4

Hence a = 2*4 = 8
and from equation B we have b = 8*2+1 = 17
(edited 3 years ago)
Original post by MarkFromWales
I've sort of fixed my method in a semi-satisfactory way.
4 + 3√2 = (24 + b√2)(3 - √a)
4 + 3√2 = 72 + 3b√2 - 24√a - b√a√2

Case I : a is not a multiple of 2
(details omitted - this doesn't lead to an integer solution)

Case 2 : a is a multiple of 2
Let a = 2m
4 + 3√2 = 72 + 3b√2 - 48√m√2 - 2√mb

Equating the non-√2 terms:
4 = 72 - 2√mb
√mb = 34
b = 34/√m (Equation A)

Equating the √2 terms:
3 = 3b - 24√m√2
1 = b - 8√m
b = 8√m + 1 (Equation B)

Substituting equation B into equation A:
34 / √m = 8√m + 1
34 = 8m + √m
8m + √m - 34 = 0
(√m - 2)(8√m + 17) = 0
√m - 2 = 0
m = 4

Hence a = 2*4 = 8
and from equation B we have b = 8*2+1 = 17√mb

Equating the integer terms:
4 = 72 - 2√mb
√mb = 34
b = 34/√m (Equation A)

Equating the √2 terms:
3 = 3b - 24√m√2
1 = b - 8√m
b = 8√m + 1 (Equation B)

Substituting equation B into equation A:
34 / √m = 8√m + 1
34 = 8m + √m
8m + √m - 34 = 0
(√m - 2)(8√m + 17) = 0
√m - 2 = 0
m = 4

Hence a = 2*4 = 8
and from equation B we have b = 8*2+1 = 17


thanks, I get it now! I really hope that isn't GCSE maths😂
photomaths
Original post by oufoufoufouf
thanks, I get it now! I really hope that isn't GCSE maths😂

Hi oufoufoufouf,
Maybe I've missed some easy method. If not, then you are right, it is way beyond GCSE.
Original post by MarkFromWales
Hi oufoufoufouf,
Maybe I've missed some easy method. If not, then you are right, it is way beyond GCSE.


A hack would be be to note that
* Both the numerator and right hand side are sqrt(2) surds, so the denominator must be (3-ksqrt(2)) as well, otherwise the surd parts won't match.
* Also the only valid values for a are 5,6,7,8 otherwise the denominator is too large to match the integer part on the right or negative.
8 is the only value for a satisfying both and then you can get b.
(edited 3 years ago)

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