# Maths yr10 Help?!

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#1
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3 weeks ago
#2
Dr Frost 🤮
Copy and paste to google. I always do that x
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#3
it didn't work nothing came up...
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3 weeks ago
#4
My first step would be to simplify root 18.
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#5
yes, 3√2, i can do that
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3 weeks ago
#6
(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.
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#7
ah I think I understand, thanks
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3 weeks ago
#8
(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.
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#9
a=8 and b=17 I think
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3 weeks ago
#10
(Original post by oufoufoufouf)
a=8 and b=17 I think
Yes.
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#11
(Original post by MarkFromWales)
Yes.
wait a minute, can you go back a step, so "equate the whole number parts", what would that be?
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3 weeks ago
#12
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.
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#13
(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?
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3 weeks ago
#14
(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
Last edited by MarkFromWales; 3 weeks ago
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#15
(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😂
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3 weeks ago
#16
photomaths
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3 weeks ago
#17
(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.
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3 weeks ago
#18
(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.
Last edited by mqb2766; 3 weeks ago
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