# Maths yr10 Help?!

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Please help urgently, I may or may not have got the right answer but I have no clue how to do it!

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#6

(Original post by

yes, 3√2, i can do that

**oufoufoufouf**)yes, 3√2, i can do that

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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#8

(Original post by

ah I think I understand, thanks

**oufoufoufouf**)ah I think I understand, thanks

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(Original post by

Yes.

**MarkFromWales**)Yes.

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#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.

Maybe someone else has a suggestion? Your answer is correct, a = 8 and b = 17 works.

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(Original post by

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.

**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.

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#14

(Original post by

yes it works, but I have no clue why. anyone else?

**oufoufoufouf**)yes it works, but I have no clue why. anyone else?

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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(Original post by

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

**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

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#17

(Original post by

thanks, I get it now! I really hope that isn't GCSE maths😂

**oufoufoufouf**)thanks, I get it now! I really hope that isn't GCSE maths😂

Maybe I've missed some easy method. If not, then you are right, it is way beyond GCSE.

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#18

(Original post by

Hi oufoufoufouf,

Maybe I've missed some easy method. If not, then you are right, it is way beyond GCSE.

**MarkFromWales**)Hi oufoufoufouf,

Maybe I've missed some easy method. If not, then you are right, it is way beyond GCSE.

* 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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