The Student Room Group

Binomial expansion with surds

a) Expand (2+√3)^5

b) Hence write (2-√3)^5 in the form a + b√3


For a) the answer is 362+209√3

For b) I guessed that it was 362-209√3 and I was right. The textbook solution just says this

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3


If there was no simplification then it would make sense to me e.g.

(2+x)^5 = x^5 + 10 x^4 + 40 x^3 + 80 x^2 + 80 x + 32

therefore

(2-x)^5 = (-x)^5 + 10 (-x)^4 + 40 (-x)^3 + 80 (-x)^2 + 80 (-x) + 32

But the fact that this expansion has been simplified first confuses me in the fact that you can just replace the positive with the negative.

Like you can't do this replacement with a different surd e.g. this is clearly not right

(2-√5)^5 = 362-209√5

So does this only work if the surd changes sign? And is it something "obvious" that doesn't need to be proved because I don't immediately see why it's true.
Original post by 0-)
a) Expand (2+√3)^5

b) Hence write (2-√3)^5 in the form a + b√3


For a) the answer is 362+209√3

For b) I guessed that it was 362-209√3 and I was right. The textbook solution just says this

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3


If there was no simplification then it would make sense to me e.g.

(2+x)^5 = x^5 + 10 x^4 + 40 x^3 + 80 x^2 + 80 x + 32

therefore

(2-x)^5 = (-x)^5 + 10 (-x)^4 + 40 (-x)^3 + 80 (-x)^2 + 80 (-x) + 32

But the fact that this expansion has been simplified first confuses me in the fact that you can just replace the positive with the negative.

Like you can't do this replacement with a different surd e.g. this is clearly not right

(2-√5)^5 = 362-209√5

So does this only work if the surd changes sign? And is it something "obvious" that doesn't need to be proved because I don't immediately see why it's true.


For even powers of sqrt(3), its no longer a surd and only odd powers contribute to the bsqrt(3) part. So thats why you can simply flip the sign as (-1)^even is positive and (-1)^odd is negative.

Just work through a simpler example like
(1+/-sqrt(2))^n
for n=1,2,3? You should see the pattern.
(edited 1 year ago)
Reply 2
Original post by mqb2766
For even powers of sqrt(5), its no longer a surd and only odd powers contribute to the bsqrt(5) part. So thats why you can simply flip the sign as (-1)^even is positive and (-1)^odd is negative.

Just work through a simpler example like
(1+/-sqrt(2))^n
for n=1,2,3? You should see the pattern.

Thanks, I understood this by doing the expansion and I could see why it works. But I think my question is more about if it's okay to quote the result without working?

Like could this be written in an A Level exam and get full marks?

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3

Is it "obvious" enough?
Original post by 0-)
Thanks, I understood this by doing the expansion and I could see why it works. But I think my question is more about if it's okay to quote the result without working?

Like could this be written in an A Level exam and get full marks?

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3

Is it "obvious" enough?


Cant say 100%. If its 1 or 2 marks, thats probably enough and the "hence" in the question tends to suggest it would be enough.
Reply 4
Original post by 0-)
a) Expand (2+√3)^5

b) Hence write (2-√3)^5 in the form a + b√3


For a) the answer is 362+209√3

For b) I guessed that it was 362-209√3 and I was right. The textbook solution just says this

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3


If there was no simplification then it would make sense to me e.g.

(2+x)^5 = x^5 + 10 x^4 + 40 x^3 + 80 x^2 + 80 x + 32

therefore

(2-x)^5 = (-x)^5 + 10 (-x)^4 + 40 (-x)^3 + 80 (-x)^2 + 80 (-x) + 32

But the fact that this expansion has been simplified first confuses me in the fact that you can just replace the positive with the negative.

Like you can't do this replacement with a different surd e.g. this is clearly not right

(2-√5)^5 = 362-209√5

So does this only work if the surd changes sign? And is it something "obvious" that doesn't need to be proved because I don't immediately see why it's true.


You can't replace 3\sqrt{3} with a different surd and quote the same coefficients because the powers of (e.g.) 5\sqrt{5} will generate different whole number multiples of the surd; what you can do is that if one part of the question asks you to evaluate (2+5)5(2 + \sqrt{5})^5 then you can write down the result for (25)5(2 - \sqrt{5})^5 because the signs of the terms involving the surd will alternate but their magnitude will be the same.
Original post by davros
You can't replace 3\sqrt{3} with a different surd and quote the same coefficients because the powers of (e.g.) 5\sqrt{5} will generate different whole number multiples of the surd; what you can do is that if one part of the question asks you to evaluate (2+5)5(2 + \sqrt{5})^5 then you can write down the result for (25)5(2 - \sqrt{5})^5 because the signs of the terms involving the surd will alternate but their magnitude will be the same.

If that was for me, I simply read a 3 as 5. Ill edit the post.
(edited 1 year ago)
Reply 6
Original post by mqb2766
If that was for me, I simply read a 3 as 5. Ill edit the post.


It was actually for the OP - I assumed you'd answered their main query about the marking, but I thought they were also asking if they could have used the "same numbers" in the root 5 expansion as in the root 3 expansion which clearly wouldn't be the case :smile:
Reply 7
Original post by davros
You can't replace 3\sqrt{3} with a different surd and quote the same coefficients because the powers of (e.g.) 5\sqrt{5} will generate different whole number multiples of the surd; what you can do is that if one part of the question asks you to evaluate (2+5)5(2 + \sqrt{5})^5 then you can write down the result for (25)5(2 - \sqrt{5})^5 because the signs of the terms involving the surd will alternate but their magnitude will be the same.

Yes I'm aware of that. My point was that if you can just quote (2-√3)^5 = (2+ (-√3))^5 = 362-209√3 without further proof then it could lead to students thinking that e.g. (2-√5)^5 = 362-209√5.

I think my confusion arises from me thinking that this line of working:

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3

was due to the solution writer saying that you can just replace √3 with -√3, but that's not the full story of what's going on.
Reply 8
Original post by 0-)
Yes I'm aware of that. My point was that if you can just quote (2-√3)^5 = (2+ (-√3))^5 = 362-209√3 without further proof then it could lead to students thinking that e.g. (2-√5)^5 = 362-209√5.

I think my confusion arises from me thinking that this line of working:

(2-√3)^5 = (2+ (-√3))^5 = 362-209√3

was due to the solution writer saying that you can just replace √3 with -√3, but that's not the full story of what's going on.


Now I'm a bit confused - possibly getting a bit old for this :biggrin: You're not quoting that bit in bold "without further proof", you're quoting it as a consequence of what you did in part (a). The point is that your second result is a direct consequence of the first one by just flipping the sign of the surd. Not sure how/why a student would be able to quote the expression 362 - 209root(3) without deriving the initial result, or why they would think that the numbers 362 and 209 would be the same for a different surd, but perhaps I'm just worrying about something that you've already sorted in your own mind :smile:
Reply 9
Original post by davros
Now I'm a bit confused - possibly getting a bit old for this :biggrin: You're not quoting that bit in bold "without further proof", you're quoting it as a consequence of what you did in part (a). The point is that your second result is a direct consequence of the first one by just flipping the sign of the surd. Not sure how/why a student would be able to quote the expression 362 - 209root(3) without deriving the initial result, or why they would think that the numbers 362 and 209 would be the same for a different surd, but perhaps I'm just worrying about something that you've already sorted in your own mind :smile:

Yes I haven't explained myself very well. To summarise:

(2+√3)^5 = 362+209√3

implying that

(2+ (-√3))^5 = 362-209√3

is not something that is obvious to me and I had to look at the expansion before the simplification to realise where it came from, as explained in the second post.

Normally for these "hence" 1 marker questions I just need to have a think without writing down any working and I can arrive at the answer.

Basically this thread is just me describing my ignorance :lol:
Original post by 0-)
Yes I haven't explained myself very well. To summarise:

(2+√3)^5 = 362+209√3

implying that

(2+ (-√3))^5 = 362-209√3

is not something that is obvious to me and I had to look at the expansion before the simplification to realise where it came from, as explained in the second post.

Normally for these "hence" 1 marker questions I just need to have a think without writing down any working and I can arrive at the answer.

Basically this thread is just me describing my ignorance :lol:


I think you're right that you have to "do the math" to understand where the numbers come from. But I also think if you write out part (a) clearly enough then it should be clear which contributions change when there is a negative sign before the surd, and which ones stay the same. If your "textbook answer" is all that's written, then it does appear a bit terse and non-obvious, but perhaps the author was equally thinking that if a student had worked out out part (a) in detail, they would see what to amend to get part (b), so no more explanation was necessary :smile:
Original post by 0-)
Yes I haven't explained myself very well. To summarise:

(2+√3)^5 = 362+209√3

implying that

(2+ (-√3))^5 = 362-209√3

is not something that is obvious to me and I had to look at the expansion before the simplification to realise where it came from, as explained in the second post.

Normally for these "hence" 1 marker questions I just need to have a think without writing down any working and I can arrive at the answer.

Basically this thread is just me describing my ignorance :lol:

It's not obvious "without the expansion", but you did do the expansion to do find (2+3)5(2+\sqrt{3})^5.

(2+x)^5 = x^5 + 10 x^4 + 40 x^3 + 80 x^2 + 80 x + 32


When you use that x=3x = \sqrt{3} to simplify this, the only "fact" you're actually using is that x2=3x^2 = 3, which is equally true if x=3x = \sqrt{-3}.

In fact, if you just look at the algebra (as opposed to considering actual numerical values (*)) it's quite hard to tell the difference between 3\sqrt{3} and 3-\sqrt{3}.

Waving my hands somewhat, this is one of the ideas behind Galois theory; looking at how you can map roots of polynomials into other roots in a way that you can't detect algebraically.

(*) e.g. it's obvious that "sqrt(3) is the one > 0", but saying something is > 0 implies knowing at least something about it's numerical value.
(edited 1 year ago)
Reply 12
Original post by DFranklin
It's not obvious "without the expansion", but you did do the expansion to do find (2+3)5(2+\sqrt{3})^5.


I see what you're saying. I've just never seen a "hence" question that isn't based solely on the answer of the previous part. Plus the working given in the solution seemed to suggest that it was just based on that.

This was just a textbook question/solution and maybe it would be worded differently in an exam.
Original post by 0-)
I see what you're saying. I've just never seen a "hence" question that isn't based solely on the answer of the previous part. Plus the working given in the solution seemed to suggest that it was just based on that.
FWIW, I've definitely seen "hence" in questions with a sign change like this (where it's mainly a matter of tracking what flipping a sign does to your previous working).

In this case, I'm reasonably sure (*) that the question is wanting is for you to have expressions for (2+3),(23)(2+\sqrt{3}), (2-\sqrt{3}) but doesn't enormously care how you get them, so the "hence" is meaning take whatever shortcuts you feel comfortable with, given you've already worked out 2+32+\sqrt{3}.

(*) There are some interesting deductions that can be made about 3622093362-209\sqrt{3} by knowing it equals (23)5(2-\sqrt{3})^5 - for example, it implies a very good rational approximation for 3\sqrt{3}.

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