This is from a past iGCSE paper. Expanding and rearranging the equation is fairly easy, but except by trial and error, or plotting it, ow can you find the integer solution for x and y?
Given that :
(5sqrt(x))^2 = y  20*sqrt(2)
where x and y are positive integers, find the value of x and the value of y.
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 06022017 18:27
Last edited by Notnek; 06022017 at 19:36. 
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 06022017 19:48
(Original post by pdk42)
This is from a past iGCSE paper. Expanding and rearranging the equation is fairly easy, but except by trial and error, or plotting it, ow can you find the integer solution for x and y?
Given that :
(5sqrt(x))^2 = y  20*sqrt(2)
where x and y are positive integers, find the value of x and the value of y.
Thread moved to Maths
If you expand you get:
Then if you try to match up the surd term on either side you get
Can you see what must be to satisfy this? You can compare terms in the rest of the equation to get once you've got .
Please post all your working if you get stuck. 
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 06022017 22:40
(Original post by notnek)
What was the full question? Do you just need to find one set of x and y that satisfy the equation?
If you expand you get:
Then if you try to match up the surd term on either side you get
Can you see what must be to satisfy this? You can compare terms in the rest of the equation to get once you've got .
Please post all your working if you get stuck.
I'm sorry for being stupid, but what do you mean by "match up the surds on either side"? I can see that is a surd, but how do you know that is? Also, why does this matching up eliminate ? Confused....
Having done what you've done, I can see that it's pretty easy to solve:
by substituting back into the equation (and it's an integer!!)Last edited by pdk42; 06022017 at 22:45. 
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 06022017 22:48
(Original post by pdk42)
Thanks for the reply. The full question was what I quoted in the first post.
I'm sorry for being stupid, but what do you mean by "match up the surds on either side"? I can see that is a surd, but how do you know that is? Also, why does this matching up eliminate ? Confused....
Firstly when I say "surd", I mean anything containing the symbol. To help you I'm going to write the equation in a different way:
Intead of rearranging to solve this like what you're used to in algebra, we're going to find a solution by matching up both sides of the equation. So what we need are the two surd terms to be the same
i.e.
And we need the rest to be the same
Does this make sense so far? Please let me know if it doesn't.
If you solve the surd equation then it just so happens that this gives you an integer soltution for . Then when you plug this into the other equation, you'll get an integer solution for also.
So can you see what must be to satisfy this equation:
? 
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 06022017 22:57
Ah, thank you. Yes, I understand that. The key is to split the equation into two separate equalities  very neat . The question is from the Pearson Edexcel Jan 14 3H paper:
http://qualifications.pearson.com/co...e_20140110.pdf
Question 17.
Thanks again  much appreciated! 
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 06022017 23:14
(Original post by pdk42)
Ah, thank you. Yes, I understand that. The key is to split the equation into two separate equalities  very neat . The question is from the Pearson Edexcel Jan 14 3H paper:
http://qualifications.pearson.com/co...e_20140110.pdf
Question 17.
Thanks again  much appreciated! 
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 10022017 18:46
Actually, on reviewing this, I do have another question....
Can you explain how the following works...(Original post by notnek)
Intead of rearranging to solve this like what you're used to in algebra, we're going to find a solution by matching up both sides of the equation. So what we need are the two surd terms to be the same
i.e.
And we need the rest to be the same
But I can't just arbitrarily convert this into:
and
So, why is it OK to match up the surd terms?
Sorry for more dumb questions... 
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 10022017 22:08
(Original post by pdk42)
Actually, on reviewing this, I do have another question....
Can you explain how the following works...
You have simply said "we need the two surd terms to be the same", but I'm not sure why, or how it's valid to do that. Being stupid for the moment, I could have this:
But I can't just arbitrarily convert this into:
and
So, why is it OK to match up the surd terms?
Sorry for more dumb questions...
If you have an equation with more than one unknown in it e.g.
then let's say you group some terms so you end up with something like this:
(stuff) + (more stuff) = (blah) + (more blah)
e.g. "stuff" could be , which would make "more stuff" .
If you can find values for your variables such that "stuff = blah" and "more stuff = more blah" then it makes sense that these values must be solutions to your equation since the equation would be balanced.
But this technique isn't guaranteed to work and most of the time it won't help you solve the equation because the values that satisfy "stuff" = "blah" won't also satisfy "more stuff" = "more blah".
Also this tecnhique doesn't always work the other way around i.e. when you've already got a solved equation. So if you have something like
then it's not necessarily the case that the two left terms (5 and 7) are equal and the two right terms are equal, and you have already shown this yourself.
Now going back to this equation:
The reason this technique was so useful is because solving the surd part gives you a single integer value for , which you can plug in to the other equation to give you an integer value for .
Also (and maybe more importantly), is irrational  let me know if you haven't heard of this word. So since and are integers, the only possible other term that can equate with to balance the equation is .
So when you have an equation that has rational and irrational parts, a useful technique is to equate the rational parts and then equate the irrational parts.Last edited by Notnek; 10022017 at 22:20. 
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 12022017 19:09
(Original post by notnek)
Good question and I'm going to have to ramble a bit to explain it
If you have an equation with more than one unknown in it e.g.
then let's say you group some terms so you end up with something like this:
(stuff) + (more stuff) = (blah) + (more blah)
e.g. "stuff" could be , which would make "more stuff" .
If you can find values for your variables such that "stuff = blah" and "more stuff = more blah" then it makes sense that these values must be solutions to your equation since the equation would be balanced.
But this technique isn't guaranteed to work and most of the time it won't help you solve the equation because the values that satisfy "stuff" = "blah" won't also satisfy "more stuff" = "more blah".
Also this tecnhique doesn't always work the other way around i.e. when you've already got a solved equation. So if you have something like
then it's not necessarily the case that the two left terms (5 and 7) are equal and the two right terms are equal, and you have already shown this yourself.
Now going back to this equation:
The reason this technique was so useful is because solving the surd part gives you a single integer value for , which you can plug in to the other equation to give you an integer value for .
Also (and maybe more importantly), is irrational  let me know if you haven't heard of this word. So since and are integers, the only possible other term that can equate with to balance the equation is .
So when you have an equation that has rational and irrational parts, a useful technique is to equate the rational parts and then equate the irrational parts.
I do understand irrational (a number which has no end when expressed as a decimal) so I can see that:
Thanks for your continued help!Last edited by pdk42; 12022017 at 19:10. 
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 12022017 19:20
(Original post by pdk42)
Is the technique of separating the rational and irrational parts like this something that will always work, or just something to try along the way?
If you have an equation
with a,b,c,d all integers and X not a perfect square, then it is always valid to deduce that
a=c and b = d.
[Proof: If X isn't a perfect square, is irrational, so it can't be written as a fraction n/m with n and m integers. But if , then . Suppose . Then we can write which is impossible, because is irrational. So we must have b=d, and then we find c=a as well, which is what we wanted to show].
But whether it's always going to be the best way to solve a problem is going to depend on the problem.
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