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Complicated square root problem.

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2

2

$begingroup$

I was wondering the general method to solve

What is the value of $sqrt{a-bsqrt{c}}?$

The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)

number-theory radicals nested-radicals

share|cite|improve this question

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
  $endgroup$
  – fleablood
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
  $endgroup$
  – JMoravitz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood I know, complex numbers. How should I write it then?
  $endgroup$
  – Max0815
  1 hour ago
  

  
  
  
  

add a comment | 

2

2

$begingroup$

I was wondering the general method to solve

What is the value of $sqrt{a-bsqrt{c}}?$

The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)

number-theory radicals nested-radicals

share|cite|improve this question

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  "The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$" and do what with it? "but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$" why not? What was supposed to happen?
  $endgroup$
  – fleablood
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
  $endgroup$
  – JMoravitz
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @fleablood I know, complex numbers. How should I write it then?
  $endgroup$
  – Max0815
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

I was wondering the general method to solve

What is the value of $sqrt{a-bsqrt{c}}?$

The basic method I learned is to set this equal to $sqrt{x-ysqrt{c}}$, but I found out that this doesn't work with $sqrt{5+2sqrt{6}}$ which equals $sqrt{3}+sqrt{2}$. What is the general method to simplify these problems?(i.e. $sqrt{a-bsqrt{c}}=?$)

number-theory radicals nested-radicals

share|cite|improve this question

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

$endgroup$

share|cite|improve this question

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

share|cite|improve this question

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

edited 9 mins ago

Michael Rozenberg

104k1891196

edited 9 mins ago

Michael Rozenberg

104k1891196

asked 1 hour ago

Max0815Max0815

66618

asked 1 hour ago

Max0815Max0815

66618

2 Answers
2

active

oldest

votes

1

$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. thanx!!!!!
  $endgroup$
  – Max0815
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
  $endgroup$
  – Stan Tendijck
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I believe so too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.

For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

share|cite|improve this answer

edited 53 mins ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is interesting. I had never seen the identities you begin with.
  $endgroup$
  – Lubin
  28 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin same with me too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We can prove it. It's not hard.
  $endgroup$
  – Michael Rozenberg
  12 mins ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. thanx!!!!!
  $endgroup$
  – Max0815
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
  $endgroup$
  – Stan Tendijck
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I believe so too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  

add a comment | 

1

$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. thanx!!!!!
  $endgroup$
  – Max0815
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work).
  $endgroup$
  – Stan Tendijck
  58 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes I believe so too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals
$$x^4 + (2-a^2) x^2 + 1$$
Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to your example, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly gicen by $sqrt{2}+sqrt{3}$. Does that make sense?

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

answered 1 hour ago

Stan TendijckStan Tendijck

1,826311

5

$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.

For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

share|cite|improve this answer

edited 53 mins ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is interesting. I had never seen the identities you begin with.
  $endgroup$
  – Lubin
  28 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin same with me too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We can prove it. It's not hard.
  $endgroup$
  – Michael Rozenberg
  12 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.

For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

share|cite|improve this answer

edited 53 mins ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is interesting. I had never seen the identities you begin with.
  $endgroup$
  – Lubin
  28 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Lubin same with me too.
  $endgroup$
  – Max0815
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We can prove it. It's not hard.
  $endgroup$
  – Michael Rozenberg
  12 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

There are the following identities.
$$sqrt{a+sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}+sqrt{frac{a-sqrt{a^2-b}}{2}}$$ and
$$sqrt{a-sqrt{b}}=sqrt{frac{a+sqrt{a^2-b}}{2}}-sqrt{frac{a-sqrt{a^2-b}}{2}},$$
where all numbers under radicals they are non-negatives.

For example:
$$sqrt{5+2sqrt6}=sqrt{5+sqrt{24}}=sqrt{frac{5+sqrt{5^2-24}}{2}}+sqrt{frac{5-sqrt{5^2-24}}{2}}=sqrt3+sqrt2.$$
This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

share|cite|improve this answer

edited 53 mins ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

$endgroup$

share|cite|improve this answer

edited 53 mins ago

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196

answered 1 hour ago

Michael RozenbergMichael Rozenberg

104k1891196