Complicated square root problem.
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
"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
$endgroup$
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
number-theory radicals nested-radicals
edited 8 mins ago
Michael Rozenberg
104k1891196
104k1891196
asked 1 hour ago
Max0815Max0815
66618
66618
$begingroup$
"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 |
$begingroup$
"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
$begingroup$
"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
$begingroup$
"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
$begingroup$
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
$begingroup$
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
$begingroup$
@fleablood I know, complex numbers. How should I write it then?
$endgroup$
– Max0815
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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?
$endgroup$
$begingroup$
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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 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.
$endgroup$
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 mins ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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?
$endgroup$
$begingroup$
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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 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?
$endgroup$
$begingroup$
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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 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?
$endgroup$
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?
edited 1 hour ago
answered 1 hour ago
Stan TendijckStan Tendijck
1,826311
1,826311
$begingroup$
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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 mins ago
add a comment |
$begingroup$
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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
Yes. thanx!!!!!
$endgroup$
– Max0815
1 hour ago
$begingroup$
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
57 mins 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
57 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
yes I believe so too.
$endgroup$
– Max0815
26 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.
$endgroup$
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 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.
$endgroup$
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 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.
$endgroup$
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.
edited 51 mins ago
answered 1 hour ago
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 mins ago
add a comment |
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 mins ago
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
This is interesting. I had never seen the identities you begin with.
$endgroup$
– Lubin
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
@Lubin same with me too.
$endgroup$
– Max0815
26 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 mins ago
$begingroup$
We can prove it. It's not hard.
$endgroup$
– Michael Rozenberg
11 mins ago
add a comment |
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"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?
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– fleablood
1 hour ago
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By the way, don't say "complex square root". "complex" has a specific mathematical meaning you didn't mean.
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– fleablood
1 hour ago
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Quite. The word "complicated" would be better to use here since "complicated" doesn't really have much mathematical use, just linguistic use.
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– JMoravitz
1 hour ago
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@fleablood I know, complex numbers. How should I write it then?
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– Max0815
1 hour ago