The number $N = sqrt2+sqrt5-sqrt6-3sqrt5+sqrt14-6sqrt5$ simplifies to? [closed]

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The number $N = sqrt2+sqrt5-sqrt6-3sqrt5+sqrt14-6sqrt5$ simplifies to.

This question is from Logarithm Chapter, I'm seriously way too confused that how can these types of questions be solved using Logarithm? There aren't any similar solved examples related to this. I even doubt that this is a Logarithm Question.

If anyone can Guide me with this question, then I think that I can solve the rest of the problems similar to this in the textbook.

Can anyone Highlight the role of Logarithm in continuous Square root Expression?

Well, the second question arises is that $sqrt5$ which is irrational. The way it is added to other numbers, it is pretty sure that it's an Irrational Expression.

Can anyone know the similar type of question or the solution to this question?

Thanks :)

logarithms

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asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

closed as off-topic by user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B Jul 28 at 14:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
  – Cornman
  Jul 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman Brain isn't working, need help with creativity
  – Abhas Kumar Sinha
  Jul 27 at 14:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can solve this by just completing the square over and over.
  – OmnipotentEntity
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OmnipotentEntity But what to do to the root 5 which is in the middle?
  – Abhas Kumar Sinha
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'll add an answer to fully explain.
  – OmnipotentEntity
  Jul 27 at 14:48
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

The number $N = sqrt2+sqrt5-sqrt6-3sqrt5+sqrt14-6sqrt5$ simplifies to.

This question is from Logarithm Chapter, I'm seriously way too confused that how can these types of questions be solved using Logarithm? There aren't any similar solved examples related to this. I even doubt that this is a Logarithm Question.

If anyone can Guide me with this question, then I think that I can solve the rest of the problems similar to this in the textbook.

Can anyone Highlight the role of Logarithm in continuous Square root Expression?

Well, the second question arises is that $sqrt5$ which is irrational. The way it is added to other numbers, it is pretty sure that it's an Irrational Expression.

Can anyone know the similar type of question or the solution to this question?

Thanks :)

logarithms

share|cite|improve this question

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

closed as off-topic by user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B Jul 28 at 14:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
  – Cornman
  Jul 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman Brain isn't working, need help with creativity
  – Abhas Kumar Sinha
  Jul 27 at 14:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can solve this by just completing the square over and over.
  – OmnipotentEntity
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OmnipotentEntity But what to do to the root 5 which is in the middle?
  – Abhas Kumar Sinha
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'll add an answer to fully explain.
  – OmnipotentEntity
  Jul 27 at 14:48
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

up vote
2
down vote

favorite

The number $N = sqrt2+sqrt5-sqrt6-3sqrt5+sqrt14-6sqrt5$ simplifies to.

This question is from Logarithm Chapter, I'm seriously way too confused that how can these types of questions be solved using Logarithm? There aren't any similar solved examples related to this. I even doubt that this is a Logarithm Question.

If anyone can Guide me with this question, then I think that I can solve the rest of the problems similar to this in the textbook.

Can anyone Highlight the role of Logarithm in continuous Square root Expression?

Well, the second question arises is that $sqrt5$ which is irrational. The way it is added to other numbers, it is pretty sure that it's an Irrational Expression.

Can anyone know the similar type of question or the solution to this question?

Thanks :)

logarithms

share|cite|improve this question

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

The number $N = sqrt2+sqrt5-sqrt6-3sqrt5+sqrt14-6sqrt5$ simplifies to.

This question is from Logarithm Chapter, I'm seriously way too confused that how can these types of questions be solved using Logarithm? There aren't any similar solved examples related to this. I even doubt that this is a Logarithm Question.

If anyone can Guide me with this question, then I think that I can solve the rest of the problems similar to this in the textbook.

Can anyone Highlight the role of Logarithm in continuous Square root Expression?

Well, the second question arises is that $sqrt5$ which is irrational. The way it is added to other numbers, it is pretty sure that it's an Irrational Expression.

Can anyone know the similar type of question or the solution to this question?

Thanks :)

logarithms

share|cite|improve this question

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

share|cite|improve this question

share|cite|improve this question

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

asked Jul 27 at 14:39

Abhas Kumar Sinha

10011

10011

closed as off-topic by user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B Jul 28 at 14:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B Jul 28 at 14:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, Did, Taroccoesbrocco, Mostafa Ayaz, John B

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
  – Cornman
  Jul 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman Brain isn't working, need help with creativity
  – Abhas Kumar Sinha
  Jul 27 at 14:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can solve this by just completing the square over and over.
  – OmnipotentEntity
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OmnipotentEntity But what to do to the root 5 which is in the middle?
  – Abhas Kumar Sinha
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'll add an answer to fully explain.
  – OmnipotentEntity
  Jul 27 at 14:48
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
  – Cornman
  Jul 27 at 14:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Cornman Brain isn't working, need help with creativity
  – Abhas Kumar Sinha
  Jul 27 at 14:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You can solve this by just completing the square over and over.
  – OmnipotentEntity
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @OmnipotentEntity But what to do to the root 5 which is in the middle?
  – Abhas Kumar Sinha
  Jul 27 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'll add an answer to fully explain.
  – OmnipotentEntity
  Jul 27 at 14:48
  

  
  
  
  

1

1

Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
– Cornman
Jul 27 at 14:43

Wolframalpha says, that it simplyfies to 2 actually. I do not think that you need the logarithm here in any way. Just some creative calculations.
– Cornman
Jul 27 at 14:43

@Cornman Brain isn't working, need help with creativity
– Abhas Kumar Sinha
Jul 27 at 14:44

@Cornman Brain isn't working, need help with creativity
– Abhas Kumar Sinha
Jul 27 at 14:44

1

1

You can solve this by just completing the square over and over.
– OmnipotentEntity
Jul 27 at 14:47

You can solve this by just completing the square over and over.
– OmnipotentEntity
Jul 27 at 14:47

@OmnipotentEntity But what to do to the root 5 which is in the middle?
– Abhas Kumar Sinha
Jul 27 at 14:47

@OmnipotentEntity But what to do to the root 5 which is in the middle?
– Abhas Kumar Sinha
Jul 27 at 14:47

I'll add an answer to fully explain.
– OmnipotentEntity
Jul 27 at 14:48

I'll add an answer to fully explain.
– OmnipotentEntity
Jul 27 at 14:48

 | 
show 1 more comment

5 Answers
5

active

oldest

votes

up vote
1
down vote



accepted

Go from inside to outside:

We have $14-6sqrt5=9-6sqrt5+5=(3-sqrt5)^2$

$sqrt(3-sqrt5)^2=3-sqrt5$

This leads to:

$6-3sqrt5+3-sqrt5=9-4sqrt5=4-4sqrt5+5=underbrace(2-sqrt5)^2_2-sqrt5<0=(sqrt5-2)^2$

We get $2+sqrt5-sqrt5+2=4$

and finally $sqrt4=2$

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

add a comment | 

up vote
6
down vote

Notice that
$$14-6sqrt5=3^2-2cdot3cdotsqrt5+(sqrt5)^2=(3-sqrt5)^2.$$

and so on...

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

add a comment | 

up vote
6
down vote

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt14-6 sqrt5$$

Let's consider just $14-6 sqrt5$. This is equal to $9 - 6 sqrt5 + 5$, which is equal to $9 - 6 sqrt5 + sqrt5^2$, this can be simplified to $(3 - sqrt5)^2$.

So we have now:

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt(3 - sqrt5)^2$$

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+3 - sqrt5$$

$$sqrt2+ sqrt5-sqrt9 - 4 sqrt5$$

Then just do the same thing again, complete the square:

$$9 - 4 sqrt5 = 4 - 4sqrt5 + sqrt5^2 = (sqrt5 - 2)^2$$

Substituting back:

$$sqrt2+ sqrt5-sqrt(sqrt5 - 2)^2$$

$$sqrt2+ sqrt5 - sqrt5 + 2$$

$$sqrt4$$

$$2$$

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

add a comment | 

up vote
3
down vote

First note that $$14-6sqrt 5=9-6sqrt 5 +5=(3-sqrt 5)^2$$

Then $$6-3sqrt 5+3-sqrt 5=5-4sqrt 5+4=(sqrt 5-2)^2$$

And $$2+sqrt 5+2-sqrt 5=4$$

How do you find these - well once I knew what I was looking for it was easy. For the first, for example, I'm looking for $$(a-bsqrt 5)^2=14-6sqrt 5$$
so that $a^2+5b^2=14$ and $2ab=6$, and these are easy to solve in integers (or to show no solution, if that is the case.) I didn't need a piece of paper o work it out.

The question says "Simplify" so I expect the simplification to go through. Some care is required with the convention on the sign of the square root.

share|cite|improve this answer

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

add a comment | 

up vote
3
down vote

Hint:

I'll give the first step. The basis is to unfold each successive radicand, showing each is a square. I'll give the first step: $;14-6sqrt5$ is a square if we can write it in the form $(a-bsqrt 5)^2 =a^2+b^2-2absqrt 5$. The simplest works:
$$(3-sqrt 5)^2=9+5-2cdot 3sqrt 5.$$
So $sqrt14-6sqrt5=3-sqrt 5 (>0)$, and one radical level above, we have
$$sqrt6-3sqrt5+sqrt14-6sqrt5=sqrt9-4sqrt 5.$$
Can you proceed?

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answered Jul 27 at 14:58

Bernard

110k635102

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  yes :) But, still do you think that there is any role of Logarithm in this question?
  – Abhas Kumar Sinha
  Jul 27 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
  – Bernard
  Jul 27 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I saw higher algebra (hall and knight), but didn't found any
  – Abhas Kumar Sinha
  Jul 27 at 15:08
  

  
  
  
  

add a comment | 

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Go from inside to outside:

We have $14-6sqrt5=9-6sqrt5+5=(3-sqrt5)^2$

$sqrt(3-sqrt5)^2=3-sqrt5$

This leads to:

$6-3sqrt5+3-sqrt5=9-4sqrt5=4-4sqrt5+5=underbrace(2-sqrt5)^2_2-sqrt5<0=(sqrt5-2)^2$

We get $2+sqrt5-sqrt5+2=4$

and finally $sqrt4=2$

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

add a comment | 

up vote
1
down vote



accepted

Go from inside to outside:

We have $14-6sqrt5=9-6sqrt5+5=(3-sqrt5)^2$

$sqrt(3-sqrt5)^2=3-sqrt5$

This leads to:

$6-3sqrt5+3-sqrt5=9-4sqrt5=4-4sqrt5+5=underbrace(2-sqrt5)^2_2-sqrt5<0=(sqrt5-2)^2$

We get $2+sqrt5-sqrt5+2=4$

and finally $sqrt4=2$

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Go from inside to outside:

We have $14-6sqrt5=9-6sqrt5+5=(3-sqrt5)^2$

$sqrt(3-sqrt5)^2=3-sqrt5$

This leads to:

$6-3sqrt5+3-sqrt5=9-4sqrt5=4-4sqrt5+5=underbrace(2-sqrt5)^2_2-sqrt5<0=(sqrt5-2)^2$

We get $2+sqrt5-sqrt5+2=4$

and finally $sqrt4=2$

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

Go from inside to outside:

We have $14-6sqrt5=9-6sqrt5+5=(3-sqrt5)^2$

$sqrt(3-sqrt5)^2=3-sqrt5$

This leads to:

$6-3sqrt5+3-sqrt5=9-4sqrt5=4-4sqrt5+5=underbrace(2-sqrt5)^2_2-sqrt5<0=(sqrt5-2)^2$

We get $2+sqrt5-sqrt5+2=4$

and finally $sqrt4=2$

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 15:00

Cornman

2,37021027

answered Jul 27 at 15:00

Cornman

2,37021027

answered Jul 27 at 15:00

Cornman

2,37021027

2,37021027

add a comment | 

add a comment | 

up vote
6
down vote

Notice that
$$14-6sqrt5=3^2-2cdot3cdotsqrt5+(sqrt5)^2=(3-sqrt5)^2.$$

and so on...

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

add a comment | 

up vote
6
down vote

Notice that
$$14-6sqrt5=3^2-2cdot3cdotsqrt5+(sqrt5)^2=(3-sqrt5)^2.$$

and so on...

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

add a comment | 

up vote
6
down vote

up vote
6
down vote

Notice that
$$14-6sqrt5=3^2-2cdot3cdotsqrt5+(sqrt5)^2=(3-sqrt5)^2.$$

and so on...

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

Notice that
$$14-6sqrt5=3^2-2cdot3cdotsqrt5+(sqrt5)^2=(3-sqrt5)^2.$$

and so on...

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 14:50

mengdie1982

2,827216

answered Jul 27 at 14:50

mengdie1982

2,827216

answered Jul 27 at 14:50

mengdie1982

2,827216

2,827216

add a comment | 

add a comment | 

up vote
6
down vote

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt14-6 sqrt5$$

Let's consider just $14-6 sqrt5$. This is equal to $9 - 6 sqrt5 + 5$, which is equal to $9 - 6 sqrt5 + sqrt5^2$, this can be simplified to $(3 - sqrt5)^2$.

So we have now:

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt(3 - sqrt5)^2$$

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+3 - sqrt5$$

$$sqrt2+ sqrt5-sqrt9 - 4 sqrt5$$

Then just do the same thing again, complete the square:

$$9 - 4 sqrt5 = 4 - 4sqrt5 + sqrt5^2 = (sqrt5 - 2)^2$$

Substituting back:

$$sqrt2+ sqrt5-sqrt(sqrt5 - 2)^2$$

$$sqrt2+ sqrt5 - sqrt5 + 2$$

$$sqrt4$$

$$2$$

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

add a comment | 

up vote
6
down vote

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt14-6 sqrt5$$

Let's consider just $14-6 sqrt5$. This is equal to $9 - 6 sqrt5 + 5$, which is equal to $9 - 6 sqrt5 + sqrt5^2$, this can be simplified to $(3 - sqrt5)^2$.

So we have now:

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt(3 - sqrt5)^2$$

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+3 - sqrt5$$

$$sqrt2+ sqrt5-sqrt9 - 4 sqrt5$$

Then just do the same thing again, complete the square:

$$9 - 4 sqrt5 = 4 - 4sqrt5 + sqrt5^2 = (sqrt5 - 2)^2$$

Substituting back:

$$sqrt2+ sqrt5-sqrt(sqrt5 - 2)^2$$

$$sqrt2+ sqrt5 - sqrt5 + 2$$

$$sqrt4$$

$$2$$

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

add a comment | 

up vote
6
down vote

up vote
6
down vote

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt14-6 sqrt5$$

Let's consider just $14-6 sqrt5$. This is equal to $9 - 6 sqrt5 + 5$, which is equal to $9 - 6 sqrt5 + sqrt5^2$, this can be simplified to $(3 - sqrt5)^2$.

So we have now:

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt(3 - sqrt5)^2$$

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+3 - sqrt5$$

$$sqrt2+ sqrt5-sqrt9 - 4 sqrt5$$

Then just do the same thing again, complete the square:

$$9 - 4 sqrt5 = 4 - 4sqrt5 + sqrt5^2 = (sqrt5 - 2)^2$$

Substituting back:

$$sqrt2+ sqrt5-sqrt(sqrt5 - 2)^2$$

$$sqrt2+ sqrt5 - sqrt5 + 2$$

$$sqrt4$$

$$2$$

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt14-6 sqrt5$$

Let's consider just $14-6 sqrt5$. This is equal to $9 - 6 sqrt5 + 5$, which is equal to $9 - 6 sqrt5 + sqrt5^2$, this can be simplified to $(3 - sqrt5)^2$.

So we have now:

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+sqrt(3 - sqrt5)^2$$

$$sqrt2+ sqrt5-sqrt6 -3 sqrt5+3 - sqrt5$$

$$sqrt2+ sqrt5-sqrt9 - 4 sqrt5$$

Then just do the same thing again, complete the square:

$$9 - 4 sqrt5 = 4 - 4sqrt5 + sqrt5^2 = (sqrt5 - 2)^2$$

Substituting back:

$$sqrt2+ sqrt5-sqrt(sqrt5 - 2)^2$$

$$sqrt2+ sqrt5 - sqrt5 + 2$$

$$sqrt4$$

$$2$$

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 14:54

OmnipotentEntity

391216

answered Jul 27 at 14:54

OmnipotentEntity

391216

answered Jul 27 at 14:54

OmnipotentEntity

391216

391216

add a comment | 

add a comment | 

up vote
3
down vote

First note that $$14-6sqrt 5=9-6sqrt 5 +5=(3-sqrt 5)^2$$

Then $$6-3sqrt 5+3-sqrt 5=5-4sqrt 5+4=(sqrt 5-2)^2$$

And $$2+sqrt 5+2-sqrt 5=4$$

How do you find these - well once I knew what I was looking for it was easy. For the first, for example, I'm looking for $$(a-bsqrt 5)^2=14-6sqrt 5$$
so that $a^2+5b^2=14$ and $2ab=6$, and these are easy to solve in integers (or to show no solution, if that is the case.) I didn't need a piece of paper o work it out.

The question says "Simplify" so I expect the simplification to go through. Some care is required with the convention on the sign of the square root.

share|cite|improve this answer

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

add a comment | 

up vote
3
down vote

First note that $$14-6sqrt 5=9-6sqrt 5 +5=(3-sqrt 5)^2$$

Then $$6-3sqrt 5+3-sqrt 5=5-4sqrt 5+4=(sqrt 5-2)^2$$

And $$2+sqrt 5+2-sqrt 5=4$$

How do you find these - well once I knew what I was looking for it was easy. For the first, for example, I'm looking for $$(a-bsqrt 5)^2=14-6sqrt 5$$
so that $a^2+5b^2=14$ and $2ab=6$, and these are easy to solve in integers (or to show no solution, if that is the case.) I didn't need a piece of paper o work it out.

The question says "Simplify" so I expect the simplification to go through. Some care is required with the convention on the sign of the square root.

share|cite|improve this answer

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

add a comment | 

up vote
3
down vote

up vote
3
down vote

First note that $$14-6sqrt 5=9-6sqrt 5 +5=(3-sqrt 5)^2$$

Then $$6-3sqrt 5+3-sqrt 5=5-4sqrt 5+4=(sqrt 5-2)^2$$

And $$2+sqrt 5+2-sqrt 5=4$$

How do you find these - well once I knew what I was looking for it was easy. For the first, for example, I'm looking for $$(a-bsqrt 5)^2=14-6sqrt 5$$
so that $a^2+5b^2=14$ and $2ab=6$, and these are easy to solve in integers (or to show no solution, if that is the case.) I didn't need a piece of paper o work it out.

The question says "Simplify" so I expect the simplification to go through. Some care is required with the convention on the sign of the square root.

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answered Jul 27 at 14:55

Mark Bennet

76.3k773170

First note that $$14-6sqrt 5=9-6sqrt 5 +5=(3-sqrt 5)^2$$

Then $$6-3sqrt 5+3-sqrt 5=5-4sqrt 5+4=(sqrt 5-2)^2$$

And $$2+sqrt 5+2-sqrt 5=4$$

How do you find these - well once I knew what I was looking for it was easy. For the first, for example, I'm looking for $$(a-bsqrt 5)^2=14-6sqrt 5$$
so that $a^2+5b^2=14$ and $2ab=6$, and these are easy to solve in integers (or to show no solution, if that is the case.) I didn't need a piece of paper o work it out.

The question says "Simplify" so I expect the simplification to go through. Some care is required with the convention on the sign of the square root.

share|cite|improve this answer

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

answered Jul 27 at 14:55

Mark Bennet

76.3k773170

76.3k773170

add a comment | 

add a comment | 

up vote
3
down vote

Hint:

I'll give the first step. The basis is to unfold each successive radicand, showing each is a square. I'll give the first step: $;14-6sqrt5$ is a square if we can write it in the form $(a-bsqrt 5)^2 =a^2+b^2-2absqrt 5$. The simplest works:
$$(3-sqrt 5)^2=9+5-2cdot 3sqrt 5.$$
So $sqrt14-6sqrt5=3-sqrt 5 (>0)$, and one radical level above, we have
$$sqrt6-3sqrt5+sqrt14-6sqrt5=sqrt9-4sqrt 5.$$
Can you proceed?

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answered Jul 27 at 14:58

Bernard

110k635102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes :) But, still do you think that there is any role of Logarithm in this question?
  – Abhas Kumar Sinha
  Jul 27 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
  – Bernard
  Jul 27 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I saw higher algebra (hall and knight), but didn't found any
  – Abhas Kumar Sinha
  Jul 27 at 15:08
  

  
  
  
  

add a comment | 

up vote
3
down vote

Hint:

I'll give the first step. The basis is to unfold each successive radicand, showing each is a square. I'll give the first step: $;14-6sqrt5$ is a square if we can write it in the form $(a-bsqrt 5)^2 =a^2+b^2-2absqrt 5$. The simplest works:
$$(3-sqrt 5)^2=9+5-2cdot 3sqrt 5.$$
So $sqrt14-6sqrt5=3-sqrt 5 (>0)$, and one radical level above, we have
$$sqrt6-3sqrt5+sqrt14-6sqrt5=sqrt9-4sqrt 5.$$
Can you proceed?

share|cite|improve this answer

answered Jul 27 at 14:58

Bernard

110k635102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes :) But, still do you think that there is any role of Logarithm in this question?
  – Abhas Kumar Sinha
  Jul 27 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
  – Bernard
  Jul 27 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I saw higher algebra (hall and knight), but didn't found any
  – Abhas Kumar Sinha
  Jul 27 at 15:08
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Hint:

I'll give the first step. The basis is to unfold each successive radicand, showing each is a square. I'll give the first step: $;14-6sqrt5$ is a square if we can write it in the form $(a-bsqrt 5)^2 =a^2+b^2-2absqrt 5$. The simplest works:
$$(3-sqrt 5)^2=9+5-2cdot 3sqrt 5.$$
So $sqrt14-6sqrt5=3-sqrt 5 (>0)$, and one radical level above, we have
$$sqrt6-3sqrt5+sqrt14-6sqrt5=sqrt9-4sqrt 5.$$
Can you proceed?

share|cite|improve this answer

answered Jul 27 at 14:58

Bernard

110k635102

Hint:

I'll give the first step. The basis is to unfold each successive radicand, showing each is a square. I'll give the first step: $;14-6sqrt5$ is a square if we can write it in the form $(a-bsqrt 5)^2 =a^2+b^2-2absqrt 5$. The simplest works:
$$(3-sqrt 5)^2=9+5-2cdot 3sqrt 5.$$
So $sqrt14-6sqrt5=3-sqrt 5 (>0)$, and one radical level above, we have
$$sqrt6-3sqrt5+sqrt14-6sqrt5=sqrt9-4sqrt 5.$$
Can you proceed?

share|cite|improve this answer

answered Jul 27 at 14:58

Bernard

110k635102

share|cite|improve this answer

share|cite|improve this answer

answered Jul 27 at 14:58

Bernard

110k635102

answered Jul 27 at 14:58

Bernard

110k635102

answered Jul 27 at 14:58

Bernard

110k635102

110k635102

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes :) But, still do you think that there is any role of Logarithm in this question?
  – Abhas Kumar Sinha
  Jul 27 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
  – Bernard
  Jul 27 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I saw higher algebra (hall and knight), but didn't found any
  – Abhas Kumar Sinha
  Jul 27 at 15:08
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  yes :) But, still do you think that there is any role of Logarithm in this question?
  – Abhas Kumar Sinha
  Jul 27 at 15:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
  – Bernard
  Jul 27 at 15:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I saw higher algebra (hall and knight), but didn't found any
  – Abhas Kumar Sinha
  Jul 27 at 15:08
  

  
  
  
  

yes :) But, still do you think that there is any role of Logarithm in this question?
– Abhas Kumar Sinha
Jul 27 at 15:02

yes :) But, still do you think that there is any role of Logarithm in this question?
– Abhas Kumar Sinha
Jul 27 at 15:02

I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
– Bernard
Jul 27 at 15:05

I don't think so, or it's well hidden :-). That's the kind of calculations I was asked to do when I entered high-school
– Bernard
Jul 27 at 15:05

I saw higher algebra (hall and knight), but didn't found any
– Abhas Kumar Sinha
Jul 27 at 15:08

I saw higher algebra (hall and knight), but didn't found any
– Abhas Kumar Sinha
Jul 27 at 15:08

add a comment |