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Evaluating expression with Integer part and Fraction part of a nested radical

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2

$begingroup$

Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$

I tried manipulating it by setting

$$ A=B+C$$

but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?

radicals

share|cite|improve this question

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Presumably you mean $C$ is the fractional part of $A$
  $endgroup$
  – Ross Millikan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by fractional part? That number is irrational.
  $endgroup$
  – Allawonder
  2 days ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$

I tried manipulating it by setting

$$ A=B+C$$

but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?

radicals

share|cite|improve this question

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Presumably you mean $C$ is the fractional part of $A$
  $endgroup$
  – Ross Millikan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by fractional part? That number is irrational.
  $endgroup$
  – Allawonder
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Let
$$A= sqrt93+28sqrt11$$
if $B$ is the integer part of $A$ and $C$ is the fraction part of $C$, what is the value of
$$B+C^2$$

I tried manipulating it by setting

$$ A=B+C$$

but I can't transform it into the expression, do I need approximate the integer part of A in order to solve this or is there another way?

radicals

share|cite|improve this question

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

$endgroup$

share|cite|improve this question

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

share|cite|improve this question

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

edited 2 days ago

Maria Mazur

48.3k1260121

edited 2 days ago

Maria Mazur

48.3k1260121

asked 2 days ago

SuperMage1SuperMage1

947211

asked 2 days ago

SuperMage1SuperMage1

947211

2 Answers
2

active

oldest

votes

4

$begingroup$

Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$

Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$

Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$

Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$

we see $B =13$...

share|cite|improve this answer

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

2

$begingroup$

Hint:

$(7+2sqrt11)^2=cdots$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

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4

$begingroup$

Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$

Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$

Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$

Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$

we see $B =13$...

share|cite|improve this answer

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

4

$begingroup$

Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$

Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$

Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$

Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$

we see $B =13$...

share|cite|improve this answer

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

add a comment | 

$begingroup$

Let's find intgers $x,y$ such that $$(x+ysqrt11)^2= 93+28sqrt11$$

Then $$x^2+11y^2 =93 ;;;wedge ;;; 2xy = 28$$

Since $11y^2<99implies y^2 <9 implies |y|leq 2$ and $xy=14$. Playing with numbers we see that $x=7$ and $y=2$ works well, so $$boxedsqrt93+28sqrt11 = 7+2sqrt11$$

Since $$13= 7+2cdot 3<7+2sqrt11 <7+2cdot sqrt49over 4 = 14$$

we see $B =13$...

share|cite|improve this answer

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

$endgroup$

share|cite|improve this answer

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

answered 2 days ago

Maria MazurMaria Mazur

48.3k1260121

2

$begingroup$

Hint:

$(7+2sqrt11)^2=cdots$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

2

$begingroup$

Hint:

$(7+2sqrt11)^2=cdots$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

add a comment | 

$begingroup$

Hint:

$(7+2sqrt11)^2=cdots$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

$endgroup$

share|cite|improve this answer

answered 2 days ago

drhabdrhab

103k545136

answered 2 days ago

drhabdrhab

103k545136

answered 2 days ago

drhabdrhab

103k545136