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












2












$begingroup$



Let
$$A= sqrt{93+28sqrt{11}}$$
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











$endgroup$












  • $begingroup$
    Presumably you mean $C$ is the fractional part of $A$
    $endgroup$
    – Ross Millikan
    Mar 23 at 14:07










  • $begingroup$
    What do you mean by fractional part? That number is irrational.
    $endgroup$
    – Allawonder
    Mar 23 at 14:10
















2












$begingroup$



Let
$$A= sqrt{93+28sqrt{11}}$$
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











$endgroup$












  • $begingroup$
    Presumably you mean $C$ is the fractional part of $A$
    $endgroup$
    – Ross Millikan
    Mar 23 at 14:07










  • $begingroup$
    What do you mean by fractional part? That number is irrational.
    $endgroup$
    – Allawonder
    Mar 23 at 14:10














2












2








2





$begingroup$



Let
$$A= sqrt{93+28sqrt{11}}$$
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











$endgroup$





Let
$$A= sqrt{93+28sqrt{11}}$$
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




radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 14:59









Maria Mazur

49.3k1360123




49.3k1360123










asked Mar 23 at 13:51









SuperMage1SuperMage1

947211




947211












  • $begingroup$
    Presumably you mean $C$ is the fractional part of $A$
    $endgroup$
    – Ross Millikan
    Mar 23 at 14:07










  • $begingroup$
    What do you mean by fractional part? That number is irrational.
    $endgroup$
    – Allawonder
    Mar 23 at 14:10


















  • $begingroup$
    Presumably you mean $C$ is the fractional part of $A$
    $endgroup$
    – Ross Millikan
    Mar 23 at 14:07










  • $begingroup$
    What do you mean by fractional part? That number is irrational.
    $endgroup$
    – Allawonder
    Mar 23 at 14:10
















$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
Mar 23 at 14:07




$begingroup$
Presumably you mean $C$ is the fractional part of $A$
$endgroup$
– Ross Millikan
Mar 23 at 14:07












$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
Mar 23 at 14:10




$begingroup$
What do you mean by fractional part? That number is irrational.
$endgroup$
– Allawonder
Mar 23 at 14:10










2 Answers
2






active

oldest

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4












$begingroup$

Let's find intgers $x,y$ such that $$(x+ysqrt{11})^2= 93+28sqrt{11}$$



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 $$boxed{sqrt{93+28sqrt{11}} = 7+2sqrt{11}}$$



Since $$13= 7+2cdot 3<7+2sqrt{11} <7+2cdot sqrt{49over 4} = 14$$



we see $B =13$...






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint:



    $(7+2sqrt{11})^2=cdots$






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Let's find intgers $x,y$ such that $$(x+ysqrt{11})^2= 93+28sqrt{11}$$



      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 $$boxed{sqrt{93+28sqrt{11}} = 7+2sqrt{11}}$$



      Since $$13= 7+2cdot 3<7+2sqrt{11} <7+2cdot sqrt{49over 4} = 14$$



      we see $B =13$...






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Let's find intgers $x,y$ such that $$(x+ysqrt{11})^2= 93+28sqrt{11}$$



        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 $$boxed{sqrt{93+28sqrt{11}} = 7+2sqrt{11}}$$



        Since $$13= 7+2cdot 3<7+2sqrt{11} <7+2cdot sqrt{49over 4} = 14$$



        we see $B =13$...






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Let's find intgers $x,y$ such that $$(x+ysqrt{11})^2= 93+28sqrt{11}$$



          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 $$boxed{sqrt{93+28sqrt{11}} = 7+2sqrt{11}}$$



          Since $$13= 7+2cdot 3<7+2sqrt{11} <7+2cdot sqrt{49over 4} = 14$$



          we see $B =13$...






          share|cite|improve this answer









          $endgroup$



          Let's find intgers $x,y$ such that $$(x+ysqrt{11})^2= 93+28sqrt{11}$$



          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 $$boxed{sqrt{93+28sqrt{11}} = 7+2sqrt{11}}$$



          Since $$13= 7+2cdot 3<7+2sqrt{11} <7+2cdot sqrt{49over 4} = 14$$



          we see $B =13$...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 23 at 14:20









          Maria MazurMaria Mazur

          49.3k1360123




          49.3k1360123























              2












              $begingroup$

              Hint:



              $(7+2sqrt{11})^2=cdots$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Hint:



                $(7+2sqrt{11})^2=cdots$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Hint:



                  $(7+2sqrt{11})^2=cdots$






                  share|cite|improve this answer









                  $endgroup$



                  Hint:



                  $(7+2sqrt{11})^2=cdots$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 23 at 14:05









                  drhabdrhab

                  104k545136




                  104k545136






























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