Simplifying a solution to $x^2 = 4 + 2sqrt{2}$












1












$begingroup$


$$x^2 = 4 + 2sqrt{2}$$



$$x = 2 + sqrt{2sqrt{2}}$$



Neglecting that the result can be negative as well, how should I continue? How can I simplify it?






algebra-precalculus radicals







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  • 6




    $begingroup$
    Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
    $endgroup$
    – Mark Bennet
    Jan 23 at 18:39






  • 3




    $begingroup$
    Why simplify the expression if it's already simple? :o)
    $endgroup$
    – Bernard
    Jan 23 at 18:43
















1












$begingroup$


$$x^2 = 4 + 2sqrt{2}$$



$$x = 2 + sqrt{2sqrt{2}}$$



Neglecting that the result can be negative as well, how should I continue? How can I simplify it?






algebra-precalculus radicals







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
    $endgroup$
    – Mark Bennet
    Jan 23 at 18:39






  • 3




    $begingroup$
    Why simplify the expression if it's already simple? :o)
    $endgroup$
    – Bernard
    Jan 23 at 18:43














1












1








1





$begingroup$


$$x^2 = 4 + 2sqrt{2}$$



$$x = 2 + sqrt{2sqrt{2}}$$



Neglecting that the result can be negative as well, how should I continue? How can I simplify it?






algebra-precalculus radicals







share|cite|improve this question











$endgroup$




$$x^2 = 4 + 2sqrt{2}$$



$$x = 2 + sqrt{2sqrt{2}}$$



Neglecting that the result can be negative as well, how should I continue? How can I simplify it?






algebra-precalculus radicals




algebra-precalculus radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 21:10









J. W. Tanner

3,2401320




3,2401320










asked Jan 23 at 18:37









wenoweno

29911




29911








  • 6




    $begingroup$
    Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
    $endgroup$
    – Mark Bennet
    Jan 23 at 18:39






  • 3




    $begingroup$
    Why simplify the expression if it's already simple? :o)
    $endgroup$
    – Bernard
    Jan 23 at 18:43














  • 6




    $begingroup$
    Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
    $endgroup$
    – Mark Bennet
    Jan 23 at 18:39






  • 3




    $begingroup$
    Why simplify the expression if it's already simple? :o)
    $endgroup$
    – Bernard
    Jan 23 at 18:43








6




6




$begingroup$
Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
$endgroup$
– Mark Bennet
Jan 23 at 18:39




$begingroup$
Note that $sqrt {a+b}neq sqrt a +sqrt b$ in general
$endgroup$
– Mark Bennet
Jan 23 at 18:39




3




3




$begingroup$
Why simplify the expression if it's already simple? :o)
$endgroup$
– Bernard
Jan 23 at 18:43




$begingroup$
Why simplify the expression if it's already simple? :o)
$endgroup$
– Bernard
Jan 23 at 18:43










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$x^2 = 4 + 2sqrt{2} rightarrow x = pm sqrt{4 + 2sqrt{2}}$






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    5












    $begingroup$

    $x^2 = 4 + 2sqrt{2} rightarrow x = pm sqrt{4 + 2sqrt{2}}$






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      $x^2 = 4 + 2sqrt{2} rightarrow x = pm sqrt{4 + 2sqrt{2}}$






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        $x^2 = 4 + 2sqrt{2} rightarrow x = pm sqrt{4 + 2sqrt{2}}$






        share|cite|improve this answer









        $endgroup$



        $x^2 = 4 + 2sqrt{2} rightarrow x = pm sqrt{4 + 2sqrt{2}}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 23 at 18:40









        lightxbulblightxbulb

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