How to simplify $sqrt{2+sqrt{3}}$ $?$ The Next CEO of Stack OverflowSimple doubt about complex...
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How to simplify $sqrt{2+sqrt{3}}$ $?$
The Next CEO of Stack OverflowSimple doubt about complex numbersRationalising the denominator: $frac{11}{3sqrt{3}+7}$Rationalize and simplify $frac{x-1}{sqrt{2sqrt{x}} + 1 - sqrt[4]{x}}$How to simplify $frac{4 + 2sqrt6}{sqrt{5 + 2sqrt{6}}}$?Proof by contradiction that $sqrt{2} + sqrt{6} < sqrt{15}$Simplify $frac {sqrt5}{sqrt3+1} - sqrtfrac{30}{8} + frac {sqrt {45}}{2}$Difficulty in “expressing radical as square”Proof that $sqrt6 - sqrt2 - sqrt3$ is irrational.Simplify $sqrt xleft(sqrt x+frac1{sqrt x}right)$Simplifying $frac { sqrt2 + sqrt 6}{sqrt2 + sqrt3}$?Simplify $frac{sqrt{24}}{8}$
$begingroup$
Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$
The answer to this question is $frac{4}{3}$ in a workbook.
How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$
But I can't simplify $sqrt{2+sqrt3}$ like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?
algebra-precalculus radicals nested-radicals
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
$endgroup$
add a comment |
$begingroup$
Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$
The answer to this question is $frac{4}{3}$ in a workbook.
How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$
But I can't simplify $sqrt{2+sqrt3}$ like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?
algebra-precalculus radicals nested-radicals
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
$endgroup$
$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45
add a comment |
$begingroup$
Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$
The answer to this question is $frac{4}{3}$ in a workbook.
How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$
But I can't simplify $sqrt{2+sqrt3}$ like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?
algebra-precalculus radicals nested-radicals
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
$endgroup$
Simplify $left(frac{2(sqrt2 + sqrt6)}{3(sqrt{2+sqrt3}}right)$
The answer to this question is $frac{4}{3}$ in a workbook.
How would I simplify $sqrt{2+sqrt3}$ $?$ If it was something like $sqrt{3 + 2sqrt2}$ , I would have simplified it as follows:
$sqrt{3 + 2sqrt2}$
$=$ $sqrt{(sqrt2)^2 + 2(sqrt2)(1) + (1)^2}$
$=$
$sqrt{(sqrt2 + 1)^2}$
$=$
$sqrt2 + 1$
But I can't simplify $sqrt{2+sqrt3}$ like that as $2+sqrt3$ is can't be written as squares of two numbers. Is there any other method?
algebra-precalculus radicals nested-radicals
algebra-precalculus radicals nested-radicals
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
edited Mar 17 at 9:14
José Carlos Santos
171k23132240
171k23132240
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
asked Mar 17 at 7:30
Sanjay SinghSanjay Singh
32
32
$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45
add a comment |
$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45
$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45
$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Hints:
$$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$
$$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$
answered Mar 17 at 7:40
learnerlearner
1,110417
$endgroup$
add a comment |
$begingroup$
$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
Note that$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)^2=frac{4left(8+4sqrt3right)}{9left(2+sqrt3right)}=frac{16}9=left(frac43right)^2.$$
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 17 at 7:35
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
Hints:
$$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$
$$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$
answered Mar 17 at 7:40
learnerlearner
1,110417
$endgroup$
add a comment |
$begingroup$
Hints:
$$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$
$$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$
answered Mar 17 at 7:40
learnerlearner
1,110417
$endgroup$
add a comment |
$begingroup$
Hints:
$$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$
$$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$
answered Mar 17 at 7:40
learnerlearner
1,110417
$endgroup$
Hints:
$$sqrt 2+sqrt 6=sqrt 2(1+sqrt 3)=sqrt{2(1+sqrt 3)^2}=sqrt{2(4+2sqrt 3)}=2sqrt{2+sqrt 3}$$
$$sqrt{2+sqrt 3}=frac 1{sqrt 2}sqrt{4+2sqrt 3}=frac 1{sqrt 2}sqrt{(sqrt 3+1)^2}=frac{sqrt 3+1}{sqrt 2}=frac{sqrt 6+sqrt 2}2$$
answered Mar 17 at 7:40
learnerlearner
1,110417
answered Mar 17 at 7:40
learnerlearner
1,110417
answered Mar 17 at 7:40
learnerlearner
1,110417
answered Mar 17 at 7:40
learnerlearner
1,110417
1,110417
add a comment |
add a comment |
$begingroup$
$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
$endgroup$
add a comment |
$begingroup$
$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
$endgroup$
add a comment |
$begingroup$
$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
$endgroup$
$$left(frac{2left(sqrt2+sqrt6right)}{3sqrt{2+sqrt3}}right)= frac{(2√2+2√6)sqrt{2+sqrt{3}}}{3(2+√3)}=frac{8+4√3}{2+√3}=frac{4}{3}.$$
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
answered Mar 17 at 8:11
saket kumarsaket kumar
168113
168113
add a comment |
add a comment |
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$begingroup$
See the latter half of this answer.
$endgroup$
– Theo Bendit
Mar 17 at 7:45