Computing $sqrt[4]{28+16 sqrt 3}$
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I want to compute following radical
$$sqrt[4]{28+16 sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
We know that $ 28 = 2 cdot 7^{frac{1}{2}}$
$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
radicals
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
$endgroup$
add a comment |
$begingroup$
I want to compute following radical
$$sqrt[4]{28+16 sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
We know that $ 28 = 2 cdot 7^{frac{1}{2}}$
$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
radicals
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
$endgroup$
$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
3
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21
add a comment |
$begingroup$
I want to compute following radical
$$sqrt[4]{28+16 sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
We know that $ 28 = 2 cdot 7^{frac{1}{2}}$
$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
radicals
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
$endgroup$
I want to compute following radical
$$sqrt[4]{28+16 sqrt 3}$$
For that, I first tried to rewrite this in terms of exponential.
$$(28+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
We know that $ 28 = 2 cdot 7^{frac{1}{2}}$
$$(2 cdot 7^{frac{1}{2}}+16cdot 3^{frac{1}{2}})^{frac{1}{4}}$$
However, I'm stuck at this step. Could you assist me?
Regards
radicals
radicals
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
asked Dec 22 '18 at 9:52
EnzoEnzo
1537
1537
$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
3
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21
add a comment |
$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
3
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21
$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
3
3
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21
add a comment |
6 Answers
6
active
oldest
votes
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$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
$endgroup$
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Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
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@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
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– Shubham Johri
Dec 22 '18 at 10:11
add a comment |
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Hint.
$28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$
$4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
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add a comment |
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Hint: Compute $$(1+sqrt{3})^4$$
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
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add a comment |
$begingroup$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.
$$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
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add a comment |
$begingroup$
Hint:
Try to write
$$28+16 sqrt 3=(a+bsqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
begin{align}
(a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
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add a comment |
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Using a calculator, you can compute $,sqrt[4]{28+16 sqrt 3} = 2.732050807...,$ which is easily recognizable as $,1+sqrt{3}.$ This is the easiest method, but,
in general, it is very much harder to simplify radicals.
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
$endgroup$
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
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@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
add a comment |
$begingroup$
$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
$endgroup$
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
$begingroup$
@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
add a comment |
$begingroup$
$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
$endgroup$
$$sqrt[4] {28+16sqrt 3}=sqrt[4] {(sqrt {12})^2 +(sqrt {16})^2 +2sqrt {16cdot 12}}=sqrt {4+2sqrt 3}=sqrt {(sqrt 3)^2 +(1)^2 +2sqrt {3cdot 1}}=sqrt 3 +1$$
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
answered Dec 22 '18 at 10:00
DigammaDigamma
6,1671440
6,1671440
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
$begingroup$
@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
add a comment |
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
$begingroup$
@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
$begingroup$
Is it a rule or something?
$endgroup$
– Enzo
Dec 22 '18 at 10:05
$begingroup$
@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
$begingroup$
@Enzo Since you are taking the fourth root, you would like to express $28+16sqrt3$ as the fourth power of some number, or the square of the square of some number. In other words, you want to find $a,b$ such that $(a+b)^2=a^2+b^2+2ab=28+16sqrt3$, and then $c,d$ such that $(c+d)^2=a+b$
$endgroup$
– Shubham Johri
Dec 22 '18 at 10:11
add a comment |
$begingroup$
Hint.
$28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$
$4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
$endgroup$
add a comment |
$begingroup$
Hint.
$28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$
$4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
$endgroup$
add a comment |
$begingroup$
Hint.
$28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$
$4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
$endgroup$
Hint.
$28+16sqrt3=12+16+2cdot2sqrt3cdot4=(2sqrt3)^2+4^2+2cdot2sqrt3cdot4=(4+2sqrt3)^2$
$4+2sqrt3=3+1+2cdot1cdotsqrt3=(sqrt3)^2+1^2+2cdot1cdotsqrt3=(sqrt3+1)^2$
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
answered Dec 22 '18 at 10:01
Shubham JohriShubham Johri
4,676717
4,676717
add a comment |
add a comment |
$begingroup$
Hint: Compute $$(1+sqrt{3})^4$$
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hint: Compute $$(1+sqrt{3})^4$$
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Hint: Compute $$(1+sqrt{3})^4$$
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Hint: Compute $$(1+sqrt{3})^4$$
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Dec 22 '18 at 9:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
add a comment |
add a comment |
$begingroup$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.
$$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.
$$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
$endgroup$
add a comment |
$begingroup$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.
$$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
$endgroup$
In general, there's not much you can do with an expression like that. But if you know that the answer is in the form $a+bsqrt 3$ for integers $a$ and $b$, then you can solve it like this:
Step 1
($a+bsqrt 3)^4=28+16sqrt 3$. So suppose ($a+bsqrt 3)^2=c+dsqrt 3$. Then $(c+dsqrt 3)^2=28+16sqrt 3$, i.e.
$$(c^2+3d^2)+2cdsqrt 3=28+16sqrt 3$$
$c^2+3d^2$ and $2cd$ are integers, so this can only be true if $c^2+3d^2=28$ and $2cd=16$.
Then $d=8/c$, so $c^2+dfrac{192}{c^2}=28$. Multiplying by $c^2$ and rearranging, we get a quadratic in $c^2$:
$$c^4-28c^2+192=0$$
This factors as $(c^2-12)(c^2-16)=0$. Since $c$ is an integer, we must have $c=pm4$, and so $d=pm2$. So we have two possible solutions: $pm(4+2sqrt 3)$. (You should try squaring this to check that you get the expected result $28+16sqrt 3$). Pick the positive solution $4+2sqrt 3$ (if you choose the negative solution, you will find that Step 2 fails).
Step 2
Now you need to solve $(a+bsqrt 3)^2=4+2sqrt 3$. You should be able to do this yourself now, using exactly the same method as Step 1.
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
answered Dec 22 '18 at 10:23
TonyKTonyK
42.1k355134
42.1k355134
add a comment |
add a comment |
$begingroup$
Hint:
Try to write
$$28+16 sqrt 3=(a+bsqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
begin{align}
(a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Hint:
Try to write
$$28+16 sqrt 3=(a+bsqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
begin{align}
(a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
$endgroup$
add a comment |
$begingroup$
Hint:
Try to write
$$28+16 sqrt 3=(a+bsqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
begin{align}
(a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
$endgroup$
Hint:
Try to write
$$28+16 sqrt 3=(a+bsqrt 3)^4$$
for suitable $a$ and $b$.
You obtain, reordering the terms
begin{align}
(a+bsqrt 3)^4&=(a^4+18a^2b^2+9b^4)+4ab(a^2+3b^2)sqrt 3.\
end{align}
Can you choose $a$ and $b$ so that
$$a^4+18a^2b^2+9b^4=28,quad ab(a^2+3b^2)=4?$$
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
edited Dec 22 '18 at 10:33
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
answered Dec 22 '18 at 10:23
BernardBernard
119k639112
119k639112
add a comment |
add a comment |
$begingroup$
Using a calculator, you can compute $,sqrt[4]{28+16 sqrt 3} = 2.732050807...,$ which is easily recognizable as $,1+sqrt{3}.$ This is the easiest method, but,
in general, it is very much harder to simplify radicals.
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
$endgroup$
add a comment |
$begingroup$
Using a calculator, you can compute $,sqrt[4]{28+16 sqrt 3} = 2.732050807...,$ which is easily recognizable as $,1+sqrt{3}.$ This is the easiest method, but,
in general, it is very much harder to simplify radicals.
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
$endgroup$
add a comment |
$begingroup$
Using a calculator, you can compute $,sqrt[4]{28+16 sqrt 3} = 2.732050807...,$ which is easily recognizable as $,1+sqrt{3}.$ This is the easiest method, but,
in general, it is very much harder to simplify radicals.
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
$endgroup$
Using a calculator, you can compute $,sqrt[4]{28+16 sqrt 3} = 2.732050807...,$ which is easily recognizable as $,1+sqrt{3}.$ This is the easiest method, but,
in general, it is very much harder to simplify radicals.
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
edited Dec 30 '18 at 2:13
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
answered Dec 22 '18 at 10:21
SomosSomos
13.1k11034
13.1k11034
add a comment |
add a comment |
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$begingroup$
Is there an exponential approach to this question?
$endgroup$
– Enzo
Dec 22 '18 at 10:08
3
$begingroup$
The statment $28 = 2 cdot 7^{frac{1}{2}}$ is false. It should be $28=2^2cdot 7$ or $28^{frac12} = 2cdot 7^{frac{1}{2}}$.
$endgroup$
– Somos
Dec 22 '18 at 10:12
$begingroup$
Read about completing a square
$endgroup$
– farruhota
Dec 30 '18 at 2:21