Mixing
Simplify $sqrt{dfrac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$ into $dfrac{sqrt{3}}{3}$
$begingroup$
I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
The solution is $displaystyle frac{sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $frac{sqrt[3]{64}}{sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $dfrac{sqrt{3}}{3}$
algebra-precalculus
edited Jan 18 at 7:01
El borito
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
$endgroup$
add a comment |
$begingroup$
I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
The solution is $displaystyle frac{sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $frac{sqrt[3]{64}}{sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $dfrac{sqrt{3}}{3}$
algebra-precalculus
edited Jan 18 at 7:01
El borito
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
$endgroup$
2
$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02
add a comment |
$begingroup$
I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
The solution is $displaystyle frac{sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $frac{sqrt[3]{64}}{sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $dfrac{sqrt{3}}{3}$
algebra-precalculus
edited Jan 18 at 7:01
El borito
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
$endgroup$
I am on the final question of a textbook chapter on radicals and this question feels more challenging, perhaps that's the idea. If you view my post history I typically make a effort to provide some working to simplify the expression to an extent, but here I am very confused about where to go or my first steps.
I am to simplify:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
The solution is $displaystyle frac{sqrt{3}}{3}$
The expression makes me "feel" like there is a rule when dividing radicals with the same radicand but different index' with different index'. Is that true? In this case, how to divide $frac{sqrt[3]{64}}{sqrt{64}}$? I know that the 3rd root and sq roots are 4 and 8 which would leave me with 1/2. using a calculator I can see that the 4th root of $256$ is 4 but I think I'm to arrive at the solution without a calculator.
Is there a prescribed approach or order of operations to simplifying an expression like this?
How can I arrive at $dfrac{sqrt{3}}{3}$
algebra-precalculus
algebra-precalculus
edited Jan 18 at 7:01
El borito
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
edited Jan 18 at 7:01
El borito
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
edited Jan 18 at 7:01
El borito
666216
edited Jan 18 at 7:01
El borito
666216
edited Jan 18 at 7:01
El borito
666216
666216
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
asked Jan 9 at 1:58
Doug FirDoug Fir
3828
3828
2
$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02
add a comment |
2
$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02
2
2
$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02
$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Note that $$sqrt[3]{64}=(64)^{frac13}=(4^3)^frac13=4$$
$$sqrt[4]{256}=(256)^{frac14}=(4^4)^{frac14}=4$$
$$sqrt{64}=8$$
$$sqrt{256}=16$$
So, we get $$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}=sqrt{dfrac{4+4}{8+16}}=sqrt{dfrac{8}{24}}=sqrt{dfrac{1}{3}}=dfrac{sqrt{3}}{3}$$
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
add a comment |
$begingroup$
You have:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
Then:
$$sqrt{frac{sqrt[3]{8^2} + sqrt[4]{16^2}}{sqrt{8^2}+sqrt{16^2}}}$$
$$sqrt{frac{sqrt[3]{2^6} + sqrt[4]{4^4}}{8 + 16}}$$
$$sqrt{frac{2^2 + 4}{24}}$$
$$sqrt{frac{8}{8 cdot 3}}$$
$$frac{1}{sqrt{3}}$$
$$frac{sqrt{3}}{sqrt{3}sqrt{3}}$$
$$frac{sqrt{3}}{3}$$
And this is the answer.
answered Jan 9 at 2:08
El boritoEl borito
666216
$endgroup$
add a comment |
$begingroup$
We know that,
$64=8×8$ or $sqrt{64}=8$
and
$256=16×16$ or $sqrt{256}=16$
Also, $64^frac{1}{3}$ and $256^frac{1}{4}$
Therefore, $64^frac{1}{3}=(8×8)^frac{1}{3} =(8^frac{1}{3})×(8^frac{1}{3})=2×2=4$
Similarly, $256^frac{1}{4}=(16×16)^frac{1}{4}=(16^frac{1}{4})×(16^frac{1}{4})=2×2=4$
Now, your expression reduces to
=$sqrt{frac{4+4}{8+16}}$
=$sqrt{frac{8}{24}}$
=$sqrt{frac{1}{3}}$
On rationalization,
$=frac{1}{sqrt{3}}×frac{sqrt{3}} {sqrt{3}}$
Hence, $frac{sqrt{3}}{3}$
Thanks.
$endgroup$
add a comment |
$begingroup$
$ 256 $ is a perfect square because $ 16 cdot 16 = 16^2 = 256 $. Thus $ sqrt{256} = 16 $. The expression $$ sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}} $$ evaluates to $$ sqrt{frac{4 + 4}{8 + 16}} $$ which simplifies to $$ sqrt{frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ sqrt{frac{8/8}{24/8}} = sqrt{frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ sqrt{frac{a}{b}} = frac{sqrt{a}}{sqrt{b}} $. Using this property,
$$ sqrt{frac{1}{3}} = frac{sqrt{1}}{sqrt{3}} = frac{1}{sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ sqrt{3} $ to achieve the desired result.
$$ frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} $$
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
$endgroup$
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $$sqrt[3]{64}=(64)^{frac13}=(4^3)^frac13=4$$
$$sqrt[4]{256}=(256)^{frac14}=(4^4)^{frac14}=4$$
$$sqrt{64}=8$$
$$sqrt{256}=16$$
So, we get $$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}=sqrt{dfrac{4+4}{8+16}}=sqrt{dfrac{8}{24}}=sqrt{dfrac{1}{3}}=dfrac{sqrt{3}}{3}$$
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
add a comment |
$begingroup$
Note that $$sqrt[3]{64}=(64)^{frac13}=(4^3)^frac13=4$$
$$sqrt[4]{256}=(256)^{frac14}=(4^4)^{frac14}=4$$
$$sqrt{64}=8$$
$$sqrt{256}=16$$
So, we get $$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}=sqrt{dfrac{4+4}{8+16}}=sqrt{dfrac{8}{24}}=sqrt{dfrac{1}{3}}=dfrac{sqrt{3}}{3}$$
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
$endgroup$
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
add a comment |
$begingroup$
Note that $$sqrt[3]{64}=(64)^{frac13}=(4^3)^frac13=4$$
$$sqrt[4]{256}=(256)^{frac14}=(4^4)^{frac14}=4$$
$$sqrt{64}=8$$
$$sqrt{256}=16$$
So, we get $$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}=sqrt{dfrac{4+4}{8+16}}=sqrt{dfrac{8}{24}}=sqrt{dfrac{1}{3}}=dfrac{sqrt{3}}{3}$$
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
$endgroup$
Note that $$sqrt[3]{64}=(64)^{frac13}=(4^3)^frac13=4$$
$$sqrt[4]{256}=(256)^{frac14}=(4^4)^{frac14}=4$$
$$sqrt{64}=8$$
$$sqrt{256}=16$$
So, we get $$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}=sqrt{dfrac{4+4}{8+16}}=sqrt{dfrac{8}{24}}=sqrt{dfrac{1}{3}}=dfrac{sqrt{3}}{3}$$
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
answered Jan 9 at 2:04
Key FlexKey Flex
8,28261233
8,28261233
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
add a comment |
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
$begingroup$
Your last step that goes from $sqrt{frac{1}{3}}$ to $frac{sqrt{3}}{3}$, how did you do that? What's the rule there?
$endgroup$
– Doug Fir
Jan 9 at 2:09
2
2
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
$begingroup$
@DougFir $dfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{sqrt{3}}cdotdfrac{1}{sqrt{3}}=dfrac{sqrt{3}}{3}$. I just multiplied the final answer with $dfrac{sqrt{3}}{sqrt{3}}$
$endgroup$
– Key Flex
Jan 9 at 2:11
add a comment |
$begingroup$
You have:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
Then:
$$sqrt{frac{sqrt[3]{8^2} + sqrt[4]{16^2}}{sqrt{8^2}+sqrt{16^2}}}$$
$$sqrt{frac{sqrt[3]{2^6} + sqrt[4]{4^4}}{8 + 16}}$$
$$sqrt{frac{2^2 + 4}{24}}$$
$$sqrt{frac{8}{8 cdot 3}}$$
$$frac{1}{sqrt{3}}$$
$$frac{sqrt{3}}{sqrt{3}sqrt{3}}$$
$$frac{sqrt{3}}{3}$$
And this is the answer.
answered Jan 9 at 2:08
El boritoEl borito
666216
$endgroup$
add a comment |
$begingroup$
You have:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
Then:
$$sqrt{frac{sqrt[3]{8^2} + sqrt[4]{16^2}}{sqrt{8^2}+sqrt{16^2}}}$$
$$sqrt{frac{sqrt[3]{2^6} + sqrt[4]{4^4}}{8 + 16}}$$
$$sqrt{frac{2^2 + 4}{24}}$$
$$sqrt{frac{8}{8 cdot 3}}$$
$$frac{1}{sqrt{3}}$$
$$frac{sqrt{3}}{sqrt{3}sqrt{3}}$$
$$frac{sqrt{3}}{3}$$
And this is the answer.
answered Jan 9 at 2:08
El boritoEl borito
666216
$endgroup$
add a comment |
$begingroup$
You have:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
Then:
$$sqrt{frac{sqrt[3]{8^2} + sqrt[4]{16^2}}{sqrt{8^2}+sqrt{16^2}}}$$
$$sqrt{frac{sqrt[3]{2^6} + sqrt[4]{4^4}}{8 + 16}}$$
$$sqrt{frac{2^2 + 4}{24}}$$
$$sqrt{frac{8}{8 cdot 3}}$$
$$frac{1}{sqrt{3}}$$
$$frac{sqrt{3}}{sqrt{3}sqrt{3}}$$
$$frac{sqrt{3}}{3}$$
And this is the answer.
answered Jan 9 at 2:08
El boritoEl borito
666216
$endgroup$
You have:
$$sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}}$$
Then:
$$sqrt{frac{sqrt[3]{8^2} + sqrt[4]{16^2}}{sqrt{8^2}+sqrt{16^2}}}$$
$$sqrt{frac{sqrt[3]{2^6} + sqrt[4]{4^4}}{8 + 16}}$$
$$sqrt{frac{2^2 + 4}{24}}$$
$$sqrt{frac{8}{8 cdot 3}}$$
$$frac{1}{sqrt{3}}$$
$$frac{sqrt{3}}{sqrt{3}sqrt{3}}$$
$$frac{sqrt{3}}{3}$$
And this is the answer.
answered Jan 9 at 2:08
El boritoEl borito
666216
answered Jan 9 at 2:08
El boritoEl borito
666216
answered Jan 9 at 2:08
El boritoEl borito
666216
answered Jan 9 at 2:08
El boritoEl borito
666216
666216
add a comment |
add a comment |
$begingroup$
We know that,
$64=8×8$ or $sqrt{64}=8$
and
$256=16×16$ or $sqrt{256}=16$
Also, $64^frac{1}{3}$ and $256^frac{1}{4}$
Therefore, $64^frac{1}{3}=(8×8)^frac{1}{3} =(8^frac{1}{3})×(8^frac{1}{3})=2×2=4$
Similarly, $256^frac{1}{4}=(16×16)^frac{1}{4}=(16^frac{1}{4})×(16^frac{1}{4})=2×2=4$
Now, your expression reduces to
=$sqrt{frac{4+4}{8+16}}$
=$sqrt{frac{8}{24}}$
=$sqrt{frac{1}{3}}$
On rationalization,
$=frac{1}{sqrt{3}}×frac{sqrt{3}} {sqrt{3}}$
Hence, $frac{sqrt{3}}{3}$
Thanks.
$endgroup$
add a comment |
$begingroup$
We know that,
$64=8×8$ or $sqrt{64}=8$
and
$256=16×16$ or $sqrt{256}=16$
Also, $64^frac{1}{3}$ and $256^frac{1}{4}$
Therefore, $64^frac{1}{3}=(8×8)^frac{1}{3} =(8^frac{1}{3})×(8^frac{1}{3})=2×2=4$
Similarly, $256^frac{1}{4}=(16×16)^frac{1}{4}=(16^frac{1}{4})×(16^frac{1}{4})=2×2=4$
Now, your expression reduces to
=$sqrt{frac{4+4}{8+16}}$
=$sqrt{frac{8}{24}}$
=$sqrt{frac{1}{3}}$
On rationalization,
$=frac{1}{sqrt{3}}×frac{sqrt{3}} {sqrt{3}}$
Hence, $frac{sqrt{3}}{3}$
Thanks.
$endgroup$
add a comment |
$begingroup$
We know that,
$64=8×8$ or $sqrt{64}=8$
and
$256=16×16$ or $sqrt{256}=16$
Also, $64^frac{1}{3}$ and $256^frac{1}{4}$
Therefore, $64^frac{1}{3}=(8×8)^frac{1}{3} =(8^frac{1}{3})×(8^frac{1}{3})=2×2=4$
Similarly, $256^frac{1}{4}=(16×16)^frac{1}{4}=(16^frac{1}{4})×(16^frac{1}{4})=2×2=4$
Now, your expression reduces to
=$sqrt{frac{4+4}{8+16}}$
=$sqrt{frac{8}{24}}$
=$sqrt{frac{1}{3}}$
On rationalization,
$=frac{1}{sqrt{3}}×frac{sqrt{3}} {sqrt{3}}$
Hence, $frac{sqrt{3}}{3}$
Thanks.
$endgroup$
We know that,
$64=8×8$ or $sqrt{64}=8$
and
$256=16×16$ or $sqrt{256}=16$
Also, $64^frac{1}{3}$ and $256^frac{1}{4}$
Therefore, $64^frac{1}{3}=(8×8)^frac{1}{3} =(8^frac{1}{3})×(8^frac{1}{3})=2×2=4$
Similarly, $256^frac{1}{4}=(16×16)^frac{1}{4}=(16^frac{1}{4})×(16^frac{1}{4})=2×2=4$
Now, your expression reduces to
=$sqrt{frac{4+4}{8+16}}$
=$sqrt{frac{8}{24}}$
=$sqrt{frac{1}{3}}$
On rationalization,
$=frac{1}{sqrt{3}}×frac{sqrt{3}} {sqrt{3}}$
Hence, $frac{sqrt{3}}{3}$
Thanks.
answered Jan 9 at 5:11
user629353
add a comment |
add a comment |
$begingroup$
$ 256 $ is a perfect square because $ 16 cdot 16 = 16^2 = 256 $. Thus $ sqrt{256} = 16 $. The expression $$ sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}} $$ evaluates to $$ sqrt{frac{4 + 4}{8 + 16}} $$ which simplifies to $$ sqrt{frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ sqrt{frac{8/8}{24/8}} = sqrt{frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ sqrt{frac{a}{b}} = frac{sqrt{a}}{sqrt{b}} $. Using this property,
$$ sqrt{frac{1}{3}} = frac{sqrt{1}}{sqrt{3}} = frac{1}{sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ sqrt{3} $ to achieve the desired result.
$$ frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} $$
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
$endgroup$
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
add a comment |
$begingroup$
$ 256 $ is a perfect square because $ 16 cdot 16 = 16^2 = 256 $. Thus $ sqrt{256} = 16 $. The expression $$ sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}} $$ evaluates to $$ sqrt{frac{4 + 4}{8 + 16}} $$ which simplifies to $$ sqrt{frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ sqrt{frac{8/8}{24/8}} = sqrt{frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ sqrt{frac{a}{b}} = frac{sqrt{a}}{sqrt{b}} $. Using this property,
$$ sqrt{frac{1}{3}} = frac{sqrt{1}}{sqrt{3}} = frac{1}{sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ sqrt{3} $ to achieve the desired result.
$$ frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} $$
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
$endgroup$
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
add a comment |
$begingroup$
$ 256 $ is a perfect square because $ 16 cdot 16 = 16^2 = 256 $. Thus $ sqrt{256} = 16 $. The expression $$ sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}} $$ evaluates to $$ sqrt{frac{4 + 4}{8 + 16}} $$ which simplifies to $$ sqrt{frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ sqrt{frac{8/8}{24/8}} = sqrt{frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ sqrt{frac{a}{b}} = frac{sqrt{a}}{sqrt{b}} $. Using this property,
$$ sqrt{frac{1}{3}} = frac{sqrt{1}}{sqrt{3}} = frac{1}{sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ sqrt{3} $ to achieve the desired result.
$$ frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} $$
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
$endgroup$
$ 256 $ is a perfect square because $ 16 cdot 16 = 16^2 = 256 $. Thus $ sqrt{256} = 16 $. The expression $$ sqrt{frac{sqrt[3]{64} + sqrt[4]{256}}{sqrt{64}+sqrt{256}}} $$ evaluates to $$ sqrt{frac{4 + 4}{8 + 16}} $$ which simplifies to $$ sqrt{frac{8}{24}} $$ Reduce the fraction inside the radical by dividing both the numerator and denominator by the greatest common divisor that divides both $ 8 $ and $ 24 $, which is $ 8 $. $$ sqrt{frac{8/8}{24/8}} = sqrt{frac{1}{3}} $$
Now from one of the basic properties involving radicals, $ sqrt{frac{a}{b}} = frac{sqrt{a}}{sqrt{b}} $. Using this property,
$$ sqrt{frac{1}{3}} = frac{sqrt{1}}{sqrt{3}} = frac{1}{sqrt{3}} $$
"Rationalize" the denominator by multiplying the numerator and denominator by $ sqrt{3} $ to achieve the desired result.
$$ frac{1 cdot sqrt{3}}{sqrt{3} cdot sqrt{3}} = frac{sqrt{3}}{3} $$
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
answered Jan 9 at 17:14
Marvin CohenMarvin Cohen
136117
136117
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
add a comment |
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
1
1
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
$begingroup$
Thanks Marvin, this is helpful
$endgroup$
– Doug Fir
Jan 9 at 18:45
add a comment |
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$begingroup$
As you said, you can just simplify all the radicals as $sqrt{64} = 8, sqrt{256} = 16, sqrt[3]{64}=4, sqrt[4]{256} = 4$ and put these back into the original equation. You may use that $64 = 2^{6}$ and $256 = 2^{8}$.
$endgroup$
– Seewoo Lee
Jan 9 at 2:02