Simplify $frac{sqrt{24}}{8}$
$begingroup$
The question I have been given is to simplify as much as possible: $frac{sqrt {24}}{8}$. I know the answer is $frac{sqrt 6}{4}$
(Note I am in a beginner math course, so go easy on me.)
My first thought was to divide to get: $frac{sqrt {12}}{4}$ and then again: $frac{sqrt 3}{1}$. However, I realized this was wrong. So I tried $frac{sqrt {24}}{2timessqrt 4}$ which would make the denominator equal to 4, which is right. So I thought I could do that to the top too, but I couldn't. I feel like I'm on the right track but not really there. Can someone help me figure out how to solve questions like these?
radicals fractions square-numbers
edited Jan 25 at 20:53
Marvin Cohen
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
$endgroup$
add a comment |
$begingroup$
The question I have been given is to simplify as much as possible: $frac{sqrt {24}}{8}$. I know the answer is $frac{sqrt 6}{4}$
(Note I am in a beginner math course, so go easy on me.)
My first thought was to divide to get: $frac{sqrt {12}}{4}$ and then again: $frac{sqrt 3}{1}$. However, I realized this was wrong. So I tried $frac{sqrt {24}}{2timessqrt 4}$ which would make the denominator equal to 4, which is right. So I thought I could do that to the top too, but I couldn't. I feel like I'm on the right track but not really there. Can someone help me figure out how to solve questions like these?
radicals fractions square-numbers
edited Jan 25 at 20:53
Marvin Cohen
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
$endgroup$
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
2
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14
add a comment |
$begingroup$
The question I have been given is to simplify as much as possible: $frac{sqrt {24}}{8}$. I know the answer is $frac{sqrt 6}{4}$
(Note I am in a beginner math course, so go easy on me.)
My first thought was to divide to get: $frac{sqrt {12}}{4}$ and then again: $frac{sqrt 3}{1}$. However, I realized this was wrong. So I tried $frac{sqrt {24}}{2timessqrt 4}$ which would make the denominator equal to 4, which is right. So I thought I could do that to the top too, but I couldn't. I feel like I'm on the right track but not really there. Can someone help me figure out how to solve questions like these?
radicals fractions square-numbers
edited Jan 25 at 20:53
Marvin Cohen
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
$endgroup$
The question I have been given is to simplify as much as possible: $frac{sqrt {24}}{8}$. I know the answer is $frac{sqrt 6}{4}$
(Note I am in a beginner math course, so go easy on me.)
My first thought was to divide to get: $frac{sqrt {12}}{4}$ and then again: $frac{sqrt 3}{1}$. However, I realized this was wrong. So I tried $frac{sqrt {24}}{2timessqrt 4}$ which would make the denominator equal to 4, which is right. So I thought I could do that to the top too, but I couldn't. I feel like I'm on the right track but not really there. Can someone help me figure out how to solve questions like these?
radicals fractions square-numbers
radicals fractions square-numbers
edited Jan 25 at 20:53
Marvin Cohen
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
edited Jan 25 at 20:53
Marvin Cohen
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
edited Jan 25 at 20:53
Marvin Cohen
160117
edited Jan 25 at 20:53
Marvin Cohen
160117
edited Jan 25 at 20:53
Marvin Cohen
160117
160117
asked Sep 11 '17 at 8:04
NiccoNicco
243
asked Sep 11 '17 at 8:04
NiccoNicco
243
asked Sep 11 '17 at 8:04
NiccoNicco
243
243
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
2
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14
add a comment |
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
2
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
2
2
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You should first simplify the square root part as much as possible, then cancel the fraction.
$$sqrt{24}=sqrt{4times 6}=sqrt{4}timessqrt{6}=2sqrt{6}$$
Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).
Now you can simplify the fraction:
$$frac{sqrt{24}}{8}=frac{2timessqrt{6}}{8}=frac{1timessqrt{6}}4=frac{sqrt{6}}4.$$
when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
add a comment |
$begingroup$
You could square the fraction, then simplify, then take the square root.
$$frac{sqrt{24}^2}{8^2}=frac{24}{64}=frac{6}{16}$$
Taking the square root of both sides leaves the desired answer. $$frac{sqrt{6}}{sqrt{16}}=frac{sqrt{6}}{4}$$
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
$endgroup$
add a comment |
$begingroup$
You can see $sqrt{24} = sqrt{4 cdot 6}$. Since $sqrt{acdot b} = sqrt{a} cdot sqrt{b}$, you can expand the nominator to result in $2cdot sqrt{6}$. You can then divide by the $2$ to get the desired result.
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
$endgroup$
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
add a comment |
$begingroup$
How would you go about simplifying $$frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$frac{24}{8} = frac{12}{4} = frac{6}{2} = frac{3}{1} = 3.$$
Believe it or not, $$frac{sqrt{24}}{2} = sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.
So, by the same process as the other one, $$frac{sqrt{24}}{8} = frac{sqrt{6}}{4}.$$
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
$endgroup$
add a comment |
$begingroup$
The radicand, $ 24 $, can be expressed as a product involving a perfect square and a non-perfect square. In this case $ 24 $ is equal to $ 4 $ (which is a perfect square) times $ 6 $ (a non-perfect square). Thus
$$ sqrt{24} = sqrt{4 times 6} = sqrt{4}sqrt{6} = 2sqrt{6} $$ and $$ frac{sqrt{24}}{8} = frac{2sqrt{6}}{8} $$
After dividing both the numerator and denominator by the common factor of $ 2 $ and $ 8 $, which is $ 2 $:
$$ frac{2sqrt{6} / 2}{8/2} = frac{sqrt{6}}{4} $$
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2424975%2fsimplify-frac-sqrt248%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You should first simplify the square root part as much as possible, then cancel the fraction.
$$sqrt{24}=sqrt{4times 6}=sqrt{4}timessqrt{6}=2sqrt{6}$$
Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).
Now you can simplify the fraction:
$$frac{sqrt{24}}{8}=frac{2timessqrt{6}}{8}=frac{1timessqrt{6}}4=frac{sqrt{6}}4.$$
when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
add a comment |
$begingroup$
You should first simplify the square root part as much as possible, then cancel the fraction.
$$sqrt{24}=sqrt{4times 6}=sqrt{4}timessqrt{6}=2sqrt{6}$$
Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).
Now you can simplify the fraction:
$$frac{sqrt{24}}{8}=frac{2timessqrt{6}}{8}=frac{1timessqrt{6}}4=frac{sqrt{6}}4.$$
when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
$endgroup$
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
add a comment |
$begingroup$
You should first simplify the square root part as much as possible, then cancel the fraction.
$$sqrt{24}=sqrt{4times 6}=sqrt{4}timessqrt{6}=2sqrt{6}$$
Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).
Now you can simplify the fraction:
$$frac{sqrt{24}}{8}=frac{2timessqrt{6}}{8}=frac{1timessqrt{6}}4=frac{sqrt{6}}4.$$
when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
$endgroup$
You should first simplify the square root part as much as possible, then cancel the fraction.
$$sqrt{24}=sqrt{4times 6}=sqrt{4}timessqrt{6}=2sqrt{6}$$
Here you should always be trying to make one of the factors the largest square number that divides the number you are taking the square root of (here $4$ is the largest square that divides $24$).
Now you can simplify the fraction:
$$frac{sqrt{24}}{8}=frac{2timessqrt{6}}{8}=frac{1timessqrt{6}}4=frac{sqrt{6}}4.$$
when doing this you can only cancel numbers from the top and bottom if they are both outside the square root (or both inside).
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
answered Sep 11 '17 at 8:11
Especially LimeEspecially Lime
22.3k22858
22.3k22858
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
add a comment |
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
$begingroup$
Great explanation, thank you! :)
$endgroup$
– Nicco
Sep 11 '17 at 8:14
add a comment |
$begingroup$
You could square the fraction, then simplify, then take the square root.
$$frac{sqrt{24}^2}{8^2}=frac{24}{64}=frac{6}{16}$$
Taking the square root of both sides leaves the desired answer. $$frac{sqrt{6}}{sqrt{16}}=frac{sqrt{6}}{4}$$
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
$endgroup$
add a comment |
$begingroup$
You could square the fraction, then simplify, then take the square root.
$$frac{sqrt{24}^2}{8^2}=frac{24}{64}=frac{6}{16}$$
Taking the square root of both sides leaves the desired answer. $$frac{sqrt{6}}{sqrt{16}}=frac{sqrt{6}}{4}$$
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
$endgroup$
add a comment |
$begingroup$
You could square the fraction, then simplify, then take the square root.
$$frac{sqrt{24}^2}{8^2}=frac{24}{64}=frac{6}{16}$$
Taking the square root of both sides leaves the desired answer. $$frac{sqrt{6}}{sqrt{16}}=frac{sqrt{6}}{4}$$
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
$endgroup$
You could square the fraction, then simplify, then take the square root.
$$frac{sqrt{24}^2}{8^2}=frac{24}{64}=frac{6}{16}$$
Taking the square root of both sides leaves the desired answer. $$frac{sqrt{6}}{sqrt{16}}=frac{sqrt{6}}{4}$$
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
answered Sep 11 '17 at 8:13
JazzachiJazzachi
8971519
8971519
add a comment |
add a comment |
$begingroup$
You can see $sqrt{24} = sqrt{4 cdot 6}$. Since $sqrt{acdot b} = sqrt{a} cdot sqrt{b}$, you can expand the nominator to result in $2cdot sqrt{6}$. You can then divide by the $2$ to get the desired result.
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
$endgroup$
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
add a comment |
$begingroup$
You can see $sqrt{24} = sqrt{4 cdot 6}$. Since $sqrt{acdot b} = sqrt{a} cdot sqrt{b}$, you can expand the nominator to result in $2cdot sqrt{6}$. You can then divide by the $2$ to get the desired result.
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
$endgroup$
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
add a comment |
$begingroup$
You can see $sqrt{24} = sqrt{4 cdot 6}$. Since $sqrt{acdot b} = sqrt{a} cdot sqrt{b}$, you can expand the nominator to result in $2cdot sqrt{6}$. You can then divide by the $2$ to get the desired result.
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
$endgroup$
You can see $sqrt{24} = sqrt{4 cdot 6}$. Since $sqrt{acdot b} = sqrt{a} cdot sqrt{b}$, you can expand the nominator to result in $2cdot sqrt{6}$. You can then divide by the $2$ to get the desired result.
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
answered Sep 11 '17 at 8:08
LarayLaray
1,746513
1,746513
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
add a comment |
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
$begingroup$
Thanks! I understand now :D
$endgroup$
– Nicco
Sep 11 '17 at 8:13
add a comment |
$begingroup$
How would you go about simplifying $$frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$frac{24}{8} = frac{12}{4} = frac{6}{2} = frac{3}{1} = 3.$$
Believe it or not, $$frac{sqrt{24}}{2} = sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.
So, by the same process as the other one, $$frac{sqrt{24}}{8} = frac{sqrt{6}}{4}.$$
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
$endgroup$
add a comment |
$begingroup$
How would you go about simplifying $$frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$frac{24}{8} = frac{12}{4} = frac{6}{2} = frac{3}{1} = 3.$$
Believe it or not, $$frac{sqrt{24}}{2} = sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.
So, by the same process as the other one, $$frac{sqrt{24}}{8} = frac{sqrt{6}}{4}.$$
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
$endgroup$
add a comment |
$begingroup$
How would you go about simplifying $$frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$frac{24}{8} = frac{12}{4} = frac{6}{2} = frac{3}{1} = 3.$$
Believe it or not, $$frac{sqrt{24}}{2} = sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.
So, by the same process as the other one, $$frac{sqrt{24}}{8} = frac{sqrt{6}}{4}.$$
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
$endgroup$
How would you go about simplifying $$frac{24}{8}?$$ You could try repeatedly halving the numerator and the denominator until arriving at the answer: $$frac{24}{8} = frac{12}{4} = frac{6}{2} = frac{3}{1} = 3.$$
Believe it or not, $$frac{sqrt{24}}{2} = sqrt 6.$$ But hey, I'm a demon, I could be trying to lead you astray. So check it on your calculator.
So, by the same process as the other one, $$frac{sqrt{24}}{8} = frac{sqrt{6}}{4}.$$
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
edited Sep 12 '17 at 22:27
Mr. Brooks
42011338
42011338
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
answered Sep 11 '17 at 21:42
The Short OneThe Short One
5541625
5541625
add a comment |
add a comment |
$begingroup$
The radicand, $ 24 $, can be expressed as a product involving a perfect square and a non-perfect square. In this case $ 24 $ is equal to $ 4 $ (which is a perfect square) times $ 6 $ (a non-perfect square). Thus
$$ sqrt{24} = sqrt{4 times 6} = sqrt{4}sqrt{6} = 2sqrt{6} $$ and $$ frac{sqrt{24}}{8} = frac{2sqrt{6}}{8} $$
After dividing both the numerator and denominator by the common factor of $ 2 $ and $ 8 $, which is $ 2 $:
$$ frac{2sqrt{6} / 2}{8/2} = frac{sqrt{6}}{4} $$
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
$endgroup$
add a comment |
$begingroup$
The radicand, $ 24 $, can be expressed as a product involving a perfect square and a non-perfect square. In this case $ 24 $ is equal to $ 4 $ (which is a perfect square) times $ 6 $ (a non-perfect square). Thus
$$ sqrt{24} = sqrt{4 times 6} = sqrt{4}sqrt{6} = 2sqrt{6} $$ and $$ frac{sqrt{24}}{8} = frac{2sqrt{6}}{8} $$
After dividing both the numerator and denominator by the common factor of $ 2 $ and $ 8 $, which is $ 2 $:
$$ frac{2sqrt{6} / 2}{8/2} = frac{sqrt{6}}{4} $$
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
$endgroup$
add a comment |
$begingroup$
The radicand, $ 24 $, can be expressed as a product involving a perfect square and a non-perfect square. In this case $ 24 $ is equal to $ 4 $ (which is a perfect square) times $ 6 $ (a non-perfect square). Thus
$$ sqrt{24} = sqrt{4 times 6} = sqrt{4}sqrt{6} = 2sqrt{6} $$ and $$ frac{sqrt{24}}{8} = frac{2sqrt{6}}{8} $$
After dividing both the numerator and denominator by the common factor of $ 2 $ and $ 8 $, which is $ 2 $:
$$ frac{2sqrt{6} / 2}{8/2} = frac{sqrt{6}}{4} $$
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
$endgroup$
The radicand, $ 24 $, can be expressed as a product involving a perfect square and a non-perfect square. In this case $ 24 $ is equal to $ 4 $ (which is a perfect square) times $ 6 $ (a non-perfect square). Thus
$$ sqrt{24} = sqrt{4 times 6} = sqrt{4}sqrt{6} = 2sqrt{6} $$ and $$ frac{sqrt{24}}{8} = frac{2sqrt{6}}{8} $$
After dividing both the numerator and denominator by the common factor of $ 2 $ and $ 8 $, which is $ 2 $:
$$ frac{2sqrt{6} / 2}{8/2} = frac{sqrt{6}}{4} $$
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
answered Jan 25 at 16:26
Marvin CohenMarvin Cohen
160117
160117
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2424975%2fsimplify-frac-sqrt248%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Dividing by 2 is ok, but then the numerator would read: $$sqrt{24}/2 = sqrt{24}/sqrt{2^2} =sqrt{24/4}.$$ Since the rule is $sqrt{ab} = sqrt{a}sqrt{b}$
$endgroup$
– P. Siehr
Sep 11 '17 at 8:12
2
$begingroup$
You cannot cancel $frac{sqrt{24}}{8}$ the same way as you do for $frac{24}{8}$. One has to respect the square root action. Note that $frac{sqrt{2}}{2}ne frac11=1$.
$endgroup$
– A.Γ.
Sep 11 '17 at 8:14