Mixing
Simplifying nested radicals? $left(sqrt{4+sqrt{16-a^2}}+sqrt{4-sqrt{16-a^2}}right)^2$
$begingroup$
I got the following term I'd like to simplify
$$left(sqrt{4+sqrt{16-a^2}}+sqrt{4-sqrt{16-a^2}}right)^2.$$
My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct):
$$8+2×left(sqrt{4+sqrt{16-a^2}}right)×left(sqrt{4-sqrt{16-a^2}}right).$$
But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.
Thanks in advance!
binomial-theorem nested-radicals
edited Mar 20 '18 at 8:26
Saad
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
$endgroup$
add a comment |
$begingroup$
I got the following term I'd like to simplify
$$left(sqrt{4+sqrt{16-a^2}}+sqrt{4-sqrt{16-a^2}}right)^2.$$
My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct):
$$8+2×left(sqrt{4+sqrt{16-a^2}}right)×left(sqrt{4-sqrt{16-a^2}}right).$$
But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.
Thanks in advance!
binomial-theorem nested-radicals
edited Mar 20 '18 at 8:26
Saad
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
$endgroup$
1
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26
add a comment |
$begingroup$
I got the following term I'd like to simplify
$$left(sqrt{4+sqrt{16-a^2}}+sqrt{4-sqrt{16-a^2}}right)^2.$$
My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct):
$$8+2×left(sqrt{4+sqrt{16-a^2}}right)×left(sqrt{4-sqrt{16-a^2}}right).$$
But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.
Thanks in advance!
binomial-theorem nested-radicals
edited Mar 20 '18 at 8:26
Saad
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
$endgroup$
I got the following term I'd like to simplify
$$left(sqrt{4+sqrt{16-a^2}}+sqrt{4-sqrt{16-a^2}}right)^2.$$
My Approach was to use the binomial formula. Therefore I'm currently at (hope thats correct):
$$8+2×left(sqrt{4+sqrt{16-a^2}}right)×left(sqrt{4-sqrt{16-a^2}}right).$$
But now I'm stuck. Any hints or suggestions how to proceed? I had a look at denesting radicals but that doesn't help me yet.
Thanks in advance!
binomial-theorem nested-radicals
binomial-theorem nested-radicals
edited Mar 20 '18 at 8:26
Saad
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
edited Mar 20 '18 at 8:26
Saad
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
edited Mar 20 '18 at 8:26
Saad
20.4k92452
edited Mar 20 '18 at 8:26
Saad
20.4k92452
edited Mar 20 '18 at 8:26
Saad
20.4k92452
20.4k92452
asked Mar 20 '18 at 8:23
klamsiklamsi
51
asked Mar 20 '18 at 8:23
klamsiklamsi
51
asked Mar 20 '18 at 8:23
klamsiklamsi
51
51
1
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26
add a comment |
1
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26
1
1
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use the fact that $sqrt{a+b}*sqrt{a-b}=sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
$endgroup$
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
add a comment |
$begingroup$
This can be done mentally.
Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $sqrt{4^2-(16-a^2)}$.
The answer is $$8+2|a|.$$
edited Jan 31 at 22:27
Math Lover
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
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%2f2699973%2fsimplifying-nested-radicals-left-sqrt4-sqrt16-a2-sqrt4-sqrt16-a2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the fact that $sqrt{a+b}*sqrt{a-b}=sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
$endgroup$
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
add a comment |
$begingroup$
Use the fact that $sqrt{a+b}*sqrt{a-b}=sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
$endgroup$
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
add a comment |
$begingroup$
Use the fact that $sqrt{a+b}*sqrt{a-b}=sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
$endgroup$
Use the fact that $sqrt{a+b}*sqrt{a-b}=sqrt{(a-b)(a+b)}$, and then use @choco_addicted's comment.
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
answered Mar 20 '18 at 8:29
MrtnyMrtny
1167
1167
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
add a comment |
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
So: $$8+2×left(sqrt{(4+sqrt{16-a^2})×(4-sqrt{16-a^2}})right).$$ $$8+2*sqrt{16-16+a^2}$$ Hence: $$8+2a$$ Seems like I missed something... as It should be $$8+2|a|$$
$endgroup$
– klamsi
Mar 20 '18 at 8:59
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
$begingroup$
This is because the square root is positive by convention. Thus, if $a$ was negative, the answer $sqrt{a^2}=a$ would be negative, going against the convention. This means $sqrt{a^2}=|a|$.
$endgroup$
– Mrtny
Mar 20 '18 at 9:19
add a comment |
$begingroup$
This can be done mentally.
Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $sqrt{4^2-(16-a^2)}$.
The answer is $$8+2|a|.$$
edited Jan 31 at 22:27
Math Lover
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
add a comment |
$begingroup$
This can be done mentally.
Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $sqrt{4^2-(16-a^2)}$.
The answer is $$8+2|a|.$$
edited Jan 31 at 22:27
Math Lover
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
$endgroup$
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
add a comment |
$begingroup$
This can be done mentally.
Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $sqrt{4^2-(16-a^2)}$.
The answer is $$8+2|a|.$$
edited Jan 31 at 22:27
Math Lover
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
$endgroup$
This can be done mentally.
Expanding the square of the sum, the individual squares indeed yield $8$, and the double product is the square root of a difference of squares $sqrt{4^2-(16-a^2)}$.
The answer is $$8+2|a|.$$
edited Jan 31 at 22:27
Math Lover
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
edited Jan 31 at 22:27
Math Lover
17410
edited Jan 31 at 22:27
Math Lover
17410
edited Jan 31 at 22:27
Math Lover
17410
17410
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
answered Mar 20 '18 at 8:29
Yves DaoustYves Daoust
132k676230
132k676230
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
add a comment |
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
So $8pm2a$ would be your final answer?
$endgroup$
– klamsi
Mar 20 '18 at 8:32
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
$begingroup$
@klamsi: mh, no, actually $8+2|a|$.
$endgroup$
– Yves Daoust
Mar 20 '18 at 8:33
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%2f2699973%2fsimplifying-nested-radicals-left-sqrt4-sqrt16-a2-sqrt4-sqrt16-a2%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
1
$begingroup$
Apply $a^2-b^2=(a+b)(a-b)$,
$endgroup$
– choco_addicted
Mar 20 '18 at 8:26