What are the possible solutions of the given equation?When do we get extraneous roots?Algebraic manipulation with square rootsFind all real numbers such that $sqrtx-frac1x + sqrt1 - frac1x = x$Transposing an equation: x = F/k + sqrt(F/c) to get F as the subjectFinding all real roots of the equation $(x+1) sqrtx+2 + (x+6)sqrtx+7 = x^2+7x+12$Justify the algorithm used to create a polynomial whose roots are squares of the roots of the given polynomial (over $mathbb C$)Integral solutions to the equation $left(frac1nright)^-1/2=sqrta+sqrt15-sqrta-sqrt15.$What rule governs $x^4=10,000$ having complex solutions?Why do I keep getting this incorrect solution when trying to find all the real solutions for $sqrt2x-3 +x=3$.Radical equation - can I square both sides with more than 1 radical on one side?
Is it normal that my co-workers at a fitness company criticize my food choices?
Why did it take so long to abandon sail after steamships were demonstrated?
PTIJ: Who should pay for Uber rides: the child or the parent?
How do anti-virus programs start at Windows boot?
Why do Australian milk farmers need to protest supermarkets' milk price?
How to answer questions about my characters?
Humanity loses the vast majority of its technology, information, and population in the year 2122. How long does it take to rebuild itself?
Good allowance savings plan?
Why do passenger jet manufacturers design their planes with stall prevention systems?
Why are there 40 737 Max planes in flight when they have been grounded as not airworthy?
How to deal with a cynical class?
Can hydraulic brake levers get hot when brakes overheat?
Does the statement `int val = (++i > ++j) ? ++i : ++j;` invoke undefined behavior?
Provisioning profile doesn't include the application-identifier and keychain-access-groups entitlements
Why would a flight no longer considered airworthy be redirected like this?
Life insurance that covers only simultaneous/dual deaths
Can the damage from a Talisman of Pure Good (or Ultimate Evil) be non-lethal?
Rules about breaking the rules. How do I do it well?
How to make healing in an exploration game interesting
How to explain that I do not want to visit a country due to personal safety concern?
Font with correct density?
How could a female member of a species produce eggs unto death?
Connecting top and bottom SMD component pads using via
An Accountant Seeks the Help of a Mathematician
What are the possible solutions of the given equation?
When do we get extraneous roots?Algebraic manipulation with square rootsFind all real numbers such that $sqrtx-frac1x + sqrt1 - frac1x = x$Transposing an equation: x = F/k + sqrt(F/c) to get F as the subjectFinding all real roots of the equation $(x+1) sqrtx+2 + (x+6)sqrtx+7 = x^2+7x+12$Justify the algorithm used to create a polynomial whose roots are squares of the roots of the given polynomial (over $mathbb C$)Integral solutions to the equation $left(frac1nright)^-1/2=sqrta+sqrt15-sqrta-sqrt15.$What rule governs $x^4=10,000$ having complex solutions?Why do I keep getting this incorrect solution when trying to find all the real solutions for $sqrt2x-3 +x=3$.Radical equation - can I square both sides with more than 1 radical on one side?
$begingroup$
I encountered a question in an exam in which we had:
Find all possible solutions of the equation $$x+y+ 1over x+1over y+4=2 (sqrt 2x+1+sqrt 2y+1) $$ where $x $ and $y$ are real numbers.
I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
I encountered a question in an exam in which we had:
Find all possible solutions of the equation $$x+y+ 1over x+1over y+4=2 (sqrt 2x+1+sqrt 2y+1) $$ where $x $ and $y$ are real numbers.
I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?
algebra-precalculus
$endgroup$
add a comment |
$begingroup$
I encountered a question in an exam in which we had:
Find all possible solutions of the equation $$x+y+ 1over x+1over y+4=2 (sqrt 2x+1+sqrt 2y+1) $$ where $x $ and $y$ are real numbers.
I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?
algebra-precalculus
$endgroup$
I encountered a question in an exam in which we had:
Find all possible solutions of the equation $$x+y+ 1over x+1over y+4=2 (sqrt 2x+1+sqrt 2y+1) $$ where $x $ and $y$ are real numbers.
I tried squaring both sides to eliminate the square roots but the number of terms became too many, making the problem very difficult to handle. I am not really able to understand how to find an easier approach or handle the terms efficiently. Would someone please help me to solve this question?
algebra-precalculus
algebra-precalculus
edited 6 hours ago
Thomas Andrews
130k12147298
130k12147298
asked 6 hours ago
Shashwat1337Shashwat1337
889
889
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$
By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have
$$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$
Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$
$endgroup$
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
add a comment |
$begingroup$
It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
$$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
$$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain:
$$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
$$x=y=1+sqrt2.$$
$endgroup$
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%2f3148348%2fwhat-are-the-possible-solutions-of-the-given-equation%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
$begingroup$
$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$
By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have
$$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$
Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$
$endgroup$
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
add a comment |
$begingroup$
$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$
By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have
$$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$
Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$
$endgroup$
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
add a comment |
$begingroup$
$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$
By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have
$$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$
Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$
$endgroup$
$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$
By Am-Gm we have $$ x+2x+1over xgeq 2sqrtx2x+1over x = 2sqrt2x+1$$ and the same for $y$, so we have
$$x+y+ 1over x+1over y+4 geq 2sqrt2x+1+2sqrt2y+1$$
Since we have equality is achieved when $x=2x+1over x$ (and the same for $y$) we have $x=y=1+sqrt2$
edited 6 hours ago
answered 6 hours ago
Maria MazurMaria Mazur
46.9k1260120
46.9k1260120
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
add a comment |
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
$begingroup$
Truly amazing!!! [+1]
$endgroup$
– Dr. Mathva
6 hours ago
2
2
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
$begingroup$
@Maria Mazur Your solution is total wrong. What happens for $xy<0?$ If you'll see down-voting it's not mine.
$endgroup$
– Michael Rozenberg
6 hours ago
2
2
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
$begingroup$
AM-GM requires terms to be positive. So your solution doesn't account for the case when terms are negative.
$endgroup$
– Anurag A
5 hours ago
add a comment |
$begingroup$
It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
$$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
$$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain:
$$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
$$x=y=1+sqrt2.$$
$endgroup$
add a comment |
$begingroup$
It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
$$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
$$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain:
$$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
$$x=y=1+sqrt2.$$
$endgroup$
add a comment |
$begingroup$
It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
$$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
$$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain:
$$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
$$x=y=1+sqrt2.$$
$endgroup$
It's $$sum_cycleft(x+frac1x+2-2sqrt2x+1right)=0$$ or
$$sum_cycfracx^2-2xsqrt2x+1+2x+1x=0$$ or
$$sum_cycfrac(x-sqrt2x+1)^2x=0,$$
which for $xy<0$ gives infinitely many solutions.
But, for $xy>0$ we obtain:
$$x=sqrt2x+1$$ and $$y=sqrt2y+1,$$ which gives
$$x=y=1+sqrt2.$$
answered 6 hours ago
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
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%2f3148348%2fwhat-are-the-possible-solutions-of-the-given-equation%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