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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?

2

$begingroup$

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

share|cite|improve this question

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

$endgroup$

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

$endgroup$

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

share|cite|improve this question

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

$endgroup$

share|cite|improve this question

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

share|cite|improve this question

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

edited 4 hours ago

Thomas Andrews

130k12147298

edited 4 hours ago

Thomas Andrews

130k12147298

asked 4 hours ago

Shashwat1337Shashwat1337

889

asked 4 hours ago

Shashwat1337Shashwat1337

889

2 Answers
2

active

oldest

votes

7

$begingroup$

$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$

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 $x=y=1+sqrt2$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Truly amazing!!! [+1]
  $endgroup$
  – Dr. Mathva
  4 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
  4 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
  4 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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
$$x=y=1+sqrt2.$$

share|cite|improve this answer

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

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7

$begingroup$

$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$

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 $x=y=1+sqrt2$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Truly amazing!!! [+1]
  $endgroup$
  – Dr. Mathva
  4 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
  4 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
  4 hours ago
  
  

  
  
  
  

add a comment | 

7

$begingroup$

$$x+y+ 1over x+1over y+4 = x+y+2x+1over x+2y+1over y $$

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 $x=y=1+sqrt2$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Truly amazing!!! [+1]
  $endgroup$
  – Dr. Mathva
  4 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
  4 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
  4 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 $x=y=1+sqrt2$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

$endgroup$

share|cite|improve this answer

edited 4 hours ago

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

answered 4 hours ago

Maria MazurMaria Mazur

46.9k1260120

2

$begingroup$

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
$$x=y=1+sqrt2.$$

share|cite|improve this answer

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

add a comment | 

2

$begingroup$

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
$$x=y=1+sqrt2.$$

share|cite|improve this answer

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

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
$$x=y=1+sqrt2.$$

share|cite|improve this answer

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200

answered 4 hours ago

Michael RozenbergMichael Rozenberg

108k1895200