The value of $a+b=ab=a^2-b^2$

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If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?

1. $(1+sqrt5)/2$


2. $(3+sqrt5)/2$


3. 2


4. $sqrt5$

Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.

algebra-precalculus

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edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

asked Jul 24 at 2:59

THEGreatGatsby

61

migrated from mathematica.stackexchange.com Jul 24 at 3:23

This question came from our site for users of Wolfram Mathematica.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It looks like this was intended for Mathematics -- would you like me to move your question to that site?
  – Mr.Wizard
  Jul 24 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry , yes please
  – THEGreatGatsby
  Jul 24 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can have $a=b=0$ which is not one of the given solutions.
  – user223391
  Jul 24 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: $a+b=(a-b)(a+b)$
  – stewbasic
  Jul 24 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
  – abiessu
  Jul 24 at 3:38
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?

1. $(1+sqrt5)/2$


2. $(3+sqrt5)/2$


3. 2


4. $sqrt5$

Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.

algebra-precalculus

share|cite|improve this question

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

asked Jul 24 at 2:59

THEGreatGatsby

61

migrated from mathematica.stackexchange.com Jul 24 at 3:23

This question came from our site for users of Wolfram Mathematica.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It looks like this was intended for Mathematics -- would you like me to move your question to that site?
  – Mr.Wizard
  Jul 24 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry , yes please
  – THEGreatGatsby
  Jul 24 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can have $a=b=0$ which is not one of the given solutions.
  – user223391
  Jul 24 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: $a+b=(a-b)(a+b)$
  – stewbasic
  Jul 24 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
  – abiessu
  Jul 24 at 3:38
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?

1. $(1+sqrt5)/2$


2. $(3+sqrt5)/2$


3. 2


4. $sqrt5$

Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.

algebra-precalculus

share|cite|improve this question

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

asked Jul 24 at 2:59

THEGreatGatsby

61

If $a+b=ab=a^2-b^2$ and $a$ and $b$ are real numbers then what do all of the expressions evaluate to?

1. $(1+sqrt5)/2$


2. $(3+sqrt5)/2$


3. 2


4. $sqrt5$

Not sure at all how to find the answer. I tried $a=ab/b$ and $a=(a^2-b^2)/b$ but after that I am lost.

algebra-precalculus

share|cite|improve this question

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

asked Jul 24 at 2:59

THEGreatGatsby

61

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

edited Jul 24 at 3:26

Parcly Taxel

33.5k136588

33.5k136588

asked Jul 24 at 2:59

THEGreatGatsby

61

asked Jul 24 at 2:59

THEGreatGatsby

61

asked Jul 24 at 2:59

THEGreatGatsby

61

61

migrated from mathematica.stackexchange.com Jul 24 at 3:23

This question came from our site for users of Wolfram Mathematica.

migrated from mathematica.stackexchange.com Jul 24 at 3:23

This question came from our site for users of Wolfram Mathematica.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It looks like this was intended for Mathematics -- would you like me to move your question to that site?
  – Mr.Wizard
  Jul 24 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry , yes please
  – THEGreatGatsby
  Jul 24 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can have $a=b=0$ which is not one of the given solutions.
  – user223391
  Jul 24 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: $a+b=(a-b)(a+b)$
  – stewbasic
  Jul 24 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
  – abiessu
  Jul 24 at 3:38
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  It looks like this was intended for Mathematics -- would you like me to move your question to that site?
  – Mr.Wizard
  Jul 24 at 3:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sorry , yes please
  – THEGreatGatsby
  Jul 24 at 3:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can have $a=b=0$ which is not one of the given solutions.
  – user223391
  Jul 24 at 3:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hint: $a+b=(a-b)(a+b)$
  – stewbasic
  Jul 24 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
  – abiessu
  Jul 24 at 3:38
  

  
  
  
  

It looks like this was intended for Mathematics -- would you like me to move your question to that site?
– Mr.Wizard
Jul 24 at 3:22

It looks like this was intended for Mathematics -- would you like me to move your question to that site?
– Mr.Wizard
Jul 24 at 3:22

sorry , yes please
– THEGreatGatsby
Jul 24 at 3:23

sorry , yes please
– THEGreatGatsby
Jul 24 at 3:23

You can have $a=b=0$ which is not one of the given solutions.
– user223391
Jul 24 at 3:29

You can have $a=b=0$ which is not one of the given solutions.
– user223391
Jul 24 at 3:29

Hint: $a+b=(a-b)(a+b)$
– stewbasic
Jul 24 at 3:38

Hint: $a+b=(a-b)(a+b)$
– stewbasic
Jul 24 at 3:38

Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
– abiessu
Jul 24 at 3:38

Note that $a^2-b^2=(a-b)(a+b)$, giving either $a-b=0$ or $a-b=1$...
– abiessu
Jul 24 at 3:38

add a comment | 

1 Answer
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Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.

If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

add a comment | 

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up vote
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Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.

If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

add a comment | 

up vote
3
down vote

Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.

If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

add a comment | 

up vote
3
down vote

up vote
3
down vote

Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.

If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

Any time you see $a^2-b^2$ you should think $(a+b)(a-b)$. Here if $a+b neq 0$ you can divide it out to get $1=a-b$ Then $a=b+1, a+b=2b+1=(b+1)b$ and feed it to the quadratic formula.

If $a+b=0$ we must have $a=b=0$ and the sum is again $0$, which should have been one of the choices.

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 3:39

Ross Millikan

275k21186351

answered Jul 24 at 3:39

Ross Millikan

275k21186351

answered Jul 24 at 3:39

Ross Millikan

275k21186351

275k21186351

add a comment | 

add a comment | 

 

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