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Square root and Cube root in the same equation [closed]

0

$begingroup$

In a recent test I took, I received a question in an awkward form:
$$sqrt y-sqrt[3]{1000-y}=16$$
How would I go about solving this?

radicals

share|cite|improve this question

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

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closed as off-topic by Javi, Saad, Leucippus, Lee David Chung Lin, Eevee Trainer Mar 24 at 4:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Saad, Leucippus, Lee David Chung Lin, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. In the future please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck (e.g. the comment under the answer by model_checker).
  $endgroup$
  – Lee David Chung Lin
  Mar 24 at 2:55
  

  
  
  
  

add a comment | 

0

$begingroup$

In a recent test I took, I received a question in an awkward form:
$$sqrt y-sqrt[3]{1000-y}=16$$
How would I go about solving this?

radicals

share|cite|improve this question

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

$endgroup$

closed as off-topic by Javi, Saad, Leucippus, Lee David Chung Lin, Eevee Trainer Mar 24 at 4:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Saad, Leucippus, Lee David Chung Lin, Eevee Trainer

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE. In the future please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck (e.g. the comment under the answer by model_checker).
  $endgroup$
  – Lee David Chung Lin
  Mar 24 at 2:55
  

  
  
  
  

add a comment | 

$begingroup$

In a recent test I took, I received a question in an awkward form:
$$sqrt y-sqrt[3]{1000-y}=16$$
How would I go about solving this?

radicals

share|cite|improve this question

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

$endgroup$

share|cite|improve this question

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

share|cite|improve this question

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

asked Mar 23 at 22:46

Kavi AgrawalKavi Agrawal

42

1 Answer
1

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oldest

votes

3

$begingroup$

Let $x = sqrt{y}$ and then you have that
$$(x-16)^3 = 1000-x^2$$
which gives you a cubic that you can solve using standard methods.

share|cite|improve this answer

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

$endgroup$

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  $begingroup$
  I have tried this, and it leads me to $x^3 - 47 x^2 + 768 x - 5096 = 0$, which according to WolframAlpha doesn't have an integer solution.
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviAgrawal: That is correct, it doesn't have an integral solution. Why do you think it should?
  $endgroup$
  – Ross Millikan
  Mar 23 at 23:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would I be able to derive the solution by hand, in that case?
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Newton Ralphson is one iterative method that could be used to approximate roots.
  $endgroup$
  – model_checker
  Mar 23 at 23:07
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $x = sqrt{y}$ and then you have that
$$(x-16)^3 = 1000-x^2$$
which gives you a cubic that you can solve using standard methods.

share|cite|improve this answer

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have tried this, and it leads me to $x^3 - 47 x^2 + 768 x - 5096 = 0$, which according to WolframAlpha doesn't have an integer solution.
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviAgrawal: That is correct, it doesn't have an integral solution. Why do you think it should?
  $endgroup$
  – Ross Millikan
  Mar 23 at 23:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would I be able to derive the solution by hand, in that case?
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Newton Ralphson is one iterative method that could be used to approximate roots.
  $endgroup$
  – model_checker
  Mar 23 at 23:07
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $x = sqrt{y}$ and then you have that
$$(x-16)^3 = 1000-x^2$$
which gives you a cubic that you can solve using standard methods.

share|cite|improve this answer

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have tried this, and it leads me to $x^3 - 47 x^2 + 768 x - 5096 = 0$, which according to WolframAlpha doesn't have an integer solution.
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @KaviAgrawal: That is correct, it doesn't have an integral solution. Why do you think it should?
  $endgroup$
  – Ross Millikan
  Mar 23 at 23:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How would I be able to derive the solution by hand, in that case?
  $endgroup$
  – Kavi Agrawal
  Mar 23 at 23:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Newton Ralphson is one iterative method that could be used to approximate roots.
  $endgroup$
  – model_checker
  Mar 23 at 23:07
  

  
  
  
  

add a comment | 

$begingroup$

Let $x = sqrt{y}$ and then you have that
$$(x-16)^3 = 1000-x^2$$
which gives you a cubic that you can solve using standard methods.

share|cite|improve this answer

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

$endgroup$

share|cite|improve this answer

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931

answered Mar 23 at 22:54

model_checkermodel_checker

4,45521931