Inequality : $sqrt x - 6 - sqrt10 -x geqslant1$ [closed]

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I have solved it by squaring both sides and got inequality $x geqslant 17/2$ but after that, the solution part have concluded on the equation

$4x^2 + 289 - 68x geqslant4(10 - x)$

How this equation is formed.

functional-inequalities

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edited Jul 19 at 4:42

Mira from Earth

1359

asked Jul 19 at 3:55

Gaurav Singh

33

closed as unclear what you're asking by abiessu, Karn Watcharasupat, Isaac Browne, Tyrone, amWhy Jul 24 at 11:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The LHS is not even defined outside of the interval $[6,10]$.
  – mweiss
  Jul 19 at 4:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Can you clarify the question you are asking?
  – abiessu
  Jul 19 at 4:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Anyone else think the title shouldn't have been changed back?
  – Sambo
  Jul 19 at 4:32
  

  
  
  
  

add a comment | 

up vote
-3
down vote

favorite

I have solved it by squaring both sides and got inequality $x geqslant 17/2$ but after that, the solution part have concluded on the equation

$4x^2 + 289 - 68x geqslant4(10 - x)$

How this equation is formed.

functional-inequalities

share|cite|improve this question

edited Jul 19 at 4:42

Mira from Earth

1359

asked Jul 19 at 3:55

Gaurav Singh

33

closed as unclear what you're asking by abiessu, Karn Watcharasupat, Isaac Browne, Tyrone, amWhy Jul 24 at 11:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The LHS is not even defined outside of the interval $[6,10]$.
  – mweiss
  Jul 19 at 4:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Can you clarify the question you are asking?
  – abiessu
  Jul 19 at 4:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Anyone else think the title shouldn't have been changed back?
  – Sambo
  Jul 19 at 4:32
  

  
  
  
  

add a comment | 

up vote
-3
down vote

favorite

up vote
-3
down vote

favorite

I have solved it by squaring both sides and got inequality $x geqslant 17/2$ but after that, the solution part have concluded on the equation

$4x^2 + 289 - 68x geqslant4(10 - x)$

How this equation is formed.

functional-inequalities

share|cite|improve this question

edited Jul 19 at 4:42

Mira from Earth

1359

asked Jul 19 at 3:55

Gaurav Singh

33

I have solved it by squaring both sides and got inequality $x geqslant 17/2$ but after that, the solution part have concluded on the equation

$4x^2 + 289 - 68x geqslant4(10 - x)$

How this equation is formed.

functional-inequalities

share|cite|improve this question

edited Jul 19 at 4:42

Mira from Earth

1359

asked Jul 19 at 3:55

Gaurav Singh

33

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 4:42

Mira from Earth

1359

edited Jul 19 at 4:42

Mira from Earth

1359

edited Jul 19 at 4:42

Mira from Earth

1359

1359

asked Jul 19 at 3:55

Gaurav Singh

33

asked Jul 19 at 3:55

Gaurav Singh

33

asked Jul 19 at 3:55

Gaurav Singh

33

33

closed as unclear what you're asking by abiessu, Karn Watcharasupat, Isaac Browne, Tyrone, amWhy Jul 24 at 11:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

closed as unclear what you're asking by abiessu, Karn Watcharasupat, Isaac Browne, Tyrone, amWhy Jul 24 at 11:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The LHS is not even defined outside of the interval $[6,10]$.
  – mweiss
  Jul 19 at 4:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Can you clarify the question you are asking?
  – abiessu
  Jul 19 at 4:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Anyone else think the title shouldn't have been changed back?
  – Sambo
  Jul 19 at 4:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The LHS is not even defined outside of the interval $[6,10]$.
  – mweiss
  Jul 19 at 4:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Can you clarify the question you are asking?
  – abiessu
  Jul 19 at 4:04
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Anyone else think the title shouldn't have been changed back?
  – Sambo
  Jul 19 at 4:32
  

  
  
  
  

The LHS is not even defined outside of the interval $[6,10]$.
– mweiss
Jul 19 at 4:03

The LHS is not even defined outside of the interval $[6,10]$.
– mweiss
Jul 19 at 4:03

2

2

Can you clarify the question you are asking?
– abiessu
Jul 19 at 4:04

Can you clarify the question you are asking?
– abiessu
Jul 19 at 4:04

1

1

Anyone else think the title shouldn't have been changed back?
– Sambo
Jul 19 at 4:32

Anyone else think the title shouldn't have been changed back?
– Sambo
Jul 19 at 4:32

add a comment | 

2 Answers
2

active

oldest

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up vote
1
down vote



accepted

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) :
$$
sqrtx-6 - sqrt10-x geq 1 iff x - 6 + (10 - x) - 2 sqrt(x-6)(10-x) geq 1 \ iff frac 32 geq sqrt(x-6)(10-x) iff 9 geq 4(x-6)(10-x) \ iff 9 geq 4(-x^2 - 60 + 16 x) iff 4x^2 - 64x + 249 geq 0 \ iff x^2 - 16x + 62.25 geq 0 iff (x-8)^2 geq 1.75 iff x notin (8 - sqrt1.75,8 + sqrt1.75)
$$

So, the answer is $8 + sqrt1.75 leq x leq 10$.
Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $sqrt10-x$ to the other side before squaring, then we get:
$$
sqrtx-6 geq 1 + sqrt10-x iff x-6 geq 1 + (10-x) + 2 sqrt10-x \ iff x - 8.5 geq sqrt10-x iff x^2 - 17x + 72.25 geq 10 - x \ iff 4x^2 - 68x + 289 geq 4(10 - x)
$$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a geq b iff a^2 geq b^2$, but it is true if $a,b$ are both positive.

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

add a comment | 

up vote
2
down vote

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with:
$$
(x-6)-(10-x)geq 1
$$
After rearranging, you get $xgeq 17/2$. But this is incorrect; squaring both sides will in fact give you:
$$
(x-6)+(10-x)-2sqrtx-6sqrt10-x geq 1
$$
You will need to then rearrange the equation and square again to get your solution.

share|cite|improve this answer

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) :
$$
sqrtx-6 - sqrt10-x geq 1 iff x - 6 + (10 - x) - 2 sqrt(x-6)(10-x) geq 1 \ iff frac 32 geq sqrt(x-6)(10-x) iff 9 geq 4(x-6)(10-x) \ iff 9 geq 4(-x^2 - 60 + 16 x) iff 4x^2 - 64x + 249 geq 0 \ iff x^2 - 16x + 62.25 geq 0 iff (x-8)^2 geq 1.75 iff x notin (8 - sqrt1.75,8 + sqrt1.75)
$$

So, the answer is $8 + sqrt1.75 leq x leq 10$.
Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $sqrt10-x$ to the other side before squaring, then we get:
$$
sqrtx-6 geq 1 + sqrt10-x iff x-6 geq 1 + (10-x) + 2 sqrt10-x \ iff x - 8.5 geq sqrt10-x iff x^2 - 17x + 72.25 geq 10 - x \ iff 4x^2 - 68x + 289 geq 4(10 - x)
$$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a geq b iff a^2 geq b^2$, but it is true if $a,b$ are both positive.

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

add a comment | 

up vote
1
down vote



accepted

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) :
$$
sqrtx-6 - sqrt10-x geq 1 iff x - 6 + (10 - x) - 2 sqrt(x-6)(10-x) geq 1 \ iff frac 32 geq sqrt(x-6)(10-x) iff 9 geq 4(x-6)(10-x) \ iff 9 geq 4(-x^2 - 60 + 16 x) iff 4x^2 - 64x + 249 geq 0 \ iff x^2 - 16x + 62.25 geq 0 iff (x-8)^2 geq 1.75 iff x notin (8 - sqrt1.75,8 + sqrt1.75)
$$

So, the answer is $8 + sqrt1.75 leq x leq 10$.
Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $sqrt10-x$ to the other side before squaring, then we get:
$$
sqrtx-6 geq 1 + sqrt10-x iff x-6 geq 1 + (10-x) + 2 sqrt10-x \ iff x - 8.5 geq sqrt10-x iff x^2 - 17x + 72.25 geq 10 - x \ iff 4x^2 - 68x + 289 geq 4(10 - x)
$$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a geq b iff a^2 geq b^2$, but it is true if $a,b$ are both positive.

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) :
$$
sqrtx-6 - sqrt10-x geq 1 iff x - 6 + (10 - x) - 2 sqrt(x-6)(10-x) geq 1 \ iff frac 32 geq sqrt(x-6)(10-x) iff 9 geq 4(x-6)(10-x) \ iff 9 geq 4(-x^2 - 60 + 16 x) iff 4x^2 - 64x + 249 geq 0 \ iff x^2 - 16x + 62.25 geq 0 iff (x-8)^2 geq 1.75 iff x notin (8 - sqrt1.75,8 + sqrt1.75)
$$

So, the answer is $8 + sqrt1.75 leq x leq 10$.
Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $sqrt10-x$ to the other side before squaring, then we get:
$$
sqrtx-6 geq 1 + sqrt10-x iff x-6 geq 1 + (10-x) + 2 sqrt10-x \ iff x - 8.5 geq sqrt10-x iff x^2 - 17x + 72.25 geq 10 - x \ iff 4x^2 - 68x + 289 geq 4(10 - x)
$$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a geq b iff a^2 geq b^2$, but it is true if $a,b$ are both positive.

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

You should have done (true on the interval $[8,10]$, where the left hand side of the inequality is positive at least) :
$$
sqrtx-6 - sqrt10-x geq 1 iff x - 6 + (10 - x) - 2 sqrt(x-6)(10-x) geq 1 \ iff frac 32 geq sqrt(x-6)(10-x) iff 9 geq 4(x-6)(10-x) \ iff 9 geq 4(-x^2 - 60 + 16 x) iff 4x^2 - 64x + 249 geq 0 \ iff x^2 - 16x + 62.25 geq 0 iff (x-8)^2 geq 1.75 iff x notin (8 - sqrt1.75,8 + sqrt1.75)
$$

So, the answer is $8 + sqrt1.75 leq x leq 10$.
Therefore, your answer is wrong, and you should use my working to verify where it is wrong. The other answer seems to point out your mistake.

Suppose we don't square immediately, but transpose , say $sqrt10-x$ to the other side before squaring, then we get:
$$
sqrtx-6 geq 1 + sqrt10-x iff x-6 geq 1 + (10-x) + 2 sqrt10-x \ iff x - 8.5 geq sqrt10-x iff x^2 - 17x + 72.25 geq 10 - x \ iff 4x^2 - 68x + 289 geq 4(10 - x)
$$

which is the equation given in the solution, and which will also lead to the right answer. The problem with squaring both sides immediately is that it is not true that $a geq b iff a^2 geq b^2$, but it is true if $a,b$ are both positive.

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

answered Jul 19 at 4:22

астон вілла олоф мэллбэрг

32k22463

32k22463

add a comment | 

add a comment | 

up vote
2
down vote

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with:
$$
(x-6)-(10-x)geq 1
$$
After rearranging, you get $xgeq 17/2$. But this is incorrect; squaring both sides will in fact give you:
$$
(x-6)+(10-x)-2sqrtx-6sqrt10-x geq 1
$$
You will need to then rearrange the equation and square again to get your solution.

share|cite|improve this answer

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

add a comment | 

up vote
2
down vote

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with:
$$
(x-6)-(10-x)geq 1
$$
After rearranging, you get $xgeq 17/2$. But this is incorrect; squaring both sides will in fact give you:
$$
(x-6)+(10-x)-2sqrtx-6sqrt10-x geq 1
$$
You will need to then rearrange the equation and square again to get your solution.

share|cite|improve this answer

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

add a comment | 

up vote
2
down vote

up vote
2
down vote

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with:
$$
(x-6)-(10-x)geq 1
$$
After rearranging, you get $xgeq 17/2$. But this is incorrect; squaring both sides will in fact give you:
$$
(x-6)+(10-x)-2sqrtx-6sqrt10-x geq 1
$$
You will need to then rearrange the equation and square again to get your solution.

share|cite|improve this answer

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

You've fallen into the old trap: we don't have $(a+b)^2 = a^2 + b^2$!! As my old math teacher would say, every time you make this mistake, a kitten dies.

To clarify, you got your answer by squaring both sides and erroneously ended up with:
$$
(x-6)-(10-x)geq 1
$$
After rearranging, you get $xgeq 17/2$. But this is incorrect; squaring both sides will in fact give you:
$$
(x-6)+(10-x)-2sqrtx-6sqrt10-x geq 1
$$
You will need to then rearrange the equation and square again to get your solution.

share|cite|improve this answer

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 4:19

edited Jul 19 at 4:19

edited Jul 19 at 4:19

answered Jul 19 at 4:14

Sambo

1,2561427

answered Jul 19 at 4:14

Sambo

1,2561427

answered Jul 19 at 4:14

Sambo

1,2561427

1,2561427

add a comment | 

add a comment |