Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$

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Let $a,b,$ and $c$ be real numbers such that

$a+b+c=2 text and a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?

$text(A) 2qquad text (B) frac103qquad text (C) 4 qquad text (D) frac163qquad text (E) frac203$

As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!

https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17

inequality systems-of-equations maxima-minima cauchy-schwarz-inequality

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edited Jul 31 at 6:16

user21820

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asked Jul 31 at 4:36

Dude156

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up vote
4
down vote

favorite

1

Let $a,b,$ and $c$ be real numbers such that

$a+b+c=2 text and a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?

$text(A) 2qquad text (B) frac103qquad text (C) 4 qquad text (D) frac163qquad text (E) frac203$

As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!

https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17

inequality systems-of-equations maxima-minima cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 31 at 6:16

user21820

35.8k440136

asked Jul 31 at 4:36

Dude156

17612

add a comment | 

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Let $a,b,$ and $c$ be real numbers such that

$a+b+c=2 text and a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?

$text(A) 2qquad text (B) frac103qquad text (C) 4 qquad text (D) frac163qquad text (E) frac203$

As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!

https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17

inequality systems-of-equations maxima-minima cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 31 at 6:16

user21820

35.8k440136

asked Jul 31 at 4:36

Dude156

17612

Let $a,b,$ and $c$ be real numbers such that

$a+b+c=2 text and a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?

$text(A) 2qquad text (B) frac103qquad text (C) 4 qquad text (D) frac163qquad text (E) frac203$

As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!

https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17

inequality systems-of-equations maxima-minima cauchy-schwarz-inequality

share|cite|improve this question

edited Jul 31 at 6:16

user21820

35.8k440136

asked Jul 31 at 4:36

Dude156

17612

share|cite|improve this question

share|cite|improve this question

edited Jul 31 at 6:16

user21820

35.8k440136

edited Jul 31 at 6:16

user21820

35.8k440136

edited Jul 31 at 6:16

user21820

35.8k440136

35.8k440136

asked Jul 31 at 4:36

Dude156

17612

asked Jul 31 at 4:36

Dude156

17612

asked Jul 31 at 4:36

Dude156

17612

17612

add a comment | 

add a comment | 

6 Answers
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2
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accepted

Hint:
$$(2-c)^2 = (a+b)^2 = (a cdot 1 + b cdot 1)^2 le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

$$c^2 - 4c + 4 le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

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edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
  – Dude156
  Jul 31 at 5:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
  – angryavian
  Jul 31 at 5:14
  

  
  
  
  

add a comment | 

up vote
5
down vote

By C-S
$$12=a^2+b^2+c^2=frac12(1^2+1^2)(a^2+b^2)+c^2geqfrac12(a+b)^2+c^2=frac12(2-c)^2+c^2,$$
which gives $$3c^2-4c-20leq0$$ or
$$(3c-10)(c+2)leq0$$ or
$$-2leq cleqfrac103.$$
Now, we get $$frac103-(-2)=frac163.$$

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answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

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up vote
3
down vote

We have $$
a+b=(a,b)cdot (1,1)leq (a^2+b^2)^1/2sqrt2, $$ which they use as $$(a+b)^2leq 2 (a^2+b^2). $$

share|cite|improve this answer

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $c^2leq3 (16-c^2),$ implies $c^2leq12$
  – user 108128
  Jul 31 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course. Thanks, I'll edit right away.
  – Martin Argerami
  Jul 31 at 4:59
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $a=mc$ and $b=nc$ then
$$m+n=dfrac2c-1~~~,~~~m^2+n^2=dfrac12c^2-1$$
by Cauchy-Schwarz
$$(m+n)^2leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20geq0$ gives $c=-2,dfrac103$ leads us to difference $dfrac163$.

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answered Jul 31 at 5:36

user 108128

18.9k41544

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up vote
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We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$begincasesa+b&=&2-c\a^2+b^2&=&12-c^2endcases$$

Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.

If that number is non-negative, then $a-b = pmsqrt20+4c-3c^2$, which together with $(1)$ gives two sets of solutions.

So it remains to solve $20 + 4c - 3c^2 ge 0$, no Cauchy-Schwarz needed.

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answered Jul 31 at 5:37

Kenny Lau

17.7k2156

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Solution

Notice that $$a+b=2-ctag1,$$and $$a^2+b^2=12-c^2.tag2$$
Since $$ 2ab leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab leq 2(a^2+b^2).tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) leq 0.$$
As a result, $$-2leq c leq frac103.$$

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answered Jul 31 at 6:08

mengdie1982

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Hint:
$$(2-c)^2 = (a+b)^2 = (a cdot 1 + b cdot 1)^2 le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

$$c^2 - 4c + 4 le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

share|cite|improve this answer

edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
  – Dude156
  Jul 31 at 5:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
  – angryavian
  Jul 31 at 5:14
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Hint:
$$(2-c)^2 = (a+b)^2 = (a cdot 1 + b cdot 1)^2 le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

$$c^2 - 4c + 4 le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

share|cite|improve this answer

edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
  – Dude156
  Jul 31 at 5:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
  – angryavian
  Jul 31 at 5:14
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Hint:
$$(2-c)^2 = (a+b)^2 = (a cdot 1 + b cdot 1)^2 le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

$$c^2 - 4c + 4 le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

share|cite|improve this answer

edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

Hint:
$$(2-c)^2 = (a+b)^2 = (a cdot 1 + b cdot 1)^2 le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.

$$c^2 - 4c + 4 le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.

share|cite|improve this answer

edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 5:05

edited Jul 31 at 5:05

edited Jul 31 at 5:05

answered Jul 31 at 4:50

angryavian

34.5k12874

answered Jul 31 at 4:50

angryavian

34.5k12874

answered Jul 31 at 4:50

angryavian

34.5k12874

34.5k12874

• 
  
  
  

  
  

  
  
  
  
  
  

  
  how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
  – Dude156
  Jul 31 at 5:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
  – angryavian
  Jul 31 at 5:14
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
  – Dude156
  Jul 31 at 5:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
  – angryavian
  Jul 31 at 5:14
  

  
  
  
  

how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
– Dude156
Jul 31 at 5:10

how is that using cauchy though? I don't understand how we can take these variables and pull out vectors. Sorry if this question is very elementary lol
– Dude156
Jul 31 at 5:10

1

1

@Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
– angryavian
Jul 31 at 5:14

@Dude156 Consider the vectors $v=(a,b)$ and $w=(1,1)$. Then $v cdot w = a + b$ while $|v| |w| = sqrt(a^2 + b^2)(1^2 + 1^2)$.
– angryavian
Jul 31 at 5:14

add a comment | 

up vote
5
down vote

By C-S
$$12=a^2+b^2+c^2=frac12(1^2+1^2)(a^2+b^2)+c^2geqfrac12(a+b)^2+c^2=frac12(2-c)^2+c^2,$$
which gives $$3c^2-4c-20leq0$$ or
$$(3c-10)(c+2)leq0$$ or
$$-2leq cleqfrac103.$$
Now, we get $$frac103-(-2)=frac163.$$

share|cite|improve this answer

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

add a comment | 

up vote
5
down vote

By C-S
$$12=a^2+b^2+c^2=frac12(1^2+1^2)(a^2+b^2)+c^2geqfrac12(a+b)^2+c^2=frac12(2-c)^2+c^2,$$
which gives $$3c^2-4c-20leq0$$ or
$$(3c-10)(c+2)leq0$$ or
$$-2leq cleqfrac103.$$
Now, we get $$frac103-(-2)=frac163.$$

share|cite|improve this answer

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

add a comment | 

up vote
5
down vote

up vote
5
down vote

By C-S
$$12=a^2+b^2+c^2=frac12(1^2+1^2)(a^2+b^2)+c^2geqfrac12(a+b)^2+c^2=frac12(2-c)^2+c^2,$$
which gives $$3c^2-4c-20leq0$$ or
$$(3c-10)(c+2)leq0$$ or
$$-2leq cleqfrac103.$$
Now, we get $$frac103-(-2)=frac163.$$

share|cite|improve this answer

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

By C-S
$$12=a^2+b^2+c^2=frac12(1^2+1^2)(a^2+b^2)+c^2geqfrac12(a+b)^2+c^2=frac12(2-c)^2+c^2,$$
which gives $$3c^2-4c-20leq0$$ or
$$(3c-10)(c+2)leq0$$ or
$$-2leq cleqfrac103.$$
Now, we get $$frac103-(-2)=frac163.$$

share|cite|improve this answer

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

answered Jul 31 at 5:16

Michael Rozenberg

87.6k1577179

87.6k1577179

add a comment | 

add a comment | 

up vote
3
down vote

We have $$
a+b=(a,b)cdot (1,1)leq (a^2+b^2)^1/2sqrt2, $$ which they use as $$(a+b)^2leq 2 (a^2+b^2). $$

share|cite|improve this answer

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $c^2leq3 (16-c^2),$ implies $c^2leq12$
  – user 108128
  Jul 31 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course. Thanks, I'll edit right away.
  – Martin Argerami
  Jul 31 at 4:59
  

  
  
  
  

add a comment | 

up vote
3
down vote

We have $$
a+b=(a,b)cdot (1,1)leq (a^2+b^2)^1/2sqrt2, $$ which they use as $$(a+b)^2leq 2 (a^2+b^2). $$

share|cite|improve this answer

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $c^2leq3 (16-c^2),$ implies $c^2leq12$
  – user 108128
  Jul 31 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course. Thanks, I'll edit right away.
  – Martin Argerami
  Jul 31 at 4:59
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

We have $$
a+b=(a,b)cdot (1,1)leq (a^2+b^2)^1/2sqrt2, $$ which they use as $$(a+b)^2leq 2 (a^2+b^2). $$

share|cite|improve this answer

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

We have $$
a+b=(a,b)cdot (1,1)leq (a^2+b^2)^1/2sqrt2, $$ which they use as $$(a+b)^2leq 2 (a^2+b^2). $$

share|cite|improve this answer

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

share|cite|improve this answer

share|cite|improve this answer

edited Jul 31 at 5:15

edited Jul 31 at 5:15

edited Jul 31 at 5:15

answered Jul 31 at 4:53

Martin Argerami

115k1071164

answered Jul 31 at 4:53

Martin Argerami

115k1071164

answered Jul 31 at 4:53

Martin Argerami

115k1071164

115k1071164

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $c^2leq3 (16-c^2),$ implies $c^2leq12$
  – user 108128
  Jul 31 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course. Thanks, I'll edit right away.
  – Martin Argerami
  Jul 31 at 4:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $c^2leq3 (16-c^2),$ implies $c^2leq12$
  – user 108128
  Jul 31 at 4:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Of course. Thanks, I'll edit right away.
  – Martin Argerami
  Jul 31 at 4:59
  

  
  
  
  

$c^2leq3 (16-c^2),$ implies $c^2leq12$
– user 108128
Jul 31 at 4:57

$c^2leq3 (16-c^2),$ implies $c^2leq12$
– user 108128
Jul 31 at 4:57

Of course. Thanks, I'll edit right away.
– Martin Argerami
Jul 31 at 4:59

Of course. Thanks, I'll edit right away.
– Martin Argerami
Jul 31 at 4:59

add a comment | 

up vote
1
down vote

Let $a=mc$ and $b=nc$ then
$$m+n=dfrac2c-1~~~,~~~m^2+n^2=dfrac12c^2-1$$
by Cauchy-Schwarz
$$(m+n)^2leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20geq0$ gives $c=-2,dfrac103$ leads us to difference $dfrac163$.

share|cite|improve this answer

answered Jul 31 at 5:36

user 108128

18.9k41544

add a comment | 

up vote
1
down vote

Let $a=mc$ and $b=nc$ then
$$m+n=dfrac2c-1~~~,~~~m^2+n^2=dfrac12c^2-1$$
by Cauchy-Schwarz
$$(m+n)^2leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20geq0$ gives $c=-2,dfrac103$ leads us to difference $dfrac163$.

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answered Jul 31 at 5:36

user 108128

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Let $a=mc$ and $b=nc$ then
$$m+n=dfrac2c-1~~~,~~~m^2+n^2=dfrac12c^2-1$$
by Cauchy-Schwarz
$$(m+n)^2leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20geq0$ gives $c=-2,dfrac103$ leads us to difference $dfrac163$.

share|cite|improve this answer

answered Jul 31 at 5:36

user 108128

18.9k41544

Let $a=mc$ and $b=nc$ then
$$m+n=dfrac2c-1~~~,~~~m^2+n^2=dfrac12c^2-1$$
by Cauchy-Schwarz
$$(m+n)^2leq2(m^2+n^2)$$
with substitution $-3c^2+4c+20geq0$ gives $c=-2,dfrac103$ leads us to difference $dfrac163$.

share|cite|improve this answer

answered Jul 31 at 5:36

user 108128

18.9k41544

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 5:36

user 108128

18.9k41544

answered Jul 31 at 5:36

user 108128

18.9k41544

answered Jul 31 at 5:36

user 108128

18.9k41544

18.9k41544

add a comment | 

add a comment | 

up vote
1
down vote

We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$begincasesa+b&=&2-c\a^2+b^2&=&12-c^2endcases$$

Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.

If that number is non-negative, then $a-b = pmsqrt20+4c-3c^2$, which together with $(1)$ gives two sets of solutions.

So it remains to solve $20 + 4c - 3c^2 ge 0$, no Cauchy-Schwarz needed.

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answered Jul 31 at 5:37

Kenny Lau

17.7k2156

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We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$begincasesa+b&=&2-c\a^2+b^2&=&12-c^2endcases$$

Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.

If that number is non-negative, then $a-b = pmsqrt20+4c-3c^2$, which together with $(1)$ gives two sets of solutions.

So it remains to solve $20 + 4c - 3c^2 ge 0$, no Cauchy-Schwarz needed.

share|cite|improve this answer

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

add a comment | 

up vote
1
down vote

up vote
1
down vote

We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$begincasesa+b&=&2-c\a^2+b^2&=&12-c^2endcases$$

Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.

If that number is non-negative, then $a-b = pmsqrt20+4c-3c^2$, which together with $(1)$ gives two sets of solutions.

So it remains to solve $20 + 4c - 3c^2 ge 0$, no Cauchy-Schwarz needed.

share|cite|improve this answer

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

We wish to find the values of $c$ such that the following simultaneous equations have real solutions to $a$ and $b$:
$$begincasesa+b&=&2-c\a^2+b^2&=&12-c^2endcases$$

Labelling the equations as $(1)$ and $(2)$ respectively, $(1)^2 - (2)$ gives $2ab
= -8-4c+2c^2$, so $(a-b)^2 = a^2+b^2 - 2ab = (12-c^2) - (-8-4c+2c^2) = 20+4c-3c^2$.

If that number is non-negative, then $a-b = pmsqrt20+4c-3c^2$, which together with $(1)$ gives two sets of solutions.

So it remains to solve $20 + 4c - 3c^2 ge 0$, no Cauchy-Schwarz needed.

share|cite|improve this answer

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

answered Jul 31 at 5:37

Kenny Lau

17.7k2156

17.7k2156

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up vote
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Solution

Notice that $$a+b=2-ctag1,$$and $$a^2+b^2=12-c^2.tag2$$
Since $$ 2ab leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab leq 2(a^2+b^2).tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) leq 0.$$
As a result, $$-2leq c leq frac103.$$

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answered Jul 31 at 6:08

mengdie1982

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Solution

Notice that $$a+b=2-ctag1,$$and $$a^2+b^2=12-c^2.tag2$$
Since $$ 2ab leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab leq 2(a^2+b^2).tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) leq 0.$$
As a result, $$-2leq c leq frac103.$$

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answered Jul 31 at 6:08

mengdie1982

2,852216

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up vote
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down vote

up vote
1
down vote

Solution

Notice that $$a+b=2-ctag1,$$and $$a^2+b^2=12-c^2.tag2$$
Since $$ 2ab leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab leq 2(a^2+b^2).tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) leq 0.$$
As a result, $$-2leq c leq frac103.$$

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answered Jul 31 at 6:08

mengdie1982

2,852216

Solution

Notice that $$a+b=2-ctag1,$$and $$a^2+b^2=12-c^2.tag2$$
Since $$ 2ab leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab leq 2(a^2+b^2).tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) leq 0.$$
As a result, $$-2leq c leq frac103.$$

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answered Jul 31 at 6:08

mengdie1982

2,852216

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 6:08

mengdie1982

2,852216

answered Jul 31 at 6:08

mengdie1982

2,852216

answered Jul 31 at 6:08

mengdie1982

2,852216

2,852216

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add a comment | 

 

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