Mixing
Don't understand solution to this mock AMC 12?
0
$begingroup$
I don't get the answer to the solution of the problem.
I understand that k=c+8, but I don't understand why the upper bound of k is when x=0, as x has to be a positive real number. Thanks in advance.
algebra-precalculus polynomials
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
$endgroup$
add a comment |
0
$begingroup$
I don't get the answer to the solution of the problem.
I understand that k=c+8, but I don't understand why the upper bound of k is when x=0, as x has to be a positive real number. Thanks in advance.
algebra-precalculus polynomials
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
$endgroup$
add a comment |
0
0
0
$begingroup$
I don't get the answer to the solution of the problem.
I understand that k=c+8, but I don't understand why the upper bound of k is when x=0, as x has to be a positive real number. Thanks in advance.
algebra-precalculus polynomials
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
$endgroup$
I don't get the answer to the solution of the problem.
I understand that k=c+8, but I don't understand why the upper bound of k is when x=0, as x has to be a positive real number. Thanks in advance.
algebra-precalculus polynomials
algebra-precalculus polynomials
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
edited Feb 3 at 0:56
J. W. Tanner
4,7871420
4,7871420
asked Feb 2 at 21:09
varunrvarunr
1
asked Feb 2 at 21:09
varunrvarunr
1
asked Feb 2 at 21:09
varunrvarunr
1
1
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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$begingroup$
If $x<0$ then $x-2<-2,$ so $(x-2)^2>4.$
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
add a comment |
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1 Answer
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1 Answer
1
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0
$begingroup$
If $x<0$ then $x-2<-2,$ so $(x-2)^2>4.$
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
add a comment |
0
$begingroup$
If $x<0$ then $x-2<-2,$ so $(x-2)^2>4.$
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
add a comment |
0
0
0
$begingroup$
If $x<0$ then $x-2<-2,$ so $(x-2)^2>4.$
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
$endgroup$
If $x<0$ then $x-2<-2,$ so $(x-2)^2>4.$
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
answered Feb 2 at 21:15
saulspatzsaulspatz
17.3k31435
17.3k31435
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
add a comment |
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
yes, but doesn't x have to be positive?
$endgroup$
– varunr
Feb 3 at 3:39
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
If there is a negative root $x,$ then $2(x-2)^2>8$ so if $k<8$ there is no negative root. This doesn't show that there exist two positive roots when $k<8.$ Is that what you're asking?
$endgroup$
– saulspatz
Feb 3 at 4:37
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
yes, and it also says that x has to positive in the problem
$endgroup$
– varunr
Feb 5 at 0:17
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
$begingroup$
@varunr Yes, but as $x$ gets close to to $0$, $k$ gets close to $8$. The length of the open interval $(0,8)$ is $8$.
$endgroup$
– saulspatz
Feb 5 at 3:54
add a comment |
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