Mixing
For which values of $m$ I get this for any $x$?
-1
$begingroup$
For which values of $m$ I get this for any $x$?
$$ (2m-4)x^2 + (m+1)x -1 > 3x - 2 $$
functions inequality
asked Jan 23 at 18:27
StackUserStackUser
61
$endgroup$
add a comment |
-1
$begingroup$
For which values of $m$ I get this for any $x$?
$$ (2m-4)x^2 + (m+1)x -1 > 3x - 2 $$
functions inequality
asked Jan 23 at 18:27
StackUserStackUser
61
$endgroup$
1
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41
add a comment |
-1
-1
-1
1
$begingroup$
For which values of $m$ I get this for any $x$?
$$ (2m-4)x^2 + (m+1)x -1 > 3x - 2 $$
functions inequality
asked Jan 23 at 18:27
StackUserStackUser
61
$endgroup$
For which values of $m$ I get this for any $x$?
$$ (2m-4)x^2 + (m+1)x -1 > 3x - 2 $$
functions inequality
functions inequality
asked Jan 23 at 18:27
StackUserStackUser
61
asked Jan 23 at 18:27
StackUserStackUser
61
asked Jan 23 at 18:27
StackUserStackUser
61
asked Jan 23 at 18:27
StackUserStackUser
61
asked Jan 23 at 18:27
StackUserStackUser
61
61
1
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41
add a comment |
1
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41
1
1
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41
add a comment |
2 Answers
2
active
oldest
votes
1
$begingroup$
Hint:
It is equivalent to
$$ (2m-4)x^2+(m-2)x+1 > 0quadtext{for all }x, $$
so it means this is a quadratic polynomial with no real root, and with a positive leading coefficient.
Can you continue?
answered Jan 23 at 18:53
BernardBernard
123k741116
$endgroup$
add a comment |
0
$begingroup$
We see that $m=2$ is valid.
For $mneq2$ we obtain the following system:
$$2m-4>0$$ and $$(m-2)^2-4(2m-4)<0.$$
Can you end it now?
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint:
It is equivalent to
$$ (2m-4)x^2+(m-2)x+1 > 0quadtext{for all }x, $$
so it means this is a quadratic polynomial with no real root, and with a positive leading coefficient.
Can you continue?
answered Jan 23 at 18:53
BernardBernard
123k741116
$endgroup$
add a comment |
1
$begingroup$
Hint:
It is equivalent to
$$ (2m-4)x^2+(m-2)x+1 > 0quadtext{for all }x, $$
so it means this is a quadratic polynomial with no real root, and with a positive leading coefficient.
Can you continue?
answered Jan 23 at 18:53
BernardBernard
123k741116
$endgroup$
add a comment |
1
1
1
$begingroup$
Hint:
It is equivalent to
$$ (2m-4)x^2+(m-2)x+1 > 0quadtext{for all }x, $$
so it means this is a quadratic polynomial with no real root, and with a positive leading coefficient.
Can you continue?
answered Jan 23 at 18:53
BernardBernard
123k741116
$endgroup$
Hint:
It is equivalent to
$$ (2m-4)x^2+(m-2)x+1 > 0quadtext{for all }x, $$
so it means this is a quadratic polynomial with no real root, and with a positive leading coefficient.
Can you continue?
answered Jan 23 at 18:53
BernardBernard
123k741116
answered Jan 23 at 18:53
BernardBernard
123k741116
answered Jan 23 at 18:53
BernardBernard
123k741116
answered Jan 23 at 18:53
BernardBernard
123k741116
123k741116
add a comment |
add a comment |
0
$begingroup$
We see that $m=2$ is valid.
For $mneq2$ we obtain the following system:
$$2m-4>0$$ and $$(m-2)^2-4(2m-4)<0.$$
Can you end it now?
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
0
$begingroup$
We see that $m=2$ is valid.
For $mneq2$ we obtain the following system:
$$2m-4>0$$ and $$(m-2)^2-4(2m-4)<0.$$
Can you end it now?
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
add a comment |
0
0
0
$begingroup$
We see that $m=2$ is valid.
For $mneq2$ we obtain the following system:
$$2m-4>0$$ and $$(m-2)^2-4(2m-4)<0.$$
Can you end it now?
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
We see that $m=2$ is valid.
For $mneq2$ we obtain the following system:
$$2m-4>0$$ and $$(m-2)^2-4(2m-4)<0.$$
Can you end it now?
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 23 at 19:59
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
add a comment |
add a comment |
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1
$begingroup$
Do you have any thoughts on how you might tackle this. Two possible strategies came to my mind, plus some simple observations. So I would suggest you think a bit about possible approaches and let us know what you have tried.
$endgroup$
– Mark Bennet
Jan 23 at 18:33
$begingroup$
As a hint, you want a quadratic function to never cross the $x$-axis, and also have it always positive. When does a function have no real roots? (From there, either the whole graph lies above or below the $x$-axis. What factor determines which of these is the case?)
$endgroup$
– KM101
Jan 23 at 18:35
$begingroup$
I don't have any idea how to solve it..
$endgroup$
– StackUser
Jan 23 at 18:39
$begingroup$
@KM101, your hint is false. And confused. My hint: put all terms to the left side. What have you learned about sign of a quadratic function ?
$endgroup$
– ama
Jan 23 at 18:39
$begingroup$
@ama How exactly is my hint false?
$endgroup$
– KM101
Jan 23 at 18:41