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Relevance of Complex roots of Quadratic Equation
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Let's say Amy is a stunt pilot, planning on doing a parabolic dive in an air show:
$y = x^2 + 4x +5$
She hopes to use this trajectory to dive close to the ground (the x-axis, height is the y-axis), and pull up just before crashing to impress everyone.
To make sure she won't die, she asks two mathematicians to ensure the lack of roots of her parabola.
The first one, Betty, arrives at $x = -2 pm sqrt-1$. She concludes that since there is no number that satisfies that statement, there is no root. Amy will never hit the ground, and everybody will be impressed.
The second one, Charlie, remembers imaginary numbers, and arrives instead at $x = -2 pm i$. Desperate, she urges Amy to cancel the air show for she will die a horrible death when her plane reaches $x = -2 pm i$.
What should Amy do?
Or, what's the point of complex roots?
I am always told that complex numbers are "true" numbers, just as much as negative numbers or fractions. Yet in this example it seems to me that complex numbers "don't exist".
complex-numbers roots quadratics
asked Jul 21 at 14:56
Juan Perez
255
add a comment |Â
up vote
2
down vote
favorite
Let's say Amy is a stunt pilot, planning on doing a parabolic dive in an air show:
$y = x^2 + 4x +5$
She hopes to use this trajectory to dive close to the ground (the x-axis, height is the y-axis), and pull up just before crashing to impress everyone.
To make sure she won't die, she asks two mathematicians to ensure the lack of roots of her parabola.
The first one, Betty, arrives at $x = -2 pm sqrt-1$. She concludes that since there is no number that satisfies that statement, there is no root. Amy will never hit the ground, and everybody will be impressed.
The second one, Charlie, remembers imaginary numbers, and arrives instead at $x = -2 pm i$. Desperate, she urges Amy to cancel the air show for she will die a horrible death when her plane reaches $x = -2 pm i$.
What should Amy do?
Or, what's the point of complex roots?
I am always told that complex numbers are "true" numbers, just as much as negative numbers or fractions. Yet in this example it seems to me that complex numbers "don't exist".
complex-numbers roots quadratics
asked Jul 21 at 14:56
Juan Perez
255
7
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
1
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let's say Amy is a stunt pilot, planning on doing a parabolic dive in an air show:
$y = x^2 + 4x +5$
She hopes to use this trajectory to dive close to the ground (the x-axis, height is the y-axis), and pull up just before crashing to impress everyone.
To make sure she won't die, she asks two mathematicians to ensure the lack of roots of her parabola.
The first one, Betty, arrives at $x = -2 pm sqrt-1$. She concludes that since there is no number that satisfies that statement, there is no root. Amy will never hit the ground, and everybody will be impressed.
The second one, Charlie, remembers imaginary numbers, and arrives instead at $x = -2 pm i$. Desperate, she urges Amy to cancel the air show for she will die a horrible death when her plane reaches $x = -2 pm i$.
What should Amy do?
Or, what's the point of complex roots?
I am always told that complex numbers are "true" numbers, just as much as negative numbers or fractions. Yet in this example it seems to me that complex numbers "don't exist".
complex-numbers roots quadratics
asked Jul 21 at 14:56
Juan Perez
255
Let's say Amy is a stunt pilot, planning on doing a parabolic dive in an air show:
$y = x^2 + 4x +5$
She hopes to use this trajectory to dive close to the ground (the x-axis, height is the y-axis), and pull up just before crashing to impress everyone.
To make sure she won't die, she asks two mathematicians to ensure the lack of roots of her parabola.
The first one, Betty, arrives at $x = -2 pm sqrt-1$. She concludes that since there is no number that satisfies that statement, there is no root. Amy will never hit the ground, and everybody will be impressed.
The second one, Charlie, remembers imaginary numbers, and arrives instead at $x = -2 pm i$. Desperate, she urges Amy to cancel the air show for she will die a horrible death when her plane reaches $x = -2 pm i$.
What should Amy do?
Or, what's the point of complex roots?
I am always told that complex numbers are "true" numbers, just as much as negative numbers or fractions. Yet in this example it seems to me that complex numbers "don't exist".
complex-numbers roots quadratics
asked Jul 21 at 14:56
Juan Perez
255
asked Jul 21 at 14:56
Juan Perez
255
asked Jul 21 at 14:56
Juan Perez
255
asked Jul 21 at 14:56
Juan Perez
255
255
7
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
1
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15
add a comment |Â
7
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
1
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15
7
7
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
1
1
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
7
down vote
accepted
And now suppose that Amy wants to compute the dimensions of a rectangular field with $48$ km2 of which she also know that the largest sides have two more kilometers than the shortest ones. This leads her to the equation $x(x+2)=48$, which has two solutions: $6$ and $-8$. And she will ignore the answer $-8$, since, although $-8$ is indeed a number, it cannot possibly be a length, which is what she's after.
A similar situation occurs with the solutions $-2pm i$. They are numbers, but these number have no meaning in this specific situation.
edited Jul 21 at 18:55
Glorfindel
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
add a comment |Â
up vote
2
down vote
Amy should find a real (pun intended) mathematician, like for instance Mabel, who knows a little calculus. What Mabel would say is this:
If the trajectory
$y(x) = x^2 + 4x + 5 tag 1$
attains a minimum value of height $y_m$, it occurs at that value $x_m$ of $x$ where
$y'(x) = 2x + 4 = 0; tag 2$
that is, at
$x = x_m = -2; tag 3$
Mabel knows $x_m$ is in fact a minimum of $y(x)$ since she evaluates
$y''(x) = 2 > 0 ; textfor all ; x; tag 4$
in particular,
$y''(x_m) = 2 > 0, tag 5$
so Mabel concludes that $y(x_m)$ the smallest possible height the plane will attain if it adheres rigorously to the dive plan $y(x)$; with just a tad more arithmetic, Mabel finds that
$y_m = y(x_m) = y(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1; tag 6$
so Mabel confidently tells Amy that her plane won't crash, since it will never fly lower than $1$ unit of height above the ground.
After living through her successful dive, Amy then decides to return to her local community college for a refresher course in basic calculus, where she comes to understand it is $x_m$ which really matters here, not the zeroes $-2 pm i$ of $y(x)$.
answered Jul 21 at 16:15
Robert Lewis
37k22255
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
add a comment |Â
up vote
0
down vote
I guess there is logical problem in your real life situation you described there. The parabola you defined by $y=x^2+4x+5$, or to put it in another form $y=(x+2)^2+1$ is function which gots no real roots since it gots a vertex at the point $(-2,1)$.Furthermore the two ways you wrote down the roots of the equation, $x_1,2=-2pmsqrt-1$ and $x_1,2=-2pm i$, both stand for the same number.
That means Amy just flips the airplane around at the point $(-2,1)$. When you consider only real roots there are 3 cases for quadratic equations: 2 dsitinctive roots, one doubled root and no real roots.
Only THESE cases are relevant for a real life situation like you described. But to go further in a mathematical context complex roots are more valuable. Without them the Fundamental Theorem of Algebra, which guarentees the existence of - at least - roots for any polynomial, would not work out.
answered Jul 21 at 15:10
mrtaurho
700219
add a comment |Â
up vote
0
down vote
Amy should go for Betty.
In the case you describe the reality is translated to a mathematical model.
Then Betty applies some mathematics in that model and reaches conclusions, and finally a converse translation of conclusions takes place from model to reality.
If this translation cannot take place then something is wrong with the model.
E.g. it can be taken "too wide".
This can be said of the model of Charlie.
This is mainly how it works between math and reality. First the (inspiring) links with reality are somehow "modded out", then (abstract) mathematics takes place leading to conclusions and finally these are projected on reality again.
answered Jul 21 at 15:27
drhab
86.5k541118
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
And now suppose that Amy wants to compute the dimensions of a rectangular field with $48$ km2 of which she also know that the largest sides have two more kilometers than the shortest ones. This leads her to the equation $x(x+2)=48$, which has two solutions: $6$ and $-8$. And she will ignore the answer $-8$, since, although $-8$ is indeed a number, it cannot possibly be a length, which is what she's after.
A similar situation occurs with the solutions $-2pm i$. They are numbers, but these number have no meaning in this specific situation.
edited Jul 21 at 18:55
Glorfindel
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
add a comment |Â
up vote
7
down vote
accepted
And now suppose that Amy wants to compute the dimensions of a rectangular field with $48$ km2 of which she also know that the largest sides have two more kilometers than the shortest ones. This leads her to the equation $x(x+2)=48$, which has two solutions: $6$ and $-8$. And she will ignore the answer $-8$, since, although $-8$ is indeed a number, it cannot possibly be a length, which is what she's after.
A similar situation occurs with the solutions $-2pm i$. They are numbers, but these number have no meaning in this specific situation.
edited Jul 21 at 18:55
Glorfindel
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
And now suppose that Amy wants to compute the dimensions of a rectangular field with $48$ km2 of which she also know that the largest sides have two more kilometers than the shortest ones. This leads her to the equation $x(x+2)=48$, which has two solutions: $6$ and $-8$. And she will ignore the answer $-8$, since, although $-8$ is indeed a number, it cannot possibly be a length, which is what she's after.
A similar situation occurs with the solutions $-2pm i$. They are numbers, but these number have no meaning in this specific situation.
edited Jul 21 at 18:55
Glorfindel
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
And now suppose that Amy wants to compute the dimensions of a rectangular field with $48$ km2 of which she also know that the largest sides have two more kilometers than the shortest ones. This leads her to the equation $x(x+2)=48$, which has two solutions: $6$ and $-8$. And she will ignore the answer $-8$, since, although $-8$ is indeed a number, it cannot possibly be a length, which is what she's after.
A similar situation occurs with the solutions $-2pm i$. They are numbers, but these number have no meaning in this specific situation.
edited Jul 21 at 18:55
Glorfindel
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
edited Jul 21 at 18:55
Glorfindel
3,19371729
edited Jul 21 at 18:55
Glorfindel
3,19371729
edited Jul 21 at 18:55
Glorfindel
3,19371729
3,19371729
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
answered Jul 21 at 15:05
José Carlos Santos
114k1698177
114k1698177
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
add a comment |Â
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
So valid answers must sometimes be discarded, if they make no sense. I though maybe there was always a meaning to them. Thanks.
– Juan Perez
Jul 21 at 18:54
1
1
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
@JuanPerez: The model would be that her curve is given by the graph of the parabola $y(x)$ where $x$ and $y$ are elements of $Bbb R$, the real numbers. This restriction is an important part of the model, although it is often just glossed over. If something was modeled by the graph of $y(x)$ as function of complex numbers $Bbb Crightarrow Bbb C$, (which would be an object in fourdimensional space of which our parabola is just a slice), then the zeros of that might be important.
– Torsten Schoeneberg
Jul 21 at 19:10
add a comment |Â
up vote
2
down vote
Amy should find a real (pun intended) mathematician, like for instance Mabel, who knows a little calculus. What Mabel would say is this:
If the trajectory
$y(x) = x^2 + 4x + 5 tag 1$
attains a minimum value of height $y_m$, it occurs at that value $x_m$ of $x$ where
$y'(x) = 2x + 4 = 0; tag 2$
that is, at
$x = x_m = -2; tag 3$
Mabel knows $x_m$ is in fact a minimum of $y(x)$ since she evaluates
$y''(x) = 2 > 0 ; textfor all ; x; tag 4$
in particular,
$y''(x_m) = 2 > 0, tag 5$
so Mabel concludes that $y(x_m)$ the smallest possible height the plane will attain if it adheres rigorously to the dive plan $y(x)$; with just a tad more arithmetic, Mabel finds that
$y_m = y(x_m) = y(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1; tag 6$
so Mabel confidently tells Amy that her plane won't crash, since it will never fly lower than $1$ unit of height above the ground.
After living through her successful dive, Amy then decides to return to her local community college for a refresher course in basic calculus, where she comes to understand it is $x_m$ which really matters here, not the zeroes $-2 pm i$ of $y(x)$.
answered Jul 21 at 16:15
Robert Lewis
37k22255
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
add a comment |Â
up vote
2
down vote
Amy should find a real (pun intended) mathematician, like for instance Mabel, who knows a little calculus. What Mabel would say is this:
If the trajectory
$y(x) = x^2 + 4x + 5 tag 1$
attains a minimum value of height $y_m$, it occurs at that value $x_m$ of $x$ where
$y'(x) = 2x + 4 = 0; tag 2$
that is, at
$x = x_m = -2; tag 3$
Mabel knows $x_m$ is in fact a minimum of $y(x)$ since she evaluates
$y''(x) = 2 > 0 ; textfor all ; x; tag 4$
in particular,
$y''(x_m) = 2 > 0, tag 5$
so Mabel concludes that $y(x_m)$ the smallest possible height the plane will attain if it adheres rigorously to the dive plan $y(x)$; with just a tad more arithmetic, Mabel finds that
$y_m = y(x_m) = y(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1; tag 6$
so Mabel confidently tells Amy that her plane won't crash, since it will never fly lower than $1$ unit of height above the ground.
After living through her successful dive, Amy then decides to return to her local community college for a refresher course in basic calculus, where she comes to understand it is $x_m$ which really matters here, not the zeroes $-2 pm i$ of $y(x)$.
answered Jul 21 at 16:15
Robert Lewis
37k22255
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Amy should find a real (pun intended) mathematician, like for instance Mabel, who knows a little calculus. What Mabel would say is this:
If the trajectory
$y(x) = x^2 + 4x + 5 tag 1$
attains a minimum value of height $y_m$, it occurs at that value $x_m$ of $x$ where
$y'(x) = 2x + 4 = 0; tag 2$
that is, at
$x = x_m = -2; tag 3$
Mabel knows $x_m$ is in fact a minimum of $y(x)$ since she evaluates
$y''(x) = 2 > 0 ; textfor all ; x; tag 4$
in particular,
$y''(x_m) = 2 > 0, tag 5$
so Mabel concludes that $y(x_m)$ the smallest possible height the plane will attain if it adheres rigorously to the dive plan $y(x)$; with just a tad more arithmetic, Mabel finds that
$y_m = y(x_m) = y(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1; tag 6$
so Mabel confidently tells Amy that her plane won't crash, since it will never fly lower than $1$ unit of height above the ground.
After living through her successful dive, Amy then decides to return to her local community college for a refresher course in basic calculus, where she comes to understand it is $x_m$ which really matters here, not the zeroes $-2 pm i$ of $y(x)$.
answered Jul 21 at 16:15
Robert Lewis
37k22255
Amy should find a real (pun intended) mathematician, like for instance Mabel, who knows a little calculus. What Mabel would say is this:
If the trajectory
$y(x) = x^2 + 4x + 5 tag 1$
attains a minimum value of height $y_m$, it occurs at that value $x_m$ of $x$ where
$y'(x) = 2x + 4 = 0; tag 2$
that is, at
$x = x_m = -2; tag 3$
Mabel knows $x_m$ is in fact a minimum of $y(x)$ since she evaluates
$y''(x) = 2 > 0 ; textfor all ; x; tag 4$
in particular,
$y''(x_m) = 2 > 0, tag 5$
so Mabel concludes that $y(x_m)$ the smallest possible height the plane will attain if it adheres rigorously to the dive plan $y(x)$; with just a tad more arithmetic, Mabel finds that
$y_m = y(x_m) = y(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1; tag 6$
so Mabel confidently tells Amy that her plane won't crash, since it will never fly lower than $1$ unit of height above the ground.
After living through her successful dive, Amy then decides to return to her local community college for a refresher course in basic calculus, where she comes to understand it is $x_m$ which really matters here, not the zeroes $-2 pm i$ of $y(x)$.
answered Jul 21 at 16:15
Robert Lewis
37k22255
answered Jul 21 at 16:15
Robert Lewis
37k22255
answered Jul 21 at 16:15
Robert Lewis
37k22255
answered Jul 21 at 16:15
Robert Lewis
37k22255
37k22255
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
add a comment |Â
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
2
2
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
It might be more instructive to find the minimum by completing the square rather than calculus here, because you can jump easily from the completed square form to the reason the complex roots were $-2 pm i$
– Silverfish
Jul 21 at 19:42
add a comment |Â
up vote
0
down vote
I guess there is logical problem in your real life situation you described there. The parabola you defined by $y=x^2+4x+5$, or to put it in another form $y=(x+2)^2+1$ is function which gots no real roots since it gots a vertex at the point $(-2,1)$.Furthermore the two ways you wrote down the roots of the equation, $x_1,2=-2pmsqrt-1$ and $x_1,2=-2pm i$, both stand for the same number.
That means Amy just flips the airplane around at the point $(-2,1)$. When you consider only real roots there are 3 cases for quadratic equations: 2 dsitinctive roots, one doubled root and no real roots.
Only THESE cases are relevant for a real life situation like you described. But to go further in a mathematical context complex roots are more valuable. Without them the Fundamental Theorem of Algebra, which guarentees the existence of - at least - roots for any polynomial, would not work out.
answered Jul 21 at 15:10
mrtaurho
700219
add a comment |Â
up vote
0
down vote
I guess there is logical problem in your real life situation you described there. The parabola you defined by $y=x^2+4x+5$, or to put it in another form $y=(x+2)^2+1$ is function which gots no real roots since it gots a vertex at the point $(-2,1)$.Furthermore the two ways you wrote down the roots of the equation, $x_1,2=-2pmsqrt-1$ and $x_1,2=-2pm i$, both stand for the same number.
That means Amy just flips the airplane around at the point $(-2,1)$. When you consider only real roots there are 3 cases for quadratic equations: 2 dsitinctive roots, one doubled root and no real roots.
Only THESE cases are relevant for a real life situation like you described. But to go further in a mathematical context complex roots are more valuable. Without them the Fundamental Theorem of Algebra, which guarentees the existence of - at least - roots for any polynomial, would not work out.
answered Jul 21 at 15:10
mrtaurho
700219
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I guess there is logical problem in your real life situation you described there. The parabola you defined by $y=x^2+4x+5$, or to put it in another form $y=(x+2)^2+1$ is function which gots no real roots since it gots a vertex at the point $(-2,1)$.Furthermore the two ways you wrote down the roots of the equation, $x_1,2=-2pmsqrt-1$ and $x_1,2=-2pm i$, both stand for the same number.
That means Amy just flips the airplane around at the point $(-2,1)$. When you consider only real roots there are 3 cases for quadratic equations: 2 dsitinctive roots, one doubled root and no real roots.
Only THESE cases are relevant for a real life situation like you described. But to go further in a mathematical context complex roots are more valuable. Without them the Fundamental Theorem of Algebra, which guarentees the existence of - at least - roots for any polynomial, would not work out.
answered Jul 21 at 15:10
mrtaurho
700219
I guess there is logical problem in your real life situation you described there. The parabola you defined by $y=x^2+4x+5$, or to put it in another form $y=(x+2)^2+1$ is function which gots no real roots since it gots a vertex at the point $(-2,1)$.Furthermore the two ways you wrote down the roots of the equation, $x_1,2=-2pmsqrt-1$ and $x_1,2=-2pm i$, both stand for the same number.
That means Amy just flips the airplane around at the point $(-2,1)$. When you consider only real roots there are 3 cases for quadratic equations: 2 dsitinctive roots, one doubled root and no real roots.
Only THESE cases are relevant for a real life situation like you described. But to go further in a mathematical context complex roots are more valuable. Without them the Fundamental Theorem of Algebra, which guarentees the existence of - at least - roots for any polynomial, would not work out.
answered Jul 21 at 15:10
mrtaurho
700219
edited Jul 21 at 15:33
answered Jul 21 at 15:10
mrtaurho
700219
answered Jul 21 at 15:10
mrtaurho
700219
answered Jul 21 at 15:10
mrtaurho
700219
700219
add a comment |Â
add a comment |Â
up vote
0
down vote
Amy should go for Betty.
In the case you describe the reality is translated to a mathematical model.
Then Betty applies some mathematics in that model and reaches conclusions, and finally a converse translation of conclusions takes place from model to reality.
If this translation cannot take place then something is wrong with the model.
E.g. it can be taken "too wide".
This can be said of the model of Charlie.
This is mainly how it works between math and reality. First the (inspiring) links with reality are somehow "modded out", then (abstract) mathematics takes place leading to conclusions and finally these are projected on reality again.
answered Jul 21 at 15:27
drhab
86.5k541118
add a comment |Â
up vote
0
down vote
Amy should go for Betty.
In the case you describe the reality is translated to a mathematical model.
Then Betty applies some mathematics in that model and reaches conclusions, and finally a converse translation of conclusions takes place from model to reality.
If this translation cannot take place then something is wrong with the model.
E.g. it can be taken "too wide".
This can be said of the model of Charlie.
This is mainly how it works between math and reality. First the (inspiring) links with reality are somehow "modded out", then (abstract) mathematics takes place leading to conclusions and finally these are projected on reality again.
answered Jul 21 at 15:27
drhab
86.5k541118
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Amy should go for Betty.
In the case you describe the reality is translated to a mathematical model.
Then Betty applies some mathematics in that model and reaches conclusions, and finally a converse translation of conclusions takes place from model to reality.
If this translation cannot take place then something is wrong with the model.
E.g. it can be taken "too wide".
This can be said of the model of Charlie.
This is mainly how it works between math and reality. First the (inspiring) links with reality are somehow "modded out", then (abstract) mathematics takes place leading to conclusions and finally these are projected on reality again.
answered Jul 21 at 15:27
drhab
86.5k541118
Amy should go for Betty.
In the case you describe the reality is translated to a mathematical model.
Then Betty applies some mathematics in that model and reaches conclusions, and finally a converse translation of conclusions takes place from model to reality.
If this translation cannot take place then something is wrong with the model.
E.g. it can be taken "too wide".
This can be said of the model of Charlie.
This is mainly how it works between math and reality. First the (inspiring) links with reality are somehow "modded out", then (abstract) mathematics takes place leading to conclusions and finally these are projected on reality again.
answered Jul 21 at 15:27
drhab
86.5k541118
edited Jul 21 at 15:42
answered Jul 21 at 15:27
drhab
86.5k541118
answered Jul 21 at 15:27
drhab
86.5k541118
answered Jul 21 at 15:27
drhab
86.5k541118
86.5k541118
add a comment |Â
add a comment |Â
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7
Well, Amy should make sure to flight that airplane in the real air and not in the complex one, lest she gets complexically crashed.
– DonAntonio
Jul 21 at 15:02
1
It is enough for Amy to check off her "imagination", i.e. the part linked with imaginary numbers.
– drhab
Jul 21 at 15:02
Amy might crash into a building though. Does she have a phobia of over engineered buildings? If so, she might have a complex complex complex...
– AccidentalFourierTransform
Jul 21 at 21:17
@AccidentalFourierTransform Worse if her fear emerges from many component emotions; then it's a complex complex-complex complex.
– Patrick Stevens
Jul 21 at 22:15