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Why does this quadratic expression have three zeroes?
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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
$endgroup$
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
2
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago
add a comment |
$begingroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
$endgroup$
Why does this quadratic expression have three zeroes? I understand that when coefficient of $x^2$, coefficient of $x$, and constant term becomes zero, it becomes an identity and has infinitely many solutions (when equates to zero).
But this quadratic expression, it’s coefficients are very small, but they are definitely not equal to zero. Why does this expression have three zeroes then?
roots quadratics
roots quadratics
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
asked 8 hours ago
π times eπ times e
1512 silver badges9 bronze badges
1512 silver badges9 bronze badges
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
2
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago
add a comment |
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
2
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
2
2
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 8 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
$endgroup$
There are zeros at $(2,0)$ and $(3,0)$
The other point of $(2.5,-0.00025)$ is not a zero, but may have been shown rounded; it is the minimum and also the vertex of the parabola
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
answered 8 hours ago
HenryHenry
105k4 gold badges85 silver badges173 bronze badges
105k4 gold badges85 silver badges173 bronze badges
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
$begingroup$
I just put 2.5 for 𝑥 and got (-0.00025) Actually I got confused when I first saw this. Our maths teacher recently taught us about identities, when all the coefficients are zero. So I was just playing with different numbers, wondering what would happen if all the coefficients became very small but still non-zero. I was so startled when I saw this expression, that I forgot to plug 2.5 for x and check if I got zero, lol. And thanks
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 8 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 8 hours ago
DavidDavid
77410 bronze badges
$endgroup$
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 8 hours ago
DavidDavid
77410 bronze badges
$endgroup$
It is not. Possibly a rounding error. Zeroes for that polynomial are at $x=2$ and $x=3$
There is a minimum at $x=2.5$, hitting the value -0.00025, which is also a possible reason why the point is highlighted
answered 8 hours ago
DavidDavid
77410 bronze badges
edited 8 hours ago
answered 8 hours ago
DavidDavid
77410 bronze badges
answered 8 hours ago
DavidDavid
77410 bronze badges
answered 8 hours ago
DavidDavid
77410 bronze badges
77410 bronze badges
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
$begingroup$
Plugging in 2.5 for $x$ gets us ($0.01225$ - $0.01250$), which is not equal to zero. I just noticed it. Thanks for clarifying
$endgroup$
– π times e
8 hours ago
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
add a comment |
$begingroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
$endgroup$
It cannot have more than 2 zeros. The fundamental theorem of algebra tells us it has exactly 2 zeros, possibly repeated.
The zeros of
$$
c(x^2-5x+6)
$$
where $c$ is any (non-zero) constant, are the same as the zeros of,
$$
(x^2-5x+6)
$$
What you are running in to is the real-life practicalities of using a computational tool to determine the zeros. The plot you show displays the $y$ values with no decimal places. Things to try
[1] Can you display the $y$ to more decimal places?
[2] Can you change the scales of the plot to get a better look?
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
answered 7 hours ago
PM.PM.
3,6182 gold badges9 silver badges25 bronze badges
3,6182 gold badges9 silver badges25 bronze badges
add a comment |
add a comment |
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$begingroup$
It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
$endgroup$
– SampleTime
8 hours ago
$begingroup$
Try $ 0.001left(xcdot x-5x+6right)=0$ instead
$endgroup$
– lhf
8 hours ago
2
$begingroup$
It doesn't. I think the software is trying to show two zeros and the vertex. But the vertex is sufficiently close to 0 (it is -0.0025), that it has rounded to 0 in the display.
$endgroup$
– Doug M
8 hours ago
$begingroup$
Yeah, thanks. I just plugged in 2.5 for $x$ and got (-0.00025) I got confused when I first saw this. Because our maths teacher recently taught us about identities (when all the coefficients are zero).
$endgroup$
– π times e
8 hours ago
$begingroup$
If in doubt, zoom in! Desmos has pretty good resolution and will warn you when you might see a rendering error.
$endgroup$
– Jam
8 hours ago