Mdthbs

Mathematica notebook opening off the screen

Optimization terminology: "Exact" v. "Approximate"

What is a solution?

Does throwing a penny at a train stop the train?

How can a dictatorship government be beneficial to a dictator in a post-scarcity society?

How to memorize multiple pieces?

How can I effectively communicate to recruiters that a phone call is not possible?

Did the Vulgar Latin verb "toccare" exist?

Why didn't Thanos kill all the Dwarves on Nidavellir?

Keep milk (or milk alternative) for a day without a fridge

Can fluent English speakers distinguish “steel”, “still” and “steal”?

Print the last, middle and first character of your code

Why does the U.S. tolerate foreign influence from Saudi Arabia and Israel on its domestic policies while not tolerating that from China or Russia?

Why can a destructor change the state of a constant object?

Is anyone advocating the promotion of homosexuality in UK schools?

How can I calculate the sum of 2 random dice out of a 3d6 pool in AnyDice?

Why do people keep referring to Leia as Princess Leia, even after the destruction of Alderaan?

Cracking the Coding Interview — 1.5 One Away

Why isn't there research to build a standard lunar, or Martian mobility platform?

How can one write good dialogue in a story without sounding wooden?

Why doesn't sea level show seasonality?

US Civil War story: man hanged from a bridge

Are neural networks prone to catastrophic forgetting?

Confirming the Identity of a (Friendly) Reviewer After the Reviews

Why does this quadratic expression have three zeroes?

ODE with complex char roots gives strange solutionsQuadratic equations: Why does factoring by grouping work?Find roots for an equation with quadratic, linear and log terms?Extracting factor from quadrinomialInfluence of small constant term on roots of polynomialtaking coefficient of highest order term zero in a polynomial and solving itQuestion related to a proof in Quadratic equationsPolynomial p(a)=1, why does it have at most 2 integer roots?Why are Quadratic progressions called so and how is general term of a quadratic sequence $an^2 + bn + c$?Roots of quadratic equation $ax^2+bx+c$ at infinity

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

2

$begingroup$

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

share|cite|improve this question

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
  $endgroup$
  – SampleTime
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try $ 0.001left(xcdot x-5x+6right)=0$ instead
  $endgroup$
  – lhf
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 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
  9 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
  9 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It doesn't: $2.5^2 - 5 times 2.5 + 6 = -1/4 neq 0$.
  $endgroup$
  – SampleTime
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try $ 0.001left(xcdot x-5x+6right)=0$ instead
  $endgroup$
  – lhf
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 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
  9 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
  9 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

share|cite|improve this question

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

share|cite|improve this question

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

asked 9 hours ago

π times eπ times e

14922 silver badges99 bronze badges

3 Answers
3

active

oldest

votes

3

$begingroup$

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

share|cite|improve this answer

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 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
  9 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

DavidDavid

7741010 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
  9 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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?

share|cite|improve this answer

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3289009%2fwhy-does-this-quadratic-expression-have-three-zeroes%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

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

share|cite|improve this answer

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 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
  9 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 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
  9 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

share|cite|improve this answer

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 bronze badges

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 bronze badges

answered 9 hours ago

HenryHenry

105k44 gold badges8585 silver badges173173 bronze badges

3

$begingroup$

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

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

DavidDavid

7741010 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
  9 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

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

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

DavidDavid

7741010 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
  9 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

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

DavidDavid

7741010 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

DavidDavid

7741010 bronze badges

answered 9 hours ago

DavidDavid

7741010 bronze badges

answered 9 hours ago

DavidDavid

7741010 bronze badges

2

$begingroup$

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?

share|cite|improve this answer

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

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?

share|cite|improve this answer

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 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?

share|cite|improve this answer

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges

answered 9 hours ago

PM.PM.

3,61822 gold badges99 silver badges2525 bronze badges