An easier way of solving a problem about the number of roots of a square equation on an interval

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

I am trying to solve the following problem: I need to find out the number of roots on the interval $[-1; 3)$ of the equation $(4-a)x^2-6ax+3=0$ depending from $a$. I know a solution, but it is too difficult to calculate. I thought I could solve that problem with the help of quadratic trinomial roots position theorem. In the end, I got a lot of inequalities and I got confused. Is there an easier way to solve this problem?

polynomials

share|cite|improve this question

edited Jul 18 at 18:02

Robert Frost

3,8821036

asked Jul 18 at 17:49

student28

13611

add a comment | 

up vote
1
down vote

favorite

I am trying to solve the following problem: I need to find out the number of roots on the interval $[-1; 3)$ of the equation $(4-a)x^2-6ax+3=0$ depending from $a$. I know a solution, but it is too difficult to calculate. I thought I could solve that problem with the help of quadratic trinomial roots position theorem. In the end, I got a lot of inequalities and I got confused. Is there an easier way to solve this problem?

polynomials

share|cite|improve this question

edited Jul 18 at 18:02

Robert Frost

3,8821036

asked Jul 18 at 17:49

student28

13611

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am trying to solve the following problem: I need to find out the number of roots on the interval $[-1; 3)$ of the equation $(4-a)x^2-6ax+3=0$ depending from $a$. I know a solution, but it is too difficult to calculate. I thought I could solve that problem with the help of quadratic trinomial roots position theorem. In the end, I got a lot of inequalities and I got confused. Is there an easier way to solve this problem?

polynomials

share|cite|improve this question

edited Jul 18 at 18:02

Robert Frost

3,8821036

asked Jul 18 at 17:49

student28

13611

I am trying to solve the following problem: I need to find out the number of roots on the interval $[-1; 3)$ of the equation $(4-a)x^2-6ax+3=0$ depending from $a$. I know a solution, but it is too difficult to calculate. I thought I could solve that problem with the help of quadratic trinomial roots position theorem. In the end, I got a lot of inequalities and I got confused. Is there an easier way to solve this problem?

polynomials

share|cite|improve this question

edited Jul 18 at 18:02

Robert Frost

3,8821036

asked Jul 18 at 17:49

student28

13611

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 18:02

Robert Frost

3,8821036

edited Jul 18 at 18:02

Robert Frost

3,8821036

edited Jul 18 at 18:02

Robert Frost

3,8821036

3,8821036

asked Jul 18 at 17:49

student28

13611

asked Jul 18 at 17:49

student28

13611

asked Jul 18 at 17:49

student28

13611

13611

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

The quadratic formula yields
$$x=frac6apmsqrt36a^2-12(4-a)2(4-a). $$
Let us set this as two functions of $a:$
$$f_pm(a)=frac6apmsqrt36a^2-12(4-a)2(4-a).$$
These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root:
$$36a^2+12a-48=12(a-1)(3a+4).$$
This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_pm$ is $(-infty,-4/3]cup[1,4)cup(4,infty). $

It remains to find the range of $f_pm,$ which is what we really care about. This will be different depending on whether we are looking at $f_+$ or $f_-.$ For $f_+,$ a simple plot reveals that $f_pm(-4/3)=-3/4,; f_pm(a)=1,$ that $$lim_ato-inftyf_+(a)=0, quad lim_ato-inftyf_-(a)=-6. $$
We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $ain(-infty,-7/5),$ there is only one root. For $ain[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $ain(-4/3,1),$ there are none. For $a=1,$ there is one. For $ain(1,13/9],$ there are two. And finally, for $ain(13/9,infty),$ there is only one.

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
  – student28
  Jul 18 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
  – Adrian Keister
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

Try using the quadratic formula. This will give $x = frac3a pm sqrt9a^2+3a-124-a$.

Now you want to find values of $x in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.

You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 geq 0$)

Using the above, you should be able to determine a range of values for $a$.

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855832%2fan-easier-way-of-solving-a-problem-about-the-number-of-roots-of-a-square-equatio%23new-answer', 'question_page');

);

Post as a guest

Name Email

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The quadratic formula yields
$$x=frac6apmsqrt36a^2-12(4-a)2(4-a). $$
Let us set this as two functions of $a:$
$$f_pm(a)=frac6apmsqrt36a^2-12(4-a)2(4-a).$$
These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root:
$$36a^2+12a-48=12(a-1)(3a+4).$$
This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_pm$ is $(-infty,-4/3]cup[1,4)cup(4,infty). $

It remains to find the range of $f_pm,$ which is what we really care about. This will be different depending on whether we are looking at $f_+$ or $f_-.$ For $f_+,$ a simple plot reveals that $f_pm(-4/3)=-3/4,; f_pm(a)=1,$ that $$lim_ato-inftyf_+(a)=0, quad lim_ato-inftyf_-(a)=-6. $$
We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $ain(-infty,-7/5),$ there is only one root. For $ain[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $ain(-4/3,1),$ there are none. For $a=1,$ there is one. For $ain(1,13/9],$ there are two. And finally, for $ain(13/9,infty),$ there is only one.

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
  – student28
  Jul 18 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
  – Adrian Keister
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The quadratic formula yields
$$x=frac6apmsqrt36a^2-12(4-a)2(4-a). $$
Let us set this as two functions of $a:$
$$f_pm(a)=frac6apmsqrt36a^2-12(4-a)2(4-a).$$
These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root:
$$36a^2+12a-48=12(a-1)(3a+4).$$
This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_pm$ is $(-infty,-4/3]cup[1,4)cup(4,infty). $

It remains to find the range of $f_pm,$ which is what we really care about. This will be different depending on whether we are looking at $f_+$ or $f_-.$ For $f_+,$ a simple plot reveals that $f_pm(-4/3)=-3/4,; f_pm(a)=1,$ that $$lim_ato-inftyf_+(a)=0, quad lim_ato-inftyf_-(a)=-6. $$
We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $ain(-infty,-7/5),$ there is only one root. For $ain[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $ain(-4/3,1),$ there are none. For $a=1,$ there is one. For $ain(1,13/9],$ there are two. And finally, for $ain(13/9,infty),$ there is only one.

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
  – student28
  Jul 18 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
  – Adrian Keister
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The quadratic formula yields
$$x=frac6apmsqrt36a^2-12(4-a)2(4-a). $$
Let us set this as two functions of $a:$
$$f_pm(a)=frac6apmsqrt36a^2-12(4-a)2(4-a).$$
These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root:
$$36a^2+12a-48=12(a-1)(3a+4).$$
This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_pm$ is $(-infty,-4/3]cup[1,4)cup(4,infty). $

It remains to find the range of $f_pm,$ which is what we really care about. This will be different depending on whether we are looking at $f_+$ or $f_-.$ For $f_+,$ a simple plot reveals that $f_pm(-4/3)=-3/4,; f_pm(a)=1,$ that $$lim_ato-inftyf_+(a)=0, quad lim_ato-inftyf_-(a)=-6. $$
We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $ain(-infty,-7/5),$ there is only one root. For $ain[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $ain(-4/3,1),$ there are none. For $a=1,$ there is one. For $ain(1,13/9],$ there are two. And finally, for $ain(13/9,infty),$ there is only one.

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

The quadratic formula yields
$$x=frac6apmsqrt36a^2-12(4-a)2(4-a). $$
Let us set this as two functions of $a:$
$$f_pm(a)=frac6apmsqrt36a^2-12(4-a)2(4-a).$$
These functions are undefined for $a=4.$ To find the exact domain, we factor what is under the square root:
$$36a^2+12a-48=12(a-1)(3a+4).$$
This is a parabola opening up, and therefore it is negative in-between the two roots of $-4/3$ and $1$. So the domain of $f_pm$ is $(-infty,-4/3]cup[1,4)cup(4,infty). $

It remains to find the range of $f_pm,$ which is what we really care about. This will be different depending on whether we are looking at $f_+$ or $f_-.$ For $f_+,$ a simple plot reveals that $f_pm(-4/3)=-3/4,; f_pm(a)=1,$ that $$lim_ato-inftyf_+(a)=0, quad lim_ato-inftyf_-(a)=-6. $$
We can piece together the graphs of these functions on the range $[-1,3),$ draw vertical lines, and see how many times we intersect a function. From this we gather that for $ain(-infty,-7/5),$ there is only one root. For $ain[-7/5,-4/3),$ there are two. For $a=-4/3,$ there is only one. For $ain(-4/3,1),$ there are none. For $a=1,$ there is one. For $ain(1,13/9],$ there are two. And finally, for $ain(13/9,infty),$ there is only one.

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 18:18

Adrian Keister

3,61721533

answered Jul 18 at 18:18

Adrian Keister

3,61721533

answered Jul 18 at 18:18

Adrian Keister

3,61721533

3,61721533

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
  – student28
  Jul 18 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
  – Adrian Keister
  Jul 18 at 20:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
  – student28
  Jul 18 at 20:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
  – Adrian Keister
  Jul 18 at 20:09
  

  
  
  
  

I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
– student28
Jul 18 at 20:04

I am sorry, but how did you piece these graphs together? Unfortunately, I didn't understand the range part.
– student28
Jul 18 at 20:04

@student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
– Adrian Keister
Jul 18 at 20:09

@student28: The range of $f_pm$ tells us where $x$ lies, because I defined $x=f_pm(a).$ As for finding values like $-7/5$ and $13/9,$ I simply "cheated" and used Mathematica, via Wolfram Dev Platform, to solve for those values. I also used the Dev Platform to do plotting.
– Adrian Keister
Jul 18 at 20:09

add a comment | 

up vote
0
down vote

Try using the quadratic formula. This will give $x = frac3a pm sqrt9a^2+3a-124-a$.

Now you want to find values of $x in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.

You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 geq 0$)

Using the above, you should be able to determine a range of values for $a$.

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

add a comment | 

up vote
0
down vote

Try using the quadratic formula. This will give $x = frac3a pm sqrt9a^2+3a-124-a$.

Now you want to find values of $x in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.

You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 geq 0$)

Using the above, you should be able to determine a range of values for $a$.

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

add a comment | 

up vote
0
down vote

up vote
0
down vote

Try using the quadratic formula. This will give $x = frac3a pm sqrt9a^2+3a-124-a$.

Now you want to find values of $x in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.

You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 geq 0$)

Using the above, you should be able to determine a range of values for $a$.

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

Try using the quadratic formula. This will give $x = frac3a pm sqrt9a^2+3a-124-a$.

Now you want to find values of $x in [-1,3)$. So try setting $x=-1$ and $x=3$ and solve for $a$.

You can get a bound on $a$ because you are looking for real roots, and the discriminant must be nonnegative ($9a^2+3a-12 geq 0$)

Using the above, you should be able to determine a range of values for $a$.

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 18:13

gd1035

29319

answered Jul 18 at 18:13

gd1035

29319

answered Jul 18 at 18:13

gd1035

29319

29319

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855832%2fan-easier-way-of-solving-a-problem-about-the-number-of-roots-of-a-square-equatio%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email