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How to prove that the quadratic equation has exactly two real solutions

prove that the equation has only on root mix: exponent, quadratic function, parameterProve that equation has exactly 2 solutionsShowing that a equation has exactly on real rootHow to show that an equation has exactly two solutions?Prove the equation $ln(x) = frac1 x-1$ has exactly 2 real solutions.Prove equation has exactly 1 real root

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3

1

$begingroup$

How to prove that the quadratic equation has exactly two real solutions when $Delta=b^2-4ac>0$?

calculus inequality

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edited 8 hours ago

Sebastiano

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• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Completing the square?
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Think of the abc-formula. The solution with the minus sign cannot coincide with the solution with the plus sign.
  $endgroup$
  – Peter
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

3

1

$begingroup$

How to prove that the quadratic equation has exactly two real solutions when $Delta=b^2-4ac>0$?

calculus inequality

share|cite|improve this question

edited 8 hours ago

Sebastiano

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asked 8 hours ago

user707254user707254

1933 bronze badges

New contributor

user707254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Completing the square?
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Think of the abc-formula. The solution with the minus sign cannot coincide with the solution with the plus sign.
  $endgroup$
  – Peter
  8 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

How to prove that the quadratic equation has exactly two real solutions when $Delta=b^2-4ac>0$?

calculus inequality

share|cite|improve this question

edited 8 hours ago

Sebastiano

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asked 8 hours ago

user707254user707254

1933 bronze badges

New contributor

user707254 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Sebastiano

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asked 8 hours ago

user707254user707254

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edited 8 hours ago

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edited 8 hours ago

Sebastiano

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edited 8 hours ago

Sebastiano

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asked 8 hours ago

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5 Answers
5

active

oldest

votes

6

$begingroup$

Let consider

$$f(x)=ax^2+bx+c implies f'(x)=2ax+b=0 ; x_0=-frac b2aimplies f''(x)=2a$$

and

$$f(x_0)=fracb^24a-fracb^22a+c=-fracb^24a+c=-frac14a(b^2-4ac)$$

then

• for $a>0$ $f(x)$ is convex and at the minimum $f(x_0)<0$
  


• for $a<0$ $f(x)$ is concave and at the minimum $f(x_0)>0$

in both cases $f(x)=0$ has exactly two solutions.

share|cite|improve this answer

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Since the function has degree $2$, there are at most $2$.

They are $dfrac-bpmsqrtb^2-4ac2a$, by the quadratic formula.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  maybe the OP is interested in a approach by calculus
  $endgroup$
  – gimusi
  8 hours ago
  

  
  
  
  

add a comment
 | 

5

$begingroup$

Let $ax^2 +bx+c=0$ be your quadratic equaltion. Suppose it have 3 roots :$alpha$ ,$beta$ ,$gamma$ All distinct.
Therefore:
$aalpha^2 +balpha+c=0$
$abeta^2 +bbeta+c=0$
$agamma^2 +bgamma+c=0$

Try to solve equation 1 and 2 you will get
$(alpha -beta)(a(alpha +beta)+b)=0$
Since $alpha neq beta$ ,
$alpha +beta=-fracba$

Now solve 2 and 3 equation
You get the same result
$(gamma -beta) neq 0$
Therfore
$(a(gamma +beta)+b)=0$
Which follows that $gamma =alpha$
Which is not possible
Thus we conclude by contradiction there are only two distict roots not more than that since leaving one root distinct all other roots will coincide.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Let $f(x)=ax^2+bx+c$ with $a>0$.

$f'(x)=2ax+b$

The extremum is attained at $x_0=frac-b2a$

and it is

$$M=afracb^24a^2-bfracb2a+c=$$
$$fracb^24a-fracb^22a+c=$$
$$c-fracb^24a=-fracb^2-4ac4a<0$$
By IVT, there is a root in $(-infty,x_0)$ and one in $(x_0,+infty)$ since
$$lim_f(x)=+infty$$

and it could not have three roots because $f'(x)=0$ has only one.

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Given $tag*ax^2+bx^2+c=0$ we have:

$$tag1 ax^2+bx=-c iff aneq 0,quad x^2+frac bax=-frac ca$$
Completing the square:

$$tag2 x^2+frac bax+fracb^24a^2=fracb^24a^2-frac ca iff left(x+fracb2aright)^2=fracb^2-4ac4a^2=fracDelta4a^2$$
If $Delta<0$ you have not real solution from $(2)$. Hence if $Deltageq 0$ from the $(2)$ you will have two real solutions: two coincident real solutions if $Delta=0$ and two distinct real solutions when $Delta >0$.

share|cite|improve this answer

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

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add a comment
 | 

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6

$begingroup$

Let consider

$$f(x)=ax^2+bx+c implies f'(x)=2ax+b=0 ; x_0=-frac b2aimplies f''(x)=2a$$

and

$$f(x_0)=fracb^24a-fracb^22a+c=-fracb^24a+c=-frac14a(b^2-4ac)$$

then

• for $a>0$ $f(x)$ is convex and at the minimum $f(x_0)<0$
  


• for $a<0$ $f(x)$ is concave and at the minimum $f(x_0)>0$

in both cases $f(x)=0$ has exactly two solutions.

share|cite|improve this answer

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

6

$begingroup$

Let consider

$$f(x)=ax^2+bx+c implies f'(x)=2ax+b=0 ; x_0=-frac b2aimplies f''(x)=2a$$

and

$$f(x_0)=fracb^24a-fracb^22a+c=-fracb^24a+c=-frac14a(b^2-4ac)$$

then

• for $a>0$ $f(x)$ is convex and at the minimum $f(x_0)<0$
  


• for $a<0$ $f(x)$ is concave and at the minimum $f(x_0)>0$

in both cases $f(x)=0$ has exactly two solutions.

share|cite|improve this answer

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let consider

$$f(x)=ax^2+bx+c implies f'(x)=2ax+b=0 ; x_0=-frac b2aimplies f''(x)=2a$$

and

$$f(x_0)=fracb^24a-fracb^22a+c=-fracb^24a+c=-frac14a(b^2-4ac)$$

then

• for $a>0$ $f(x)$ is convex and at the minimum $f(x_0)<0$
  


• for $a<0$ $f(x)$ is concave and at the minimum $f(x_0)>0$

in both cases $f(x)=0$ has exactly two solutions.

share|cite|improve this answer

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

answered 8 hours ago

gimusigimusi

94k88 gold badges4646 silver badges9595 bronze badges

5

$begingroup$

Since the function has degree $2$, there are at most $2$.

They are $dfrac-bpmsqrtb^2-4ac2a$, by the quadratic formula.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  maybe the OP is interested in a approach by calculus
  $endgroup$
  – gimusi
  8 hours ago
  

  
  
  
  

add a comment
 | 

5

$begingroup$

Since the function has degree $2$, there are at most $2$.

They are $dfrac-bpmsqrtb^2-4ac2a$, by the quadratic formula.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  maybe the OP is interested in a approach by calculus
  $endgroup$
  – gimusi
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Since the function has degree $2$, there are at most $2$.

They are $dfrac-bpmsqrtb^2-4ac2a$, by the quadratic formula.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Chris CusterChris Custer

20.5k33 gold badges88 silver badges3232 bronze badges

5

$begingroup$

Let $ax^2 +bx+c=0$ be your quadratic equaltion. Suppose it have 3 roots :$alpha$ ,$beta$ ,$gamma$ All distinct.
Therefore:
$aalpha^2 +balpha+c=0$
$abeta^2 +bbeta+c=0$
$agamma^2 +bgamma+c=0$

Try to solve equation 1 and 2 you will get
$(alpha -beta)(a(alpha +beta)+b)=0$
Since $alpha neq beta$ ,
$alpha +beta=-fracba$

Now solve 2 and 3 equation
You get the same result
$(gamma -beta) neq 0$
Therfore
$(a(gamma +beta)+b)=0$
Which follows that $gamma =alpha$
Which is not possible
Thus we conclude by contradiction there are only two distict roots not more than that since leaving one root distinct all other roots will coincide.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

$endgroup$

add a comment
 | 

5

$begingroup$

Let $ax^2 +bx+c=0$ be your quadratic equaltion. Suppose it have 3 roots :$alpha$ ,$beta$ ,$gamma$ All distinct.
Therefore:
$aalpha^2 +balpha+c=0$
$abeta^2 +bbeta+c=0$
$agamma^2 +bgamma+c=0$

Try to solve equation 1 and 2 you will get
$(alpha -beta)(a(alpha +beta)+b)=0$
Since $alpha neq beta$ ,
$alpha +beta=-fracba$

Now solve 2 and 3 equation
You get the same result
$(gamma -beta) neq 0$
Therfore
$(a(gamma +beta)+b)=0$
Which follows that $gamma =alpha$
Which is not possible
Thus we conclude by contradiction there are only two distict roots not more than that since leaving one root distinct all other roots will coincide.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let $ax^2 +bx+c=0$ be your quadratic equaltion. Suppose it have 3 roots :$alpha$ ,$beta$ ,$gamma$ All distinct.
Therefore:
$aalpha^2 +balpha+c=0$
$abeta^2 +bbeta+c=0$
$agamma^2 +bgamma+c=0$

Try to solve equation 1 and 2 you will get
$(alpha -beta)(a(alpha +beta)+b)=0$
Since $alpha neq beta$ ,
$alpha +beta=-fracba$

Now solve 2 and 3 equation
You get the same result
$(gamma -beta) neq 0$
Therfore
$(a(gamma +beta)+b)=0$
Which follows that $gamma =alpha$
Which is not possible
Thus we conclude by contradiction there are only two distict roots not more than that since leaving one root distinct all other roots will coincide.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

answered 8 hours ago

Akshaj BansalAkshaj Bansal

30288 bronze badges

4

$begingroup$

Let $f(x)=ax^2+bx+c$ with $a>0$.

$f'(x)=2ax+b$

The extremum is attained at $x_0=frac-b2a$

and it is

$$M=afracb^24a^2-bfracb2a+c=$$
$$fracb^24a-fracb^22a+c=$$
$$c-fracb^24a=-fracb^2-4ac4a<0$$
By IVT, there is a root in $(-infty,x_0)$ and one in $(x_0,+infty)$ since
$$lim_f(x)=+infty$$

and it could not have three roots because $f'(x)=0$ has only one.

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Let $f(x)=ax^2+bx+c$ with $a>0$.

$f'(x)=2ax+b$

The extremum is attained at $x_0=frac-b2a$

and it is

$$M=afracb^24a^2-bfracb2a+c=$$
$$fracb^24a-fracb^22a+c=$$
$$c-fracb^24a=-fracb^2-4ac4a<0$$
By IVT, there is a root in $(-infty,x_0)$ and one in $(x_0,+infty)$ since
$$lim_f(x)=+infty$$

and it could not have three roots because $f'(x)=0$ has only one.

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let $f(x)=ax^2+bx+c$ with $a>0$.

$f'(x)=2ax+b$

The extremum is attained at $x_0=frac-b2a$

and it is

$$M=afracb^24a^2-bfracb2a+c=$$
$$fracb^24a-fracb^22a+c=$$
$$c-fracb^24a=-fracb^2-4ac4a<0$$
By IVT, there is a root in $(-infty,x_0)$ and one in $(x_0,+infty)$ since
$$lim_f(x)=+infty$$

and it could not have three roots because $f'(x)=0$ has only one.

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

39.6k22 gold badges1717 silver badges3535 bronze badges

3

$begingroup$

Given $tag*ax^2+bx^2+c=0$ we have:

$$tag1 ax^2+bx=-c iff aneq 0,quad x^2+frac bax=-frac ca$$
Completing the square:

$$tag2 x^2+frac bax+fracb^24a^2=fracb^24a^2-frac ca iff left(x+fracb2aright)^2=fracb^2-4ac4a^2=fracDelta4a^2$$
If $Delta<0$ you have not real solution from $(2)$. Hence if $Deltageq 0$ from the $(2)$ you will have two real solutions: two coincident real solutions if $Delta=0$ and two distinct real solutions when $Delta >0$.

share|cite|improve this answer

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Given $tag*ax^2+bx^2+c=0$ we have:

$$tag1 ax^2+bx=-c iff aneq 0,quad x^2+frac bax=-frac ca$$
Completing the square:

$$tag2 x^2+frac bax+fracb^24a^2=fracb^24a^2-frac ca iff left(x+fracb2aright)^2=fracb^2-4ac4a^2=fracDelta4a^2$$
If $Delta<0$ you have not real solution from $(2)$. Hence if $Deltageq 0$ from the $(2)$ you will have two real solutions: two coincident real solutions if $Delta=0$ and two distinct real solutions when $Delta >0$.

share|cite|improve this answer

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Given $tag*ax^2+bx^2+c=0$ we have:

$$tag1 ax^2+bx=-c iff aneq 0,quad x^2+frac bax=-frac ca$$
Completing the square:

$$tag2 x^2+frac bax+fracb^24a^2=fracb^24a^2-frac ca iff left(x+fracb2aright)^2=fracb^2-4ac4a^2=fracDelta4a^2$$
If $Delta<0$ you have not real solution from $(2)$. Hence if $Deltageq 0$ from the $(2)$ you will have two real solutions: two coincident real solutions if $Delta=0$ and two distinct real solutions when $Delta >0$.

share|cite|improve this answer

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges

answered 8 hours ago

SebastianoSebastiano

1811010 bronze badges