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Proving a result by making discriminant zero

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Proving a result by making discriminant zero

$y =f(x) =(ax^2 + bx +c)/(dx^2+ex+f)$ We have to find the conditions for this it takes all real values.Equation with the discriminantWhat is $textitthe$ discriminant of a degree $n$ polynomial?Question on integral values of a quadraticDifference between “two equal roots” and “one root”?Theory of Equations and co primeWhere am I making the mistake?Intuitive meaning behind the DiscriminantWhat causes the extraneous root when intersecting parabola $y^2=4ax$ with circle $x^2+y^2=9a^2/4$?

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

6

3

$begingroup$

If the roots of given Quadratic equation
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$
are equal, prove the following:
$$frac2b=frac1a+frac1c$$.

MY approach:

Method 1: put Discriminant=0 and get stuck.

Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.

But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous.
Thank you.

algebra-precalculus quadratics

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate you trying to work it directly using discriminant, but your method2 is rly clever!
  $endgroup$
  – ganeshie8
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ganeshie8 grace of ganesh hehehe
  $endgroup$
  – George Carlin
  8 hours ago
  

  
  
  
  

add a comment | 

6

3

$begingroup$

If the roots of given Quadratic equation
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$
are equal, prove the following:
$$frac2b=frac1a+frac1c$$.

MY approach:

Method 1: put Discriminant=0 and get stuck.

Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.

But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous.
Thank you.

algebra-precalculus quadratics

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I appreciate you trying to work it directly using discriminant, but your method2 is rly clever!
  $endgroup$
  – ganeshie8
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ganeshie8 grace of ganesh hehehe
  $endgroup$
  – George Carlin
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

If the roots of given Quadratic equation
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0$$
are equal, prove the following:
$$frac2b=frac1a+frac1c$$.

MY approach:

Method 1: put Discriminant=0 and get stuck.

Method2:add $acx$ on both sides and get $(x-1)$ as factor ,so other root is also 1 and hence the result.

But if someone can prove the result by making discriminant zero (method 1), that would be more rigorous.
Thank you.

algebra-precalculus quadratics

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

share|cite|improve this question

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

edited 8 hours ago

Robert Z

112k1010 gold badges7979 silver badges153153 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

asked 8 hours ago

George CarlinGeorge Carlin

14455 bronze badges

3 Answers
3

active

oldest

votes

9

$begingroup$

Following your first approach, we have that the discriminant is
$$beginalign
Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\
&=(b(a+c)-2ac)^2
endalign.$$
and by letting $Delta=0$, it is easy to show that $frac2b=frac1a+frac1c$.

As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $frac2b=frac1a+frac1c$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that in just one step? I mean... is it an obvious identity or result?
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ab123 I was actually writing up something more or less along these lines. I didn't bother doing the tedious stuff, I just dumped the quadratic into Wolfram, told it to find the discriminant, and lo and behold one of the alternative forms worked: link
  $endgroup$
  – Eevee Trainer
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer wow! Wolfram is quite intelligent
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeCarlin Because $a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2$ is a polynomial identity which holds for any $x$. I edited my answer.
  $endgroup$
  – Robert Z
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What you wrote is not an identity. The identity is $$ a(b-c)x^2+b(c-a)x+c(a-b)-a(b-c)(x-1)^2=(x-1)(ab+bc-2ac)$$ for all $,a,b,c,x.,$ Now if the left side is $0$ and $,xne 1,$ then we get $,ab+bc-2ac=0.$
  $endgroup$
  – Somos
  3 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

If you label:
$$begincasesa(b-c)=p\
b(c-a)=q\
c(a-b)=rendcases Rightarrow p+q+r=0$$
then the equation becomes:
$$px^2+qx+r=0,\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 Rightarrow p=rRightarrow \
a(b-c)=c(a-b) Rightarrow 2ac=ab+bc stackrelcdot frac1abcRightarrow frac2b=frac1a+frac1c.$$

share|cite|improve this answer

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.

With synthetic division by $x-1$, we get, assuming $ane0$ and $bne c$,
$$
beginarraycc
& ab-ac & bc-ab & ac-bc \
1 & downarrow & ab-ac & bc-ac \
hline
& ab-ac & bc-ac & 0
endarray
$$
The other root is
$$
-fracbc-acab-ac
$$
Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes
$$
2ac=bc+ab
$$
and, dividing by $abc$, we get
$$
frac2b=frac1a+frac1c
$$

If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $cne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.

share|cite|improve this answer

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

$endgroup$

add a comment | 

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9

$begingroup$

Following your first approach, we have that the discriminant is
$$beginalign
Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\
&=(b(a+c)-2ac)^2
endalign.$$
and by letting $Delta=0$, it is easy to show that $frac2b=frac1a+frac1c$.

As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $frac2b=frac1a+frac1c$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that in just one step? I mean... is it an obvious identity or result?
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ab123 I was actually writing up something more or less along these lines. I didn't bother doing the tedious stuff, I just dumped the quadratic into Wolfram, told it to find the discriminant, and lo and behold one of the alternative forms worked: link
  $endgroup$
  – Eevee Trainer
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer wow! Wolfram is quite intelligent
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeCarlin Because $a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2$ is a polynomial identity which holds for any $x$. I edited my answer.
  $endgroup$
  – Robert Z
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What you wrote is not an identity. The identity is $$ a(b-c)x^2+b(c-a)x+c(a-b)-a(b-c)(x-1)^2=(x-1)(ab+bc-2ac)$$ for all $,a,b,c,x.,$ Now if the left side is $0$ and $,xne 1,$ then we get $,ab+bc-2ac=0.$
  $endgroup$
  – Somos
  3 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

9

$begingroup$

Following your first approach, we have that the discriminant is
$$beginalign
Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\
&=(b(a+c)-2ac)^2
endalign.$$
and by letting $Delta=0$, it is easy to show that $frac2b=frac1a+frac1c$.

As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $frac2b=frac1a+frac1c$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that in just one step? I mean... is it an obvious identity or result?
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ab123 I was actually writing up something more or less along these lines. I didn't bother doing the tedious stuff, I just dumped the quadratic into Wolfram, told it to find the discriminant, and lo and behold one of the alternative forms worked: link
  $endgroup$
  – Eevee Trainer
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer wow! Wolfram is quite intelligent
  $endgroup$
  – ab123
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GeorgeCarlin Because $a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2$ is a polynomial identity which holds for any $x$. I edited my answer.
  $endgroup$
  – Robert Z
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What you wrote is not an identity. The identity is $$ a(b-c)x^2+b(c-a)x+c(a-b)-a(b-c)(x-1)^2=(x-1)(ab+bc-2ac)$$ for all $,a,b,c,x.,$ Now if the left side is $0$ and $,xne 1,$ then we get $,ab+bc-2ac=0.$
  $endgroup$
  – Somos
  3 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Following your first approach, we have that the discriminant is
$$beginalign
Delta&=b^2(c−a)^2-4a(b−c)c(a−b)\
&=b^2(c−a)^2+4ac(b^2-b(a+c)+ac)\
&=b^2(c+a)^2+(2ac)^2-2b(a+c)(2ac)\
&=(b(a+c)-2ac)^2
endalign.$$
and by letting $Delta=0$, it is easy to show that $frac2b=frac1a+frac1c$.

As regards your second approach, since you already noted that $x=1$ is a root, we have a double root if and only if the following polynomial identity holds (polynomial factorization is unique):
$$a(b-c)x^2 +b(c-a)x + c(a-b)=0=a(b-c)(x-1)^2.$$
Now by letting $x=0$ we find $$c(a-b)=a(b-c)$$ which implies $frac2b=frac1a+frac1c$.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

answered 8 hours ago

Robert ZRobert Z

112k1010 gold badges7979 silver badges153153 bronze badges

2

$begingroup$

If you label:
$$begincasesa(b-c)=p\
b(c-a)=q\
c(a-b)=rendcases Rightarrow p+q+r=0$$
then the equation becomes:
$$px^2+qx+r=0,\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 Rightarrow p=rRightarrow \
a(b-c)=c(a-b) Rightarrow 2ac=ab+bc stackrelcdot frac1abcRightarrow frac2b=frac1a+frac1c.$$

share|cite|improve this answer

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

If you label:
$$begincasesa(b-c)=p\
b(c-a)=q\
c(a-b)=rendcases Rightarrow p+q+r=0$$
then the equation becomes:
$$px^2+qx+r=0,\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 Rightarrow p=rRightarrow \
a(b-c)=c(a-b) Rightarrow 2ac=ab+bc stackrelcdot frac1abcRightarrow frac2b=frac1a+frac1c.$$

share|cite|improve this answer

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

$endgroup$

add a comment | 

$begingroup$

If you label:
$$begincasesa(b-c)=p\
b(c-a)=q\
c(a-b)=rendcases Rightarrow p+q+r=0$$
then the equation becomes:
$$px^2+qx+r=0,\
D=q^2-4pr=(-p-r)^2-4pr=(p-r)^2=0 Rightarrow p=rRightarrow \
a(b-c)=c(a-b) Rightarrow 2ac=ab+bc stackrelcdot frac1abcRightarrow frac2b=frac1a+frac1c.$$

share|cite|improve this answer

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

$endgroup$

share|cite|improve this answer

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

answered 6 hours ago

farruhotafarruhota

25.4k22 gold badges1010 silver badges4646 bronze badges

1

$begingroup$

Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.

With synthetic division by $x-1$, we get, assuming $ane0$ and $bne c$,
$$
beginarraycc
& ab-ac & bc-ab & ac-bc \
1 & downarrow & ab-ac & bc-ac \
hline
& ab-ac & bc-ac & 0
endarray
$$
The other root is
$$
-fracbc-acab-ac
$$
Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes
$$
2ac=bc+ab
$$
and, dividing by $abc$, we get
$$
frac2b=frac1a+frac1c
$$

If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $cne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.

share|cite|improve this answer

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.

With synthetic division by $x-1$, we get, assuming $ane0$ and $bne c$,
$$
beginarraycc
& ab-ac & bc-ab & ac-bc \
1 & downarrow & ab-ac & bc-ac \
hline
& ab-ac & bc-ac & 0
endarray
$$
The other root is
$$
-fracbc-acab-ac
$$
Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes
$$
2ac=bc+ab
$$
and, dividing by $abc$, we get
$$
frac2b=frac1a+frac1c
$$

If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $cne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.

share|cite|improve this answer

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

$endgroup$

add a comment | 

$begingroup$

Since $a(b-c)+b(c-a)+c(a-b)=0$, you know that $x=1$ is a root, as you observed.

With synthetic division by $x-1$, we get, assuming $ane0$ and $bne c$,
$$
beginarraycc
& ab-ac & bc-ab & ac-bc \
1 & downarrow & ab-ac & bc-ac \
hline
& ab-ac & bc-ac & 0
endarray
$$
The other root is
$$
-fracbc-acab-ac
$$
Therefore $bc-ac=-ab+ac$ is the condition that the equation has two equal roots. This becomes
$$
2ac=bc+ab
$$
and, dividing by $abc$, we get
$$
frac2b=frac1a+frac1c
$$

If $a=0$ or $b=c$ the equation is not a quadratic. Can it be $b=0$? In this case we get $-acx^2+ac=0$ and we need $cne0$; the roots are $1$ and $-1$. With $c=0$ it would be the same.

share|cite|improve this answer

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

$endgroup$

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answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges

answered 7 hours ago

egregegreg

192k1414 gold badges9191 silver badges217217 bronze badges