Mixing
Quadratic Equations - Proof of Common Root [on hold]
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Prove that the equations (q-r)x^2 + (r-p)x + p-q = 0 and (r-p)x^2 + (p-q)x + q-r = 0 have a common root.
algebra-precalculus roots
asked Aug 3 at 18:28
Gaurav Choudhury
62
put on hold as off-topic by amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus Aug 3 at 20:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus
add a comment |Â
up vote
-2
down vote
favorite
Prove that the equations (q-r)x^2 + (r-p)x + p-q = 0 and (r-p)x^2 + (p-q)x + q-r = 0 have a common root.
algebra-precalculus roots
asked Aug 3 at 18:28
Gaurav Choudhury
62
put on hold as off-topic by amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus Aug 3 at 20:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus
4
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
1
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
3
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Prove that the equations (q-r)x^2 + (r-p)x + p-q = 0 and (r-p)x^2 + (p-q)x + q-r = 0 have a common root.
algebra-precalculus roots
asked Aug 3 at 18:28
Gaurav Choudhury
62
Prove that the equations (q-r)x^2 + (r-p)x + p-q = 0 and (r-p)x^2 + (p-q)x + q-r = 0 have a common root.
algebra-precalculus roots
asked Aug 3 at 18:28
Gaurav Choudhury
62
edited Aug 3 at 20:30
asked Aug 3 at 18:28
Gaurav Choudhury
62
asked Aug 3 at 18:28
Gaurav Choudhury
62
asked Aug 3 at 18:28
Gaurav Choudhury
62
62
put on hold as off-topic by amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus Aug 3 at 20:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus
put on hold as off-topic by amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus Aug 3 at 20:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, John Ma, Mike Miller, Xander Henderson, Leucippus
4
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
1
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
3
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38
add a comment |Â
4
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
1
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
3
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38
4
4
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
1
1
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
3
3
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
0
down vote
Hint: Computing the roots, we get
$$x_1=1,x_2=fracp-qq-r,x_3=1,x_4=fracr-qp-r$$
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
add a comment |Â
up vote
0
down vote
Consider the polinomial of degree 2 vector space. Let $a in mathbbR$ such as $(q - r, r - p, p - q) = a(r - p, p - q, q - r) iff begincases
q - r - ar + ap = 0\
r - p -ap + aq = 0\
p - q -aq + ar = 0
endcases
$. Then
$left(beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright)left(beginarrayccc
p\
q\
r
endarrayright) = left(beginarrayccc
0\
0\
0
endarrayright)$ and $left|beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright| = 0$
Hence,the set $S = (q - r)x^2 + (r - p)x + p - q, (r - p)x^2 + (p - q)x + (q - r)$ is linear dependent. The dimension of space generated by $S$ is least equal 1 and the kernel dimension is great or equal 2. Hence, the equation system are at least one common solution.
answered Aug 3 at 19:16
GinoCHJ
12
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Computing the roots, we get
$$x_1=1,x_2=fracp-qq-r,x_3=1,x_4=fracr-qp-r$$
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
add a comment |Â
up vote
0
down vote
Hint: Computing the roots, we get
$$x_1=1,x_2=fracp-qq-r,x_3=1,x_4=fracr-qp-r$$
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Computing the roots, we get
$$x_1=1,x_2=fracp-qq-r,x_3=1,x_4=fracr-qp-r$$
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
Hint: Computing the roots, we get
$$x_1=1,x_2=fracp-qq-r,x_3=1,x_4=fracr-qp-r$$
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
answered Aug 3 at 18:52
Dr. Sonnhard Graubner
66.6k32659
66.6k32659
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
add a comment |Â
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
What happens if $q=r$?
– Michael Rozenberg
Aug 3 at 19:04
1
1
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
If $q=r$ we get no quadratic equation.
– Dr. Sonnhard Graubner
Aug 3 at 19:07
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
Then if $$rne p$$ we get $$x=1$$ and $$x=0$$ or $$x=1$$
– Dr. Sonnhard Graubner
Aug 3 at 19:11
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
I think it's enough $x=1$.
– Michael Rozenberg
Aug 3 at 19:12
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
Ok, so i will sleep better Michael.
– Dr. Sonnhard Graubner
Aug 3 at 19:18
add a comment |Â
up vote
0
down vote
Consider the polinomial of degree 2 vector space. Let $a in mathbbR$ such as $(q - r, r - p, p - q) = a(r - p, p - q, q - r) iff begincases
q - r - ar + ap = 0\
r - p -ap + aq = 0\
p - q -aq + ar = 0
endcases
$. Then
$left(beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright)left(beginarrayccc
p\
q\
r
endarrayright) = left(beginarrayccc
0\
0\
0
endarrayright)$ and $left|beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright| = 0$
Hence,the set $S = (q - r)x^2 + (r - p)x + p - q, (r - p)x^2 + (p - q)x + (q - r)$ is linear dependent. The dimension of space generated by $S$ is least equal 1 and the kernel dimension is great or equal 2. Hence, the equation system are at least one common solution.
answered Aug 3 at 19:16
GinoCHJ
12
add a comment |Â
up vote
0
down vote
Consider the polinomial of degree 2 vector space. Let $a in mathbbR$ such as $(q - r, r - p, p - q) = a(r - p, p - q, q - r) iff begincases
q - r - ar + ap = 0\
r - p -ap + aq = 0\
p - q -aq + ar = 0
endcases
$. Then
$left(beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright)left(beginarrayccc
p\
q\
r
endarrayright) = left(beginarrayccc
0\
0\
0
endarrayright)$ and $left|beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright| = 0$
Hence,the set $S = (q - r)x^2 + (r - p)x + p - q, (r - p)x^2 + (p - q)x + (q - r)$ is linear dependent. The dimension of space generated by $S$ is least equal 1 and the kernel dimension is great or equal 2. Hence, the equation system are at least one common solution.
answered Aug 3 at 19:16
GinoCHJ
12
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider the polinomial of degree 2 vector space. Let $a in mathbbR$ such as $(q - r, r - p, p - q) = a(r - p, p - q, q - r) iff begincases
q - r - ar + ap = 0\
r - p -ap + aq = 0\
p - q -aq + ar = 0
endcases
$. Then
$left(beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright)left(beginarrayccc
p\
q\
r
endarrayright) = left(beginarrayccc
0\
0\
0
endarrayright)$ and $left|beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright| = 0$
Hence,the set $S = (q - r)x^2 + (r - p)x + p - q, (r - p)x^2 + (p - q)x + (q - r)$ is linear dependent. The dimension of space generated by $S$ is least equal 1 and the kernel dimension is great or equal 2. Hence, the equation system are at least one common solution.
answered Aug 3 at 19:16
GinoCHJ
12
Consider the polinomial of degree 2 vector space. Let $a in mathbbR$ such as $(q - r, r - p, p - q) = a(r - p, p - q, q - r) iff begincases
q - r - ar + ap = 0\
r - p -ap + aq = 0\
p - q -aq + ar = 0
endcases
$. Then
$left(beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright)left(beginarrayccc
p\
q\
r
endarrayright) = left(beginarrayccc
0\
0\
0
endarrayright)$ and $left|beginarrayccc
a & 1 & -1-a\
-1-a & a & 1\
1 & -1-a & a
endarrayright| = 0$
Hence,the set $S = (q - r)x^2 + (r - p)x + p - q, (r - p)x^2 + (p - q)x + (q - r)$ is linear dependent. The dimension of space generated by $S$ is least equal 1 and the kernel dimension is great or equal 2. Hence, the equation system are at least one common solution.
answered Aug 3 at 19:16
GinoCHJ
12
edited Aug 3 at 19:30
answered Aug 3 at 19:16
GinoCHJ
12
answered Aug 3 at 19:16
GinoCHJ
12
answered Aug 3 at 19:16
GinoCHJ
12
12
add a comment |Â
add a comment |Â
4
Add some of your thoughts, what have you tried so far?
– zzuussee
Aug 3 at 18:30
1
@GoodDeeds I think you assumed x*2* = $x^2,$ etc. Let's ask the asker, first and foremost, before assuming we know what they are asking and editing (with the possibility of having changed the question entirely).
– amWhy
Aug 3 at 18:33
@amWhy Ok, sorry, should I rollback? I assumed looking at the title
– GoodDeeds
Aug 3 at 18:33
@GoodDeeds: No, let's just ask Gaurav if the edited question is what they intended to ask. If they fail to respond yes or no, we'll just leave the question as you formatted it.
– amWhy
Aug 3 at 18:34
3
Plug in $x=1$ and see what happens.
– Anurag A
Aug 3 at 18:38