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Using polynomial $x^n+ax+b=0$ and proving an equation
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If $ a_1$, $ a_2$, $ a_3$,..., $ a_n$ are roots of the equation $x^n+ax+b=0$, then prove that $( a_1- a_2)( a_1- a_3)( a_1- a_4)..( a_1- a_n)=n a_1^n-1+a$.
By taking n=3 and using the roots 2,4 &-6 i am able to prove the answer but not able to do via substitution.
polynomials
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
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up vote
0
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If $ a_1$, $ a_2$, $ a_3$,..., $ a_n$ are roots of the equation $x^n+ax+b=0$, then prove that $( a_1- a_2)( a_1- a_3)( a_1- a_4)..( a_1- a_n)=n a_1^n-1+a$.
By taking n=3 and using the roots 2,4 &-6 i am able to prove the answer but not able to do via substitution.
polynomials
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If $ a_1$, $ a_2$, $ a_3$,..., $ a_n$ are roots of the equation $x^n+ax+b=0$, then prove that $( a_1- a_2)( a_1- a_3)( a_1- a_4)..( a_1- a_n)=n a_1^n-1+a$.
By taking n=3 and using the roots 2,4 &-6 i am able to prove the answer but not able to do via substitution.
polynomials
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
If $ a_1$, $ a_2$, $ a_3$,..., $ a_n$ are roots of the equation $x^n+ax+b=0$, then prove that $( a_1- a_2)( a_1- a_3)( a_1- a_4)..( a_1- a_n)=n a_1^n-1+a$.
By taking n=3 and using the roots 2,4 &-6 i am able to prove the answer but not able to do via substitution.
polynomials
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
asked Jul 24 at 1:55
Samar Imam Zaidi
1,063316
1,063316
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Since $a_1 ldots a_n$ are roots of $x^n + ax + b$, then
$$x^n + ax + b = (x-a_1)ldots(x-a_n)$$
Derive both sides w.r.t $x$ you get:
beginequation
beginsplit
nx^n-1 + a
&= (x-a_2)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_2)ldots(x-a_n) \
&+ vdots \
& + (x-a_1)(x-a_2)ldots(x-a_n-1)
endsplit
endequation
Plug $x = a_1$ on both sides you get
beginequation
beginsplit
na_1^n-1 + a
&= (a_1-a_2)(a_1-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_2)ldots(a_1-a_n) \
&+ vdots \
& + (a_1-a_1)(a_1-a_2)ldots(a_1-a_n-1)
endsplit
endequation
Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get
$$na_1^n-1 + a = (a_1-a_2)(a_1-a_3)ldots(a_1-a_n)$$
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
add a comment |Â
up vote
4
down vote
If you are using $a$ as a coefficient, then using $a_j$ for a typical
root is not a good choice of notation.
You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)=
nx^n-1+a$, so your RHS is $f'(a_1)$. Is that significant?
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Since $a_1 ldots a_n$ are roots of $x^n + ax + b$, then
$$x^n + ax + b = (x-a_1)ldots(x-a_n)$$
Derive both sides w.r.t $x$ you get:
beginequation
beginsplit
nx^n-1 + a
&= (x-a_2)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_2)ldots(x-a_n) \
&+ vdots \
& + (x-a_1)(x-a_2)ldots(x-a_n-1)
endsplit
endequation
Plug $x = a_1$ on both sides you get
beginequation
beginsplit
na_1^n-1 + a
&= (a_1-a_2)(a_1-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_2)ldots(a_1-a_n) \
&+ vdots \
& + (a_1-a_1)(a_1-a_2)ldots(a_1-a_n-1)
endsplit
endequation
Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get
$$na_1^n-1 + a = (a_1-a_2)(a_1-a_3)ldots(a_1-a_n)$$
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
add a comment |Â
up vote
4
down vote
accepted
Since $a_1 ldots a_n$ are roots of $x^n + ax + b$, then
$$x^n + ax + b = (x-a_1)ldots(x-a_n)$$
Derive both sides w.r.t $x$ you get:
beginequation
beginsplit
nx^n-1 + a
&= (x-a_2)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_2)ldots(x-a_n) \
&+ vdots \
& + (x-a_1)(x-a_2)ldots(x-a_n-1)
endsplit
endequation
Plug $x = a_1$ on both sides you get
beginequation
beginsplit
na_1^n-1 + a
&= (a_1-a_2)(a_1-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_2)ldots(a_1-a_n) \
&+ vdots \
& + (a_1-a_1)(a_1-a_2)ldots(a_1-a_n-1)
endsplit
endequation
Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get
$$na_1^n-1 + a = (a_1-a_2)(a_1-a_3)ldots(a_1-a_n)$$
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Since $a_1 ldots a_n$ are roots of $x^n + ax + b$, then
$$x^n + ax + b = (x-a_1)ldots(x-a_n)$$
Derive both sides w.r.t $x$ you get:
beginequation
beginsplit
nx^n-1 + a
&= (x-a_2)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_2)ldots(x-a_n) \
&+ vdots \
& + (x-a_1)(x-a_2)ldots(x-a_n-1)
endsplit
endequation
Plug $x = a_1$ on both sides you get
beginequation
beginsplit
na_1^n-1 + a
&= (a_1-a_2)(a_1-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_2)ldots(a_1-a_n) \
&+ vdots \
& + (a_1-a_1)(a_1-a_2)ldots(a_1-a_n-1)
endsplit
endequation
Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get
$$na_1^n-1 + a = (a_1-a_2)(a_1-a_3)ldots(a_1-a_n)$$
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
Since $a_1 ldots a_n$ are roots of $x^n + ax + b$, then
$$x^n + ax + b = (x-a_1)ldots(x-a_n)$$
Derive both sides w.r.t $x$ you get:
beginequation
beginsplit
nx^n-1 + a
&= (x-a_2)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_3)ldots(x-a_n) \
&+ (x-a_1)(x-a_2)ldots(x-a_n) \
&+ vdots \
& + (x-a_1)(x-a_2)ldots(x-a_n-1)
endsplit
endequation
Plug $x = a_1$ on both sides you get
beginequation
beginsplit
na_1^n-1 + a
&= (a_1-a_2)(a_1-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_3)ldots(a_1-a_n) \
&+ (a_1-a_1)(x-a_2)ldots(a_1-a_n) \
&+ vdots \
& + (a_1-a_1)(a_1-a_2)ldots(a_1-a_n-1)
endsplit
endequation
Notice that all terms from the second to the last rows above are zeros due to $a_1-a_1$, hence we get
$$na_1^n-1 + a = (a_1-a_2)(a_1-a_3)ldots(a_1-a_n)$$
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
answered Jul 24 at 2:06
Ahmad Bazzi
2,6271417
2,6271417
add a comment |Â
add a comment |Â
up vote
4
down vote
If you are using $a$ as a coefficient, then using $a_j$ for a typical
root is not a good choice of notation.
You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)=
nx^n-1+a$, so your RHS is $f'(a_1)$. Is that significant?
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
add a comment |Â
up vote
4
down vote
If you are using $a$ as a coefficient, then using $a_j$ for a typical
root is not a good choice of notation.
You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)=
nx^n-1+a$, so your RHS is $f'(a_1)$. Is that significant?
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
add a comment |Â
up vote
4
down vote
up vote
4
down vote
If you are using $a$ as a coefficient, then using $a_j$ for a typical
root is not a good choice of notation.
You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)=
nx^n-1+a$, so your RHS is $f'(a_1)$. Is that significant?
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
If you are using $a$ as a coefficient, then using $a_j$ for a typical
root is not a good choice of notation.
You might want to observe that if $f(x)=x^n+ax+b$, then $f'(x)=
nx^n-1+a$, so your RHS is $f'(a_1)$. Is that significant?
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
answered Jul 24 at 1:59
Lord Shark the Unknown
85.2k950111
85.2k950111
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
add a comment |Â
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
I copied this question as it appear in question paper, i used roots and then it worked but not able to rove it
– Samar Imam Zaidi
Jul 24 at 2:01
add a comment |Â
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