Solving polynominals equations (relationship of roots)
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$
So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac{11}{3}$
polynomials roots
$endgroup$
add a comment |
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$
So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac{11}{3}$
polynomials roots
$endgroup$
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
$begingroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$
So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac{11}{3}$
polynomials roots
$endgroup$
The roots of $x^3-4x^2+x+6$ are $alpha$, $beta$, and $omega$.
Find (evaluate):
$$frac{alpha+beta}{omega}+frac{alpha+omega}{beta}+frac{beta+omega}{alpha}$$
So far I have found:
$$alpha+beta+omega=frac{-b}{a} = 4 \
alphabeta+betaomega+alphaomega=frac{c}{a} = 1 \
alpha×beta×omega=frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$frac{alpha^2beta+alphabeta^2+alpha^2omega+alphaomega^2+beta^2omega+betaomega^2}{alphabetaomega}$$
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-dfrac{11}{3}$
polynomials roots
polynomials roots
edited 43 mins ago
Lee David Chung Lin
4,51851342
4,51851342
asked 1 hour ago
Alex Alex
186
186
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
1
1
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago
add a comment |
4 Answers
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$begingroup$
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$
$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$
$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$
$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$
$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
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$begingroup$
Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
frac13-5+1=\
-frac{11}{3}.$$
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add a comment |
$begingroup$
That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
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active
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active
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votes
$begingroup$
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$
$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$
$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$
$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$
$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
$endgroup$
add a comment |
$begingroup$
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$
$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$
$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$
$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$
$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
$endgroup$
add a comment |
$begingroup$
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$
$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$
$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$
$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$
$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
$endgroup$
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}$$
$$= frac{alpha + beta + omega - omega}{omega} + frac{beta + omega + alpha - alpha}{alpha} + frac{alpha + omega + beta - beta}{beta}$$
$$ = (alpha + beta + omega) left(frac{1}{alpha} + frac{1}{beta} + frac{1}{omega}right) - 3$$
$$ = (alpha + beta + omega) left(frac{betaomega}{alphabetaomega} + frac{alphaomega}{alphabetaomega} + frac{alphabeta}{alphabetaomega}right) - 3$$
$$ = frac{alpha + beta + omega}{alphabetaomega}(betaomega + alphaomega + alphabeta) - 3$$
I think you should be able to take it from there.
answered 1 hour ago
user1952500user1952500
1,5351016
1,5351016
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add a comment |
$begingroup$
Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$
$endgroup$
add a comment |
$begingroup$
Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$
$endgroup$
add a comment |
$begingroup$
Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$
$endgroup$
Hint: We can write $$frac{4-w}{w}+frac{4-beta}{beta}+frac{4-alpha}{alpha}$$ and this is $$4left(frac{alphabeta+alpha w+wbeta}{alpha beta w}right)-3$$ and this is $$-frac{2}{3}left(1-beta w-alpha w+alpha w+beta wright)$$
This simplifies to $$-frac{2}{3}-3=-frac{11}{3}$$
answered 1 hour ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.7k42867
79.7k42867
add a comment |
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
frac13-5+1=\
-frac{11}{3}.$$
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
frac13-5+1=\
-frac{11}{3}.$$
$endgroup$
add a comment |
$begingroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
frac13-5+1=\
-frac{11}{3}.$$
$endgroup$
Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 Rightarrow (x+1)(x-2)(x-3)=0 Rightarrow \
alpha =-1, beta =2,omega=3.$$
Hence:
$$frac{alpha + beta}{omega} + frac{beta + omega}{alpha} + frac{alpha + omega}{beta}=\
frac{-1+ 2}{3} + frac{2 + 3}{-1} + frac{-1 + 3}{2}=\
frac13-5+1=\
-frac{11}{3}.$$
answered 19 mins ago
farruhotafarruhota
22.5k2942
22.5k2942
add a comment |
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$begingroup$
That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.
$endgroup$
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.
$endgroup$
add a comment |
$begingroup$
That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.
$endgroup$
That follows from your results, since we get: $dfrac{4-omega}{omega}+dfrac{4-beta}{beta}+dfrac{4-alpha}{alpha}=dfrac{4(omegabeta+omega alpha+betaalpha)-3omegabetaalpha}{omega beta alpha}=dfrac{4+18}{-6}=-dfrac{11}3$.
answered 1 hour ago
Chris CusterChris Custer
14.7k3827
14.7k3827
add a comment |
add a comment |
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$begingroup$
For latex, you use instead of /.
$endgroup$
– BadAtGeometry
1 hour ago