Roots of a Polynomial Question
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If $alpha$ and $beta$ are roots of the equation $x^2+mx+n=0$ , find the roots of $nx^2+(2n-m^2)x+n=0$ in terms of $alpha$ and $beta$.
I really need help with this problem. I've started by finding the sum and products of roots of the first equation and don't know what to do next.
polynomials
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
asked Jan 22 at 11:01
nicknick
313
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add a comment |
$begingroup$
If $alpha$ and $beta$ are roots of the equation $x^2+mx+n=0$ , find the roots of $nx^2+(2n-m^2)x+n=0$ in terms of $alpha$ and $beta$.
I really need help with this problem. I've started by finding the sum and products of roots of the first equation and don't know what to do next.
polynomials
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
asked Jan 22 at 11:01
nicknick
313
$endgroup$
$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
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– Bill Dubuque
Jan 22 at 18:26
add a comment |
$begingroup$
If $alpha$ and $beta$ are roots of the equation $x^2+mx+n=0$ , find the roots of $nx^2+(2n-m^2)x+n=0$ in terms of $alpha$ and $beta$.
I really need help with this problem. I've started by finding the sum and products of roots of the first equation and don't know what to do next.
polynomials
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
asked Jan 22 at 11:01
nicknick
313
$endgroup$
If $alpha$ and $beta$ are roots of the equation $x^2+mx+n=0$ , find the roots of $nx^2+(2n-m^2)x+n=0$ in terms of $alpha$ and $beta$.
I really need help with this problem. I've started by finding the sum and products of roots of the first equation and don't know what to do next.
polynomials
polynomials
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
asked Jan 22 at 11:01
nicknick
313
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
asked Jan 22 at 11:01
nicknick
313
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
edited Jan 22 at 11:03
Thomas Shelby
2,9771421
2,9771421
asked Jan 22 at 11:01
nicknick
313
asked Jan 22 at 11:01
nicknick
313
asked Jan 22 at 11:01
nicknick
313
313
$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
$endgroup$
– Bill Dubuque
Jan 22 at 18:26
add a comment |
$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
$endgroup$
– Bill Dubuque
Jan 22 at 18:26
$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
$endgroup$
– Bill Dubuque
Jan 22 at 18:26
$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
$endgroup$
– Bill Dubuque
Jan 22 at 18:26
add a comment |
3 Answers
3
active
oldest
votes
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Substituting $m,n$ obtained from Vieta, the second equation is
$$alphabeta x^2+(2alphabeta-(alpha+beta)^2)x+alphabeta=0$$
or
$$x^2-left(fracalphabeta+fracbetaalpharight)x+1=0.$$
Again by Vieta, the roots are obvious.
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
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@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
add a comment |
$begingroup$
Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(alpha+beta)^2(alpha-beta)^2$.
answered Jan 22 at 11:11
lhflhf
165k10171396
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add a comment |
$begingroup$
Try writing $m$ and $n$ in terms of $alpha$ and $beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-alpha)(x-beta)$.
answered Jan 22 at 11:08
KlausKlaus
1,5599
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Substituting $m,n$ obtained from Vieta, the second equation is
$$alphabeta x^2+(2alphabeta-(alpha+beta)^2)x+alphabeta=0$$
or
$$x^2-left(fracalphabeta+fracbetaalpharight)x+1=0.$$
Again by Vieta, the roots are obvious.
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
add a comment |
$begingroup$
Substituting $m,n$ obtained from Vieta, the second equation is
$$alphabeta x^2+(2alphabeta-(alpha+beta)^2)x+alphabeta=0$$
or
$$x^2-left(fracalphabeta+fracbetaalpharight)x+1=0.$$
Again by Vieta, the roots are obvious.
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
$endgroup$
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
add a comment |
$begingroup$
Substituting $m,n$ obtained from Vieta, the second equation is
$$alphabeta x^2+(2alphabeta-(alpha+beta)^2)x+alphabeta=0$$
or
$$x^2-left(fracalphabeta+fracbetaalpharight)x+1=0.$$
Again by Vieta, the roots are obvious.
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
$endgroup$
Substituting $m,n$ obtained from Vieta, the second equation is
$$alphabeta x^2+(2alphabeta-(alpha+beta)^2)x+alphabeta=0$$
or
$$x^2-left(fracalphabeta+fracbetaalpharight)x+1=0.$$
Again by Vieta, the roots are obvious.
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
answered Jan 22 at 11:15
Yves DaoustYves Daoust
127k673226
127k673226
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
add a comment |
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
How did you arrive to the second equation?
$endgroup$
– nick
Jan 22 at 11:36
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
$begingroup$
@nick: obviously by dividing by $alphabeta$.
$endgroup$
– Yves Daoust
Jan 22 at 12:46
add a comment |
$begingroup$
Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(alpha+beta)^2(alpha-beta)^2$.
answered Jan 22 at 11:11
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(alpha+beta)^2(alpha-beta)^2$.
answered Jan 22 at 11:11
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(alpha+beta)^2(alpha-beta)^2$.
answered Jan 22 at 11:11
lhflhf
165k10171396
$endgroup$
Hint: The discriminant of $nx^2+(2n-m^2)x+n$ is $m^4 - 4 m^2 n=m^2(m^2-4n)=(alpha+beta)^2(alpha-beta)^2$.
answered Jan 22 at 11:11
lhflhf
165k10171396
answered Jan 22 at 11:11
lhflhf
165k10171396
answered Jan 22 at 11:11
lhflhf
165k10171396
answered Jan 22 at 11:11
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
$begingroup$
Try writing $m$ and $n$ in terms of $alpha$ and $beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-alpha)(x-beta)$.
answered Jan 22 at 11:08
KlausKlaus
1,5599
$endgroup$
add a comment |
$begingroup$
Try writing $m$ and $n$ in terms of $alpha$ and $beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-alpha)(x-beta)$.
answered Jan 22 at 11:08
KlausKlaus
1,5599
$endgroup$
add a comment |
$begingroup$
Try writing $m$ and $n$ in terms of $alpha$ and $beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-alpha)(x-beta)$.
answered Jan 22 at 11:08
KlausKlaus
1,5599
$endgroup$
Try writing $m$ and $n$ in terms of $alpha$ and $beta$. Then solve the second equation in the usual way. Hint: $x^2+mx+n = (x-alpha)(x-beta)$.
answered Jan 22 at 11:08
KlausKlaus
1,5599
answered Jan 22 at 11:08
KlausKlaus
1,5599
answered Jan 22 at 11:08
KlausKlaus
1,5599
answered Jan 22 at 11:08
KlausKlaus
1,5599
1,5599
add a comment |
add a comment |
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$begingroup$
Hint $ $ Its root product $=1$ and root sum $= mleft[dfracmnright]-2, =, (alpha+beta)left[dfrac1alpha+dfrac1betaright] -2, =, dfracalphabeta+dfracbetaalpha$ $qquadqquadqquad $
$endgroup$
– Bill Dubuque
Jan 22 at 18:26