• Happy 20 YEARS to us!

Polynomial Roots (1 Viewer)

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
If a and b are roots of the equantion x^2 + mx + n = 0, find the roots of
nx^2 + (2n-m^2)x + n = 0

Let roots of equn nx^2 + (2n-m^2)x + n = 0 be z, y

i know that

zy = 1
and
z+y = (2n-m^2)/n

but what i cant figure out is how to get a and b from z+y
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,030
Location
Guess
Gender
Male
HSC
2009
If a and b are roots of the equantion x^2 + mx + n = 0, find the roots of
nx^2 + (2n-m^2)x + n = 0

Let roots of equn nx^2 + (2n-m^2)x + n = 0 be z, y

i know that

zy = 1
and
z+y = (2n-m^2)/n

but what i cant figure out is how to get a and b from z+y
for x^2+mx+n=0:
a+b=-m
ab=n
Also,
x=[-m+-sqrt(m^2-4n)]/2 ....1

for nx^2+(2n-m^2)x+n=0:
zy=1

z+y=(m^2-2n)/n
=[(a+b)^2-2ab]/ab
=[a^2+b^2]/ab

roots for last equ are:
x=[m^2-2n+-msqrt(m^2-4n)]/2n

But it's easy to solve the last equ using quadratic formula instead of all the algebra with the 1st one.
 
Last edited:

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
the answer was:

z = a/b
y = b/a

did u get that?
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,030
Location
Guess
Gender
Male
HSC
2009
the answer was:

z = a/b
y = b/a

did u get that?
Yeah, just sub in what i got for my first post and u should get it. ie sub in a+b=-m and ab=n into x=[m^2-2n+-msqrt(m^2-4n)]/2n, it ends up working out :)
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
What about this question

If the roots of the equation x^3 +3x^2 -2x + 1 =0 are a,b,c, find the value of

a^2(b+c) + b^2(a+c) + c^2(a+b)
 

gurmies

Drover
Joined
Mar 20, 2008
Messages
1,219
Location
North Bondi
Gender
Male
HSC
2009
Use the fact that a^2(b+c)+b^2(a+c)+c^2(a+b) = (a+b+c)(ab + ac + bc) - 3abc
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
Thanks so much guys.

Study, can i just ask how did you know to do that?

Also gurmies, where did you learn that expansion from?
 

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
Thanks so much guys.

Study, can i just ask how did you know to do that?

Also gurmies, where did you learn that expansion from?
I knew that I needed a^3+b^3+c^3 and so I tried to find it and while doing so, I realised that I don't even need to find the value of a^3+b^3+c^3 because I could eliminate it together with 3(a^2+b^2+c^2).
 
Last edited:

gurmies

Drover
Joined
Mar 20, 2008
Messages
1,219
Location
North Bondi
Gender
Male
HSC
2009
Initially I did it the way study-freak did (in year 11). The next day at school my friend showed me that expansion :)
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
Yeah, just sub in what i got for my first post and u should get it. ie sub in a+b=-m and ab=n into x=[m^2-2n+-msqrt(m^2-4n)]/2n, it ends up working out :)
I guess its my mistake, but i left out a part of the q, where it said, find in terms of a and b

So thats where i am stuck, how would you get from (a^2+b^2)/ab = z+y

to getting z=a/b and y=b/a
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
and thxs study freak and gurmies, for your excellent help
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,030
Location
Guess
Gender
Male
HSC
2009
I guess its my mistake, but i left out a part of the q, where it said, find in terms of a and b

So thats where i am stuck, how would you get from (a^2+b^2)/ab = z+y

to getting z=a/b and y=b/a
1) The easiest way is to solve the longer quadratic equation using the quadratic formula.
2) Using the fact that a+b=-m and ab=n, sub these into x=[m^2-2n+-msqrt(m^2-4n)]/2n which you attained in 1)

It works, trust me. To use z+y=[a^2+b^2]/ab would be taking the longer solution ie don't use z+y=[a^2+b^2]/ab
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,467
Gender
Male
HSC
2010
1) The easiest way is to solve the longer quadratic equation using the quadratic formula.
2) Using the fact that a+b=-m and ab=n, sub these into x=[m^2-2n+-msqrt(m^2-4n)]/2n which you attained in 1)

It works, trust me. To use z+y=[a^2+b^2]/ab would be taking the longer solution ie don't use z+y=[a^2+b^2]/ab

O thank you.

Gurmies, your solution is way to elegant for simpletons like me.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top