Polynomial Roots (2 Viewers)

[Lukybear]

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]

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.

 

[Lukybear]

the answer was:

z = a/b
y = b/a

did u get that?

 

[shaon0]

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]

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)

 

[study-freak]

 

[gurmies]

Use the fact that a^2(b+c)+b^2(a+c)+c^2(a+b) = (a+b+c)(ab + ac + bc) - 3abc

 

[Lukybear]

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]

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).

 

[gurmies]

Initially I did it the way study-freak did (in year 11). The next day at school my friend showed me that expansion

 

[Lukybear]

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]

and thxs study freak and gurmies, for your excellent help

 

[shaon0]

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

 

[gurmies]

 

[shaon0]

Nice work Gurmies

 

[Lukybear]

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.

 

[Lukybear]

Also 3rd last step. What did you do there? I know you completed the square, but where did that one go?

 

[shaon0]

(a^2+b^2/2ab)^2-(a^2-b^2/2ab)^2=(b/a)(a/b)=1

 

[gurmies]

O thank you.

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

Haha - it wasn't so much elegant, as it was me skipping steps

 

[shaon0]

Haha - it wasn't so much elegant, as it was me skipping steps

did you know the answers for y and z before you wrote your solution?