# Polynomial Roots (1 Viewer)

#### Lukybear

##### Active Member
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.

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z = a/b
y = b/a

did u get that?

#### shaon0

##### ...

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

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

##### Bored of
$\bg_white \\a+b+c=-3 \\Sub a, b, c into the polynomial \\a^3+3a^2-2a+1=0 \\b^3+3b^2-2b+1=0 \\c^3+3c^2-2c+1=0 \\Add \\(a^3+b^3+c^3)+3(a^2+b^2+c^2)-2(a+b+c)+3=0 \\ \\a^2(b+c) + b^2(a+c) + c^2(a+b) \\=a^2(-3-a)+b^2(-3-b)+c^2(-3-c) \\=-3(a^2+b^2+c^2)-(a^3+b^3+c^3) \\=-2(a+b+c)+3 \\=6+3 \\=9$

#### gurmies

##### Drover
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
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
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).

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#### gurmies

##### Drover
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
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
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

##### Drover
$\bg_white \sum a_{i}=a+b=-m \\\\ \prod a_{i}=ab=n \\\\ \sum y_{i}=y+z=\frac{m^{2}-2n}{n}=\frac{\left ( a+b \right )^{2}-2ab}{ab}=\frac{a^{2}+b^{2}}{ab} \\\\ \prod y_{i}=yz=1 \\\\ \therefore y^{2}-\left ( \frac{a^{2}+b^{2}}{ab} \right )y+1=0 \\\\ \left ( y-\frac{a^{2}+b^{2}}{2ab} \right )^{2}=\left ( \frac{a^{2}-b^{2}}{2ab} \right )^{2} \\\\ y=\frac{a}{b} \Rightarrow z=\frac{b}{a}$

#### shaon0

##### ...
$\bg_white \sum a_{i}=a+b=-m \\\\ \prod a_{i}=ab=n \\\\ \sum y_{i}=y+z=\frac{m^{2}-2n}{n}=\frac{\left ( a+b \right )^{2}-2ab}{ab}=\frac{a^{2}+b^{2}}{ab} \\\\ \prod y_{i}=yz=1 \\\\ \therefore y^{2}-\left ( \frac{a^{2}+b^{2}}{ab} \right )y+1=0 \\\\ \left ( y-\frac{a^{2}+b^{2}}{2ab} \right )^{2}=\left ( \frac{a^{2}-b^{2}}{2ab} \right )^{2} \\\\ y=\frac{a}{b} \Rightarrow z=\frac{b}{a}$
Nice work Gurmies

#### Lukybear

##### Active Member
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

##### Active Member
$\bg_white \sum a_{i}=a+b=-m \\\\ \prod a_{i}=ab=n \\\\ \sum y_{i}=y+z=\frac{m^{2}-2n}{n}=\frac{\left ( a+b \right )^{2}-2ab}{ab}=\frac{a^{2}+b^{2}}{ab} \\\\ \prod y_{i}=yz=1 \\\\ \therefore y^{2}-\left ( \frac{a^{2}+b^{2}}{ab} \right )y+1=0 \\\\ \left ( y-\frac{a^{2}+b^{2}}{2ab} \right )^{2}=\left ( \frac{a^{2}-b^{2}}{2ab} \right )^{2} \\\\ y=\frac{a}{b} \Rightarrow z=\frac{b}{a}$
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

##### Drover
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?