One cubic root is reciprocal of other question

Question is:

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

I am not really sure how to approach this one. Here is how I started.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \beta = \frac{-b}{a}$

$\beta = \frac{-b}{a} - 1 = \frac{-b}{a} - \frac{a}{a} = \frac{a-b}{a}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be

$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$

Hence

$\frac{a-b}{a} = \frac{-d}{a}$

$a-b = -d$

Is this going in the right direction. How can I get unstuck?

Reply 1

TeeEm

Original post

by acomber

Question is:

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

I am not really sure how to approach this one. Here is how I started.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \beta = \frac{-b}{a}$

$\beta = \frac{-b}{a} - 1 = \frac{a-b}{a}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be

$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$

Hence

$\frac{a-b}{a} = \frac{-d}{a}$

$a-b = -d$

Is this going in the right direction. How can I get unstuck?

I am not following the very first manipulation for -b/a

Reply 2

acomber

The sum of the roots = -b/a ? In this case the roots are alpha reciprocal of alpha and then I call the 3rd root beta. The three roots summed = -b/a.

Thank you, I see the mistake now.

So I now have:

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

I am not really sure how to approach this one. Here is how I started.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \beta = \frac{-b}{a}$

$\beta = \frac{-b}{a} - 1 = \frac{-b}{a} - \frac{a}{a} = \frac{-a-b}{a}$

$\beta = \frac{-a-b}{a}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be

$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$

Hence

$\beta = \frac{-a-b}{a} = \frac{-d}{a}$

Multiply by -a

$a+b = d$

Is this going in the right direction. How can I get unstuck?

Original post

by TeeEm

I am not following the very first manipulation for -b/a

(edited 11 years ago)

Reply 3

TeeEm

Original post

by acomber

The sum of the roots = -b/a ? In this case the roots are alpha reciprocal of alpha and then I call the 3rd root beta. The three roots summed = -b/a.

how is a +1/a +b = 1+ b?

a=alpha
b = beta

Reply 4

acomber

Ah yes - must have been tired. I was thinking multiplication.

So by fixing the errors I can now say.

Show that one root of az^3 + bz^2 + cz + d = 0 is the reciprocal of another root if and only if a^2 - d^2 = ac - bd.

We can say one root is $\alpha$ and the other is $\frac{1}{\alpha}$

The other root can be $\beta$

So the three roots of the cubic are $\alpha, \frac{1}{\alpha}, \beta$

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

Not sure what I can do with this:

$\beta = \frac{-b}{a} - \alpha - \frac{1}{\alpha}$

The c/a is trickier

$\alpha . \frac{1}{\alpha} + \alpha . \beta + \frac{1}{\alpha} . \beta = \frac{c}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

Not quite sure how to proceed with this because I am not sure how to work out what $\alpha\beta$ would be

$(\alpha)(\frac{1}{\alpha})(\beta) = \beta = \frac{-d}{a}$

What could I do next?

Original post

by TeeEm

how is a +1/a +b = 1+ b?

a=alpha
b = beta

Reply 5

TeeEm

Original post

by acomber

you have 3 "parametric equations" and two "parameters", alpha and beta.

you eliminate the same way if you had 2 parametric equations and 1 parameter.

It may involve tricks.

I will have a look at it a bit later if you have not managed to bet rid of alpha and beta

Reply 6

TeeEm

Original post

by acomber

It is not that hard.

Just did it rough

Try it first youself

Reply 7

acomber

So far then I have these three equations.

$\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

$1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

$\beta = \frac{-d}{a}$

So, I have the \beta on its own. But I don't have an eqn with just \alpha. So stuck really.

Original post

by TeeEm

It is not that hard.

Just did it rough

Try it first youself

Reply 8

TeeEm

Original post

by acomber

a=alpha
b=beta

A=a
B=b
C=c
D=d

factorize the b in the second equation so it leaves (a+1/a)

then plug the b=-D/A into the first 2 equations

solve the fist 2 equations for a+1/a

then set them equal

then answer in about 2 lines

Reply 9

acomber

Ahhhhh, thank you sooo much.

I didn't think about considering alpha + 1/alpha as a unit.

(1) $\alpha + \frac{1}{\alpha} + \beta = \frac{-b}{a}$

(2) $1 + \alpha\beta + \frac{\beta}{\alpha} = \frac{c}{a}$

(3) $\beta = \frac{-d}{a}$

(2)factorise beta
$1+ \beta(\alpha + \frac{1}{\alpha}) = \frac{c}{a}$

substitute -d/a for beta for (1) and (2)
$\alpha + \frac{1}{\alpha} + \frac{-d}{a} = \frac{-b}{a}$

$\alpha + \frac{1}{\alpha} = \frac{-b}{a} - \frac{-d}{a} = \frac{d-b}{a}$

$1+ \frac{-d}{a}(\alpha + \frac{1}{\alpha}) = \frac{c}{a}$

$\frac{-d}{a}(\alpha + \frac{1}{\alpha}) = \frac{c}{a} = \frac{a}{a} = \frac{c-a}{a}$

$-d(\alpha + \frac{1}{\alpha}) = c-a$

$\alpha + \frac{1}{\alpha} = \frac{c-a}{-d} = \frac{a-c}{d}$

Therefore

$\frac{d-b}{a} = \frac{a-c}{d}$

$d(d-b) = a(a-c)$

$d^2-db = a^2-ac$

$a^2-d^2 = ac-bd$

Hurray

Original post

by TeeEm

Reply 10

TeeEm

Original post

by acomber

well done

(Excellent LaTex skills)

Reply 11

aliovais

This question is two way, one side we are assuming that roots reciprocal relation and proved a^2-d^2=ac-bd. But we for the converse we have to assume the relation a^2-d^2=ac-bd and have to show two roots are reciprocal. BECAUSE IT IS MENTIONED THAT IF AND ONLY IF.

Best Regards
Ali

Reply 12