Thread: Roots of cubic equations and new equation

1. Roots of cubic equations and new equation

The equation $x^3 + 4x^2 + 3x + 2 = 0$ has roots $\alpha , \beta, \gamma$.

Find the equation with roots $\beta \gamma , \gamma \alpha , \alpha \beta$

For this I so far have:

$\alpha + \beta + \gamma$ = -4
$\beta \alpha + \gamma \alpha + \alpha \gamma$ = 3
$\alpha \beta \gamma$ = -2

How would I progress from here?

2. work out $(x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = x^3-(\beta\gamma+\alpha\gamma+\alpha\beta)x^2 +(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\b eta\gamma^2)x - \alpha^2\beta^2\gamma^2$ with the relations you have.

(Then $(x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = 0$ is your desired equation)

3. Originally Posted by db5vry
The equation $x^3 + 4x^2 + 3x + 2 = 0$ has roots $\alpha , \beta, \gamma$.

Find the equation with roots $\beta \gamma , \gamma \alpha , \alpha \beta$
I would've made $x^3 + 4x^2 + 3x + 2 = (x-\alpha)(x-\beta)(x-\gamma)$ and then factorised the LHS to find $\beta \gamma , \alpha$

Originally Posted by db5vry

$\alpha + \beta + \gamma$ = -4
$\beta \alpha + \gamma \alpha + \alpha \gamma$ = 3
$\alpha \beta \gamma$ = -2

How would I progress from here?
You should be able to solve it by substitution and elimination

for example

$\beta \alpha + \gamma \alpha + \alpha \gamma = 3$

Is the same as

$\alpha (\beta + 2\gamma )= 3$

$\alpha = \frac{3}{\beta + 2\gamma }$

4. Originally Posted by Dinkydoe
work out $(x-\alpha\gamma)(x-\alpha\beta)(x-\beta\gamma) = x^3-(\beta\gamma+\alpha\gamma+\alpha\beta)x^2 +(\alpha^2\beta\gamma+\alpha\beta^2\gamma+\alpha\b eta\gamma^2)x - \alpha^2\beta^2\gamma^2$ with the relations you have.
How does this work? I can see why you substitute $\beta \gamma + \gamma \alpha + \alpha \beta$ into the coeffient space but why does it become $\alpha^2\beta^2\gamma^2$ at the end? I don't understand that (yet).

5. Originally Posted by db5vry
but why does it become $\alpha^2\beta^2\gamma^2$ at the end?
It is the product of the last term in each factor

$-\alpha\gamma\times -\alpha\beta\times -\beta\gamma = -\alpha^2\beta^2\gamma^2$

6. Right, I found an answer for this question in detail now but I'm now sure how this works either:

It says consider $\beta\gamma + \gamma\alpha + \alpha\beta$ = 3

then

$\beta\gamma^2\alpha + \gamma\alpha^2\beta\gamma + \alpha\beta^2\gamma = \alpha\beta\gamma(\alpha + \beta + \gamma)$

= -2 multiplied by -4 = 8

How does the equation with the squares originate? What do you calculate in order to get to that?

7. How does this work? I can see why you substitute into the coeffient space but why does it become at the end? I don't understand that (yet).
There was asked for an equation with roots $\alpha\gamma,\beta\gamma, \alpha\beta$.

$(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0$ is ofcourse such an equation.

If you multiply all the factors you see the result is what I've written out.

the last coefficient is $-\alpha\gamma\cdot - \alpha\beta\cdot - \beta\gamma = -\alpha^2\beta^2\gamma^2$

8. Originally Posted by Dinkydoe
There was asked for an equation with roots $\alpha\gamma,\beta\gamma, \alpha\beta$.

$(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0$ is ofcourse such an equation.

If you multiply all the factors you see the result is what I've written out.

the last coefficient is $-\alpha\gamma\cdot - \alpha\beta\cdot - \beta\gamma = -\alpha^2\beta^2\gamma^2$
I now understand the last coefficient but I'm not sure about the one that I described in another post. Is there some way you can help me with the other one please

9. Have you actually tried to write out the expression $(x-\alpha\beta)(x-\beta\gamma)(x-\alpha\gamma)$ ?

You can multiply do you? Then collect the coefficients for the powers of x and decide what their values are on the basis of the relations you have.

10. Originally Posted by Dinkydoe
Have you actually tried to write out the expression $(x-\alpha\beta)(x-\beta\gamma)(x-\alpha\gamma)$ ?

You can multiply do you? Then collect the coefficients for the powers of x and decide what their values are on the basis of the relations you have.
I have multiplied that out but it doesn't make anything easier or make absolutely any sense to me!

My mark scheme says:

= -4
= 3
= -2

then consider = 3 [I understand this all up to here]

= -2 multiplied by -4 = 8 [I do not know how this works - where is the above line coming from?]

$\beta\gamma . \gamma\alpha . \alpha\beta$ = $\alpha^2\beta^2\gamma^2$

= 4 [I understand this]

then the required equation is

$x^3 - 3x^2 + 8x - 4 = 0$ [I can see how it got here considering the previous working]

All I really want to know is how the mark scheme has come to this! I understand most of it, barring the line which results in "=8".

11. Then I wonder how you came to these relations

= -4
= 3
= -2
The only way you could have found these relations is by considering $(x-\alpha)(x-\beta)(x-\gamma) = x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma +\beta\gamma)x - \alpha\beta\gamma = x^3 + 4x^2 + 3x + 2$

Then the relations you start out with follow.

After that you want an equation with roots $\alpha\gamma,\beta\gamma,\alpha\beta$
An equation that satisfies this property is: $(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0$

But we like to know explicitly how this polynomial looks like from the relations we derived.

Thus we write out this factored form: This can be tenacious work but it's not that hard actually. Mathematics is 10 % inspiration, 90% perspiration.

12. Originally Posted by Dinkydoe
Then I wonder how you came to these relations

The only way you could have found these relations is by considering $(x-\alpha)(x-\beta)(x-\gamma) = x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma +\beta\gamma)x - \alpha\beta\gamma = x^3 + 4x^2 + 3x + 2$

Then the relations you start out with follow.

After that you want an equation with roots $\alpha\gamma,\beta\gamma,\alpha\beta$
An equation that satisfies this property is: $(x-\alpha\gamma)(x-\beta\gamma)(x-\alpha\beta) = 0$

But we like to know explicitly how this polynomial looks like from the relations we derived.

Thus we write out this factored form: This can be tenacious work but it's not that hard actually. Mathematics is 10 % inspiration, 90% perspiration.

I came to the relations that you quoted by using: -b/a to get -4, c/a to get 3 and -d/a to get -2 which was simple enough.

I agree with what you said about mathematics, except with me much of the 90% perspiration is lost in confusion.

Please could you write out a solution? I have tortured myself too long over this question and have tried what you said - expanding those brackets but I can't do that without obtaining $x^2$ and $x$ in a number of terms and it doesn't add up to what my mark scheme details.