Using Vieta's formula on a polynomial equation

[Aion]

Let $x_1,x_2,x_3,x_4$ be roots to the equation $x^4+x^3-2x-4=0$. Determine the equation with roots $x_1^2,x_2^2,x_3^2,x_4^2$.

My attempt:

Expanding the product
$$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=x^4-\left(\sum_{i=1}^4x_i\right)x^3+\left(\sum_{1 \leq i < j \leq 4} x_i x_j\right)x^2 -\left(\sum_{1 \leq i < j < u \leq 4} x_i x_j x_u\right)x+x_1x_2x_3x_4$$
By comparing with $x^4+x^3-2x-4=0$ we see that

$$x_1+x_2+x_3+x_4=-1$$$$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=0$$$$x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=2$$$$x_1x_2x_3x_4=4$$
We aim to find the polynomial whose roots are $x_1^2,x_2^2,x_3^2,x_4^2$. Let's write it as

$$x^4-a_1x^3+a_2x^2-a_3x+a_4$$
Now notice that
$$a_1=x_1^2+x_2^2+x_3^2+x_4^2=(x_1+x_2+x_3+x_4)^2-2\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 \right)=(-1)^2-2(0)=1$$$$a_2=x_1^2x_2^2+x_1^2x_3^2+x_1^2x_4^2+x_2^2x_3^2+x_2^2x_4^2+x_3^2x_4^2=\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 \right)^2-2\left(x_1x_2x_3+x_1x_2x_3+x_1x_3x_4+x_2x_3x_4 \right)=-4$$$$a_3=x_1^2x_2^2x_3^2+x_1^2x_2^2x_4^2+x_1^2x_3^2x_4^2+x_2^2x_3^2x_4^2=\left(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 \right)^2-2\left(x_1x_2x_3x_4(x_1+x_2+x_3+x_4) \right)=12$$$$a_4=\left(x_1x_2x_3x_4\right)^2=4^2=16$$
Hence the equation with roots $x_1^2,x_2^2,x_3^2,x_4^2$ is
$$x^4-x^3-4x^2-12x+16=0$$
Is this solution correct? It took me a while to verify the identities, and I’m wondering if there’s a simpler way to solve this problem.

 

[Aion]

Let $x_1,x_2,x_3,x_4$ be roots to the equation $x^4+x^3-2x-4=0$. Determine the equation with roots $x_1^2,x_2^2,x_3^2,x_4^2$.

My attempt:

Expanding the product
$$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=x^4-\left(\sum_{i=1}^4x_i\right)x^3+\left(\sum_{1 \leq i < j \leq 4} x_i x_j\right)x^2 -\left(\sum_{1 \leq i < j < u \leq 4} x_i x_j x_u\right)x+x_1x_2x_3x_4$$
By comparing with $x^4+x^3-2x-4=0$ we see that

$$x_1+x_2+x_3+x_4=-1$$$$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=0$$$$x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4=2$$$$x_1x_2x_3x_4=4$$
We aim to find the polynomial whose roots are $x_1^2,x_2^2,x_3^2,x_4^2$. Let's write it as

$$x^4-a_1x^3+a_2x^2-a_3x+a_4$$
Now notice that
$$a_1=x_1^2+x_2^2+x_3^2+x_4^2=(x_1+x_2+x_3+x_4)^2-2\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 \right)=(-1)^2-2(0)=1$$$$a_2=x_1^2x_2^2+x_1^2x_3^2+x_1^2x_4^2+x_2^2x_3^2+x_2^2x_4^2+x_3^2x_4^2=\left(x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 \right)^2-2\left(x_1x_2x_3+x_1x_2x_3+x_1x_3x_4+x_2x_3x_4 \right)=-4$$$$a_3=x_1^2x_2^2x_3^2+x_1^2x_2^2x_4^2+x_1^2x_3^2x_4^2+x_2^2x_3^2x_4^2=\left(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 \right)^2-2\left(x_1x_2x_3x_4(x_1+x_2+x_3+x_4) \right)=12$$$$a_4=\left(x_1x_2x_3x_4\right)^2=4^2=16$$
Hence the equation with roots $x_1^2,x_2^2,x_3^2,x_4^2$ is
$$x^4-x^3-4x^2-12x+16=0$$
Is this solution correct? It took me a while to verify the identities, and I’m wondering if there’s a simpler way to solve this problem.

I made a mistake, $x_1x_2x_3x_4=-4$ however it doesn't affect the solution.

 

[blamocur]

I get a different coefficient at 'x'.

I find another approach slightly easier. Hint: $(x-x_k)(x+x_k) = x^2 - x_k^2$.

 

[Aion]

I get a different coefficient at 'x'.

I find another approach slightly easier. Hint: $(x-x_k)(x+x_k) = x^2 - x_k^2$.

Are you suggesting that we multiply the polynomial $P(x)$ and $P(-x)$, the roots of the resulting polynomial will be $x_k^2$?

 

[Aion]

Are you suggesting that we multiply the polynomial $P(x)$ and $P(-x)$, the roots of the resulting polynomial will be $x_k^2$?

$$P(x)\cdot P(-x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(-x-x_1)(-x-x_2)(-x-x_3)(-x-x_4)$$Using the identity $(x-x_k)(-x-x_k)=-(x^2-x_k^2)$. Then the product becomes

$$P(x)\cdot P(-x)=(x^2-x_1^2)(x^2-x_2^2)(x^2-x_3^2)(x^2-x_4^2)$$
And it has roots $\pm \sqrt{x_1^2},\pm \sqrt{x_2^2},\pm \sqrt{x_3^2}\pm \sqrt{x_4^2}$. And we have

$$P(x)\cdot P(-x)=(x^4+x^3-2x-4)(x^4-x^3+2x-4)=x^8-x^6-4x^4-4x^2+16$$
If we let $y=x^2$, then the polynomial will have roots $x_1^2,x_2^2,x_3^2,x_4^2$

$$(y-x_1^2)(y-x_2^2)(y-x_3^2)(y-x_4^2)=0$$And the polynomial simplifies into
$$y^4-y^3-4y^2-4y+16=0$$