Using Vieta's formula on a polynomial equation

[Aion]

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

My attempt:

Expanding the product
[math](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[/math]
By comparing with [imath]x^4+x^3-2x-4=0[/imath] we see that

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

[math]x^4-a_1x^3+a_2x^2-a_3x+a_4[/math]
Now notice that
[math]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[/math][math]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[/math][math]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[/math][math]a_4=\left(x_1x_2x_3x_4\right)^2=4^2=16[/math]
Hence the equation with roots [imath]x_1^2,x_2^2,x_3^2,x_4^2[/imath] is
[math]x^4-x^3-4x^2-12x+16=0[/math]
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 [imath]x_1,x_2,x_3,x_4[/imath] be roots to the equation [imath]x^4+x^3-2x-4=0[/imath]. Determine the equation with roots [imath]x_1^2,x_2^2,x_3^2,x_4^2[/imath].

My attempt:

Expanding the product
[math](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[/math]
By comparing with [imath]x^4+x^3-2x-4=0[/imath] we see that

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

[math]x^4-a_1x^3+a_2x^2-a_3x+a_4[/math]
Now notice that
[math]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[/math][math]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[/math][math]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[/math][math]a_4=\left(x_1x_2x_3x_4\right)^2=4^2=16[/math]
Hence the equation with roots [imath]x_1^2,x_2^2,x_3^2,x_4^2[/imath] is
[math]x^4-x^3-4x^2-12x+16=0[/math]
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, [imath]x_1x_2x_3x_4=-4[/imath] however it doesn't affect the solution.

 

[blamocur]

I get a different coefficient at 'x'.

I find another approach slightly easier. Hint: [imath](x-x_k)(x+x_k) = x^2 - x_k^2[/imath].

 

[Aion]

I get a different coefficient at 'x'.

I find another approach slightly easier. Hint: [imath](x-x_k)(x+x_k) = x^2 - x_k^2[/imath].

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

 

[Aion]

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

[math]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)[/math]Using the identity [imath](x-x_k)(-x-x_k)=-(x^2-x_k^2)[/imath]. Then the product becomes

[math]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)[/math]
And it has roots [imath]\pm \sqrt{x_1^2},\pm \sqrt{x_2^2},\pm \sqrt{x_3^2}\pm \sqrt{x_4^2}[/imath]. And we have

[math]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[/math]
If we let [imath]y=x^2[/imath], then the polynomial will have roots [imath]x_1^2,x_2^2,x_3^2,x_4^2[/imath]

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