Find value of $\prod(1+\alpha^2)$ if $\alpha,\beta,\gamma,\delta$ are the roots of equation $x^4+4x^3-6x^2+7x-9$ then find $\prod(1+\alpha^2)$
I know


*

*$\sum \alpha=-4$

*$\sum \alpha\beta=-6$

*$\sum \alpha\beta\gamma=-7$

*$\alpha\beta\gamma\delta=-9$


I know the sum and products and other things about the roots but I am not being to rearrange them such that I am able to get the required answer

Transforming roots from $\alpha$ to $1+\alpha^2$,
$$y=1+\alpha^2$$
$$\alpha=\sqrt{y-1}$$
As $\alpha$ satisfies our given polynomial, Substitute $\alpha$ into that.
We get new equation with roots $1+\alpha^2,1+\beta^2,\ldots$
$$(x-1)^2+4(x-1)\sqrt{x-1}-6(x-1)+7\sqrt{x-1}-9=0$$
Rearranging and squaring,
$$(x-1)^2(x-7)^2+81-18(x-1)(x-7)=(x-1)(4x+3)^2$$
$$(x^2+1-2x)(x^2+49-14x)+81-18(x^2+7-8x)=(x-1)(16x^2+9+24x)$$
$$x^4-32x^3+34x^2+45x+13=0$$
$$\text{Product of roots}=\frac{13}{1}=13$$
