Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$

Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$ Are Vieta's formulas appropriate?

Solution 1:

There is no need to use Vieta's formulas. Let $$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$ Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that $$(\alpha^2+1)(\beta ^2+1)(\gamma^2+1)(\delta^2+1)=f(i)f(-i)=|f(i)|^2=|-7+3i|^2=49+9=58.$$

Solution 2:

Rearrange the equation to $z(2z^2-1)=z^4+z^2-7$ square this & substitute $y=z^2+1$ \begin{eqnarray*} (y-1)(2y-3)^2=((y-1)^2+(y-1)-7)^2 \\ y^4+\cdots +58 =0. \end{eqnarray*} Note that the roots of this polynomial will be $\alpha^2+1,\beta^2+1,\gamma^2+1,\delta^2+1$ and we have not calculated all the terms explicitly (only the ones required to obtain the product of these roots.) So $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\color{blue}{58}$.

Solution 3:

transform the equation so that you get the desired roots.

so the given equation has roosts $\alpha, \ \beta,\ \gamma, \ \delta $ first transform $x \to \sqrt{x}$ so the new equation ha roots $\alpha^2, \ \beta^2,\ \gamma^2, \ \delta^2 $ then transform $x \to x-1$ to get the equation whose roots are $\alpha^2+1, \ \beta^2+1,\ \gamma^2+1, \ \delta^2+1 $

now the answer is product of roots of new equation.