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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?

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4 Answers 4

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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 $$(\alpha^2+1)(\beta ^2+1)(\gamma^2+1)(\delta^2+1)=f(i)f(-i)=|f(i)|^2=|-7+3i|^2=49+9=58.$$

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    $\begingroup$ I loved your answer as it used the concept of complex numbers in a very easy way. $\endgroup$ Oct 9, 2017 at 2:19
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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 $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)=\color{blue}{58}$.

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    $\begingroup$ [+1] One should say to the OP that forming the equation with transformed roots is a standard method for this kind of problems (substitution/elimination method), here done with the less calculation possible! $\endgroup$
    – Jean Marie
    Oct 8, 2017 at 16:58
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You can do it by the Viete's theorem!

We obtain: $$\sum_{cyc}\alpha=2;$$ $$\frac{1}{4}\sum_{sym}\alpha\beta=1;$$ $$\sum_{cyc}\alpha\beta\gamma=-1$$ and $$\alpha\beta\gamma\delta=-7.$$ Thus, $$\prod_{cyc}(1+\alpha^2)=1+\alpha^2\beta^2\gamma^2\delta^2+\sum_{cyc}\alpha^2\beta^2\gamma^2+\frac{1}{4}\sum_{sym}\alpha^2\beta^2+\sum_{cyc}\alpha^2=$$ $$=1+49+\left(\sum_{cyc}\alpha\beta\gamma\right)^2-2\alpha\beta\gamma\delta\cdot\frac{1}{4}\sum_{sym}\alpha\beta+\left(\frac{1}{4}\sum_{sym}\alpha\beta\right)^2-2\sum_{cyc}\alpha\sum_{cyc}\alpha\beta\gamma+2\alpha\beta\gamma\delta+$$ $$+\left(\sum_{cyc}\alpha\right)^2-2\cdot\frac{1}{4}\sum_{sym}\alpha\beta=$$ $$=50+1-2\cdot(-7)\cdot1+1^2-2\cdot2\cdot(-1)+2\cdot(-7)+2^2-2=58.$$ Also we can use the following. $$(a^2+b^2)(x^2+y^2)=(ax+by)^2+(ay-bx)^2=(ax-by)^2+(ay+bx)^2.$$ Indeed, $$\prod_{cyc}(\alpha^2+1^2)=((\alpha+\beta)^2+(\alpha\beta-1)^2)((\gamma+\delta)^2+(\gamma\delta-1)^2)=$$ $$=((\alpha+\beta)(\gamma+\delta)-(\alpha\beta-1)(\gamma\delta-1))^2+((\alpha+\beta)(\gamma\delta-1)+(\alpha\beta-1)(\delta+\gamma))^2=$$ $$=\left(\frac{1}{4}\sum_{sym}\alpha\beta-\alpha\beta\gamma\delta-1\right)^2+\left(\sum_{cyc}\alpha\beta\gamma-\sum_{cyc}\alpha\right)^2=(1+7-1)^2+(-1-2)^2=58.$$ Done!

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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.

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