Question

If the roots of the equation $x^4-6x^3+18x^2-30x+25=0$ are of the form $\alpha \pm i\beta$ and $\beta \pm i\alpha$, then $(\alpha ,\beta )=$

• A. $(-1,-2)$
• B. $(1,2)$
• C. $(1,-2)$
• D. $(5,1)$

Answer 1

The correct option is B $(1,2)$

As roots of $x^4-6x^3+18x^2-30x+25=0$ are $\alpha \pm i\beta$ and $\beta \pm i\alpha ,$ then

$S_1=\alpha +i\beta +\alpha -i\beta +\beta +i\alpha +\beta -i\alpha =6$$\Rightarrow 2\alpha +2\beta =6\Rightarrow \alpha +\beta =3$ ...(1)

$S_2=(\alpha +i\beta )(\alpha -i\beta )+(\alpha +i\beta )(\beta +i\alpha )$$+(\alpha +i\beta )(\beta -i\alpha )+(\alpha -i\beta )(\beta +i\alpha )$

$+(\alpha -i\beta )(\beta -i\alpha )+(\beta +i\alpha )(\beta -i\alpha )=18$

$\Rightarrow {\alpha }^2+{\beta }^2+\alpha \beta +i{\alpha }^2+i{\beta }^2-\alpha \beta +\alpha \beta -i{\alpha }^2$$+i{\beta }^2+\alpha \beta +\alpha \beta +i{\alpha }^2-i{\beta }^2+\alpha \beta$

$+\alpha \beta -i{\alpha }^2-i{\beta }^2-\alpha \beta +{\beta }^2+{\alpha }^2=18$$\Rightarrow 2{\alpha }^2+2{\beta }^2+4\alpha \beta =18$

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =9\Rightarrow {(\alpha +\beta )}^2=9$ ...(2)

$S_3=(\alpha +i\beta )(\alpha -i\beta )(\beta +i\alpha )$$+(\alpha +i\beta )(\alpha -i\beta )(\beta -i\alpha )$$+(\alpha +i\beta )(\beta +i\alpha )(\beta -i\alpha )$

$+(\alpha -i\beta )(\beta +i\alpha )(\beta -i\alpha )=30$

$\Rightarrow ({\alpha }^2+{\beta }^2)(\beta +i\alpha )+({\alpha }^2+{\beta }^2)(\beta -i\alpha )$$+({\alpha }^2+{\beta }^2)(\alpha +i\beta )+({\alpha }^2+{\beta }^2)(\alpha -i\beta )=30$ ...(3)

$\Rightarrow ({\alpha }^2+{\beta }^2)(2\beta +2\alpha )=30$ ...(4)

From (1) $({\alpha }^2+{\beta }^2)=5$

$S_4=(\alpha +i\beta )(\alpha -i\beta )(\beta -i\alpha )(\beta +i\alpha )=25$

$\Rightarrow ({\alpha }^2+{\beta }^2)({\alpha }^2+{\beta }^2)=25$

$\Rightarrow {({\alpha }^2+{\beta }^2)}^2=25$ ...(5)

From (1), (4), (5)

$\alpha -\beta =1$

$\therefore \alpha =1$ and $\beta =2$

Hence $(\alpha ,\beta )=(1,2)$