Question

If the equation ${\text{x}}^{\text{4}}\text{+ a}{\text{x}}^{\text{3}}\text{+ b}{\text{x}}^{\text{2}}\text{+ cx + d = 0}\left(\text{a,b,c,d}\in \text{R}\right)$ has four imaginary roots, two with sum 3 +4i and the other two with product 13+i, then then the value of $\frac{b}{17}$ is:

Answer 1

Given equation $x^4+a{x}^2+b{x}^2+cx+d=0$

has four imaginary roots, Let it be $\alpha ,\overline{\alpha }$ and

$\beta$ and $\overline{\beta }$

Such that, $\alpha +\beta =3+4i$

and $\alpha \beta =13+i$

Now $\alpha \overline{\alpha }+\overline{\alpha }\beta +\beta \overline{\beta }+\overline{\beta }\alpha +\alpha \beta +\overline{\alpha }\overline{\beta }$$=\frac{coefficientof{x}^2}{coefficientof{x}^4}$$=b$

$\Rightarrow \alpha \overline{\alpha }+\overline{\alpha }\beta +\beta \overline{\beta }+\overline{\beta }\alpha +\alpha \beta +\overline{\alpha \beta }=b$

$\Rightarrow \overline{\alpha }(\alpha +\beta )+\overline{\beta }(\alpha +\beta )+\alpha \beta +\overline{\alpha \beta }=b$

$\Rightarrow (\alpha +\beta )(\overline{\alpha }+\overline{\beta })+13-i+13+i=b$

$\Rightarrow (\alpha +\beta )\left(\overline{\alpha +\beta }\right)+26=b$

$\Rightarrow (\alpha +\beta ).\left(\overline{\alpha +\beta }\right)=b-26$

$\Rightarrow (3-4i)(3+4i)+26=b$

$\Rightarrow 9-16i^2+26=b$

$\Rightarrow 9+16+26=b$

$\Rightarrow b=51$

Now, $\frac{b}{17}=\frac{51}{7}=3$