If the biquadratic $x^4+ax^3+bx^2+cx+d=0(a,b,c,d\in R)$ has $4$ non real roots,two with sum $3+4i$ and the other two with product $13+i$.Find the value of $b$.
Since there are four non real roots,so i let the roots as $\alpha,\bar{\alpha},\beta,\bar{\beta}$.
Then i let $\alpha+\beta=3+4i,\bar{\alpha}\bar{\beta}=13+i$
$\therefore\alpha+\beta=3+4i,\alpha\beta=13-i$
But i am stuck now,how should i proceed now and find the value of $b$.Please help me.
Thanks.
Best Answer
By Vieta’s theorem, $b$ is the sum of all products of pairwise distinct roots, i. e. $$ \begin{align*} b &= \alpha\overline\alpha + \alpha\beta + \alpha\overline\beta + \overline{\beta\alpha} + \beta\overline\alpha + \beta\overline\beta\\ &= \overline{(\alpha+\beta)}\cdot (\alpha+\beta) + \alpha\beta + \overline{\alpha\beta}\\ &= (3-4i)(3+4i) + 13-i + 13+i\\ &= 9 + 16 + 26 = 51, \end{align*} $$ so $b=51$.