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 $13+i$ 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.
 A: 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 $b=51$.
A: You now have two simultaneous equations in the two variables $\alpha$ and $\beta$. There are several ways to solve those equations: the most straightforward is substitution.
From your first simultaneous equation,
$$\beta=(3+4i)-\alpha$$
Substituting that into the second simultaneous equation,
$$\alpha(3+4i-\alpha)=13-i$$
$$(3+4i)\alpha-\alpha^2=13-i$$
$$\alpha^2+(-3-4i)\alpha+(13-i)=0$$
Using the quadratic formula,
$$\begin{align}
\alpha &= \frac{-(-3-4i)\pm\sqrt{(-3-4i)^2-4\cdot 1\cdot(13-i)}}{2\cdot 1} \\
 &= \frac 12\left(3+4i\pm\sqrt{-59+28i}\right)
\end{align}$$
The expression for $\beta$ is the same, except using $\mp$ rather than $\pm$. I have been unable to simplify that square root. Since the norm of the number inside the square root is the product of two distinct primes, I doubt that there is a good simplification.
You should be able to finish from there. Let me know if you need more help.
