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 $13+i$

Question

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 $b$.

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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.

Answer 1

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.

Answer 2

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 $b=51$.