$x^4 + 4rx + 3s = 0$ has no real roots. Relate $r, s$.

Question

It is given that $x^4 + 4rx + 3s = 0$ has no real roots. What can be said about r and s?

a) $r^2 < s^3$
b) $r^2 > s^3$
c) $r^4 < s^3$
d) $r^4 > s^3$

How to even begin??

Answer 1

The equation has no real roots if and only if the minimum value of the function $f(x)=x^4+4rx+3s$ is positive.

Set the derivative of $f(x)$ equal to $0$ and solve, to find an expression for the $x$ at which the minimum occurs.

Plug this into $f(x)$ to find the minimum value of our function.

Write down the condition that this minimum value is positive.

Answer 2

Hint:

Assuming $\,r,s\in\Bbb R\, $ , the roots of the polynomial are $\,z,\bar z,w,\bar w\;,\;\;z,w\in\Bbb C-\Bbb R\,$ (why?) , so:

$$z\bar zw\bar w=|z|^2|w|^2=3s$$

$$z+\bar z+w+\bar w=0\implies \operatorname{Re}(z+w)=0$$

$$|z|^2(w+\bar w)+|w|^2(z+\bar z)=-4r\;,\;\;\text{so}\ldots$$