Suppose $p, q, r, s\in R$ and $\alpha, \beta$ be the roots of $x^{2}+px+q=0$ and $\alpha^{4}, \beta^{4}$ be the roots of $x^{2}-rx+s=0$, then the equation $x^{2}-4qx+2q^{2}-r=0$ has always

The discriminant of the equation x^2 - 4qx + 2q^2 - r = 0 is calculated as 16q^2 - 4(2q^2 - r), which simplifies to 8q^2 + 4r. Using the identity a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2 = (p^2 - 2q)^2 - 2q^2 = p^4 - 4p^2q + 2q^2, we know r = p^4 - 4p^2q + 2q^2. Substituting r into the discriminant gives 16q^2 - 4p^4 + 16p^2q, which equals -4(p^4 - 4p^2q + 4q^2) or -4(p^2 - 2q)^2. Since this discriminant is strictly negative, the quadratic equation has two imaginary roots.

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