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(C) Given the quadratic equation $px^2 + qx + r = 0$ where $p, q, r \in \mathbb{R}$ and $r > p > 0.$Since the roots $\alpha$ and $\beta$ are complex,t

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greater than $2$

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(C) Given the quadratic equation $px^2 + qx + r = 0$ where $p, q, r \in \mathbb{R}$ and $r > p > 0.$Since the roots $\alpha$ and $\beta$ are complex,they must be conjugates of each other,i.e.,$\beta = \bar{\alpha}.$This implies $|\beta| = |\bar{\alpha}| = |\alpha|.$From the properties of quadratic equations,the product of the roots is $\alpha \beta = \frac{r}{p}.$Since $\beta = \bar{\alpha},$ we have $\alpha \bar{\alpha} = |\alpha|^2 = \frac{r}{p}.$Given $r > p > 0,$ we have $\frac{r}{p} > 1.$Therefore,$|\alpha|^2 > 1,$ which implies $|\alpha| > 1.$Now,consider $|\alpha| + |\beta| = |\alpha| + |\alpha| = 2|\alpha|.$Since $|\alpha| > 1,$ it follows that $2|\alpha| > 2.$Thus,$|\alpha| + |\beta| > 2.$

Let $p, q, r \in \mathbb{R}$ and $r > p > 0.$ If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta,$ then $|\alpha| + |\beta|$ is:

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