Let p, q, r in mathbb{R} and r p 0. If th | vedclass

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

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greater than $2$

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(C) The given quadratic equation is $px^2 + qx + r = 0$ with $p, q, r \in \mathbb{R}$ and $r > p > 0.$Since the roots $\alpha$ and $\beta$ are complex,the discriminant $D = q^2 - 4pr This implies $q^2 Since the coefficients are real,the complex roots must be conjugates,i.e.,$\beta = \bar{\alpha}.$Thus,$|\alpha| = |\beta| = \sqrt{\alpha \bar{\alpha}} = \sqrt{\alpha \beta}.$From the properties of quadratic equations,the product of the roots is $\alpha \beta = \frac{r}{p}.$Therefore,$|\alpha| = |\beta| = \sqrt{\frac{r}{p}}.$Given $r > p > 0,$ we have $\frac{r}{p} > 1,$ which implies $\sqrt{\frac{r}{p}} > 1.$Then,$|\alpha| + |\beta| = 2|\alpha| = 2\sqrt{\frac{r}{p}}.$Since $\sqrt{\frac{r}{p}} > 1,$ it follows that $2\sqrt{\frac{r}{p}} > 2.$Hence,$|\alpha| + |\beta| > 2.$

List-$I$	List-$II$
$(i)$ $\alpha = \beta$	$(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii)$ $\alpha = 2\beta$	$(B)$ $2b^2 = 9ac$
$(iii)$ $\alpha = 3\beta$	$(C)$ $b^2 = 6ac$
$(iv)$ $\alpha = \beta^2$	$(D)$ $3b^2 = 16ac$
	$(E)$ $b^2 = 4ac$
	$(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$

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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