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(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$. By Vieta's formulas,$\alpha_1 + \alpha_2 = -b/a$ and $\alpha_1 \alpha_2 = c/

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$b^2 pr = q^2 ac$

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(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$. By Vieta's formulas,$\alpha_1 + \alpha_2 = -b/a$ and $\alpha_1 \alpha_2 = c/a$. Similarly,$\beta_1, \beta_2$ are the roots of $px^2 + qx + r = 0$,so $\beta_1 + \beta_2 = -q/p$ and $\beta_1 \beta_2 = r/p$. The system $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution if the determinant of the coefficient matrix is zero: $\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0$,which implies $\alpha_1 / \beta_1 = \alpha_2 / \beta_2 = k$ (say). Then $\alpha_1 = k \beta_1$ and $\alpha_2 = k \beta_2$. From the sum and product of roots: $\alpha_1 + \alpha_2 = k(\beta_1 + \beta_2) \implies -b/a = k(-q/p) \implies b/a = k(q/p) \implies k = bp/aq$. $\alpha_1 \alpha_2 = k^2(\beta_1 \beta_2) \implies c/a = k^2(r/p)$. Substituting $k = bp/aq$: $c/a = (b^2 p^2 / a^2 q^2) \cdot (r/p) = (b^2 pr) / (a^2 q^2)$. $c/a = (b^2 pr) / (a^2 q^2) \implies c = (b^2 pr) / (a q^2) \implies a q^2 c = b^2 pr$. Thus,$b^2 pr = q^2 ac$.

If $\alpha_1, \alpha_2$ and $\beta_1, \beta_2$ are the roots of the equations $ax^2 + bx + c = 0$ and $px^2 + qx + r = 0$ respectively,and the system of equations $\alpha_1 y + \alpha_2 z = 0$ and $\beta_1 y + \beta_2 z = 0$ has a non-zero solution,then:

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