If alpha_1, alpha_2 and beta_1, beta_2 are th | vedclass

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(B) Given that $\alpha_1, \alpha_2$ are the roots of $ax^2 + bx + c = 0$. So,$\alpha_1 + \alpha_2 = -\frac{b}{a}$ and $\alpha_1 \alpha_2 = \frac{c}{a}

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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 $ax^2 + bx + c = 0$. So,$\alpha_1 + \alpha_2 = -\frac{b}{a}$ and $\alpha_1 \alpha_2 = \frac{c}{a}$. Now,$\beta_1, \beta_2$ are the roots of $px^2 + qx + r = 0$. So,$\beta_1 + \beta_2 = -\frac{q}{p}$ and $\beta_1 \beta_2 = \frac{r}{p}$. 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: $\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0 \implies \frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} = k$ (say). Then $\alpha_1 = k \beta_1$ and $\alpha_2 = k \beta_2$. From the sum and product of roots: $\frac{\alpha_1 + \alpha_2}{\beta_1 + \beta_2} = \frac{k(\beta_1 + \beta_2)}{\beta_1 + \beta_2} = k = \frac{-b/a}{-q/p} = \frac{bp}{aq}$. Also,$\frac{\alpha_1 \alpha_2}{\beta_1 \beta_2} = \frac{k^2 \beta_1 \beta_2}{\beta_1 \beta_2} = k^2 = \frac{c/a}{r/p} = \frac{cp}{ar}$. Equating $k^2$: $\left(\frac{bp}{aq}\right)^2 = \frac{cp}{ar} \implies \frac{b^2 p^2}{a^2 q^2} = \frac{cp}{ar}$. Simplifying: $b^2 p^2 ar = cp a^2 q^2 \implies b^2 pr = q^2 ac$.

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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