Consider the equation x^2 + alpha x + beta = | vedclass

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(A) Since $\alpha$ and $\beta$ are roots of $x^2 + \alpha x + \beta = 0$,by Vieta's formulas:$\alpha + \beta = -\alpha \implies \beta = -2\alpha$  ...

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The inequality is satisfied by exactly two integral values of $y$.

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(A) Since $\alpha$ and $\beta$ are roots of $x^2 + \alpha x + \beta = 0$,by Vieta's formulas:$\alpha + \beta = -\alpha \implies \beta = -2\alpha$  .......$(1)$$\alpha \beta = \beta$  .......$(2)$From $(2)$,$\beta(\alpha - 1) = 0$. Since $\alpha 
eq \beta$,$\beta$ cannot be $0$ (if $\beta=0$,then $\alpha=0$,but $\alpha 
eq \beta$). Thus,$\alpha = 1$.Substituting $\alpha = 1$ into $(1)$,we get $\beta = -2$.Now,the inequality is $||y - (-2)| - 1| This implies $-1 This means $-2 Thus,$y \in (-4, -2) \cup (-2, 0)$.The integral values of $y$ are $-3$ and $-1$,which are exactly two values. Therefore,option $A$ is correct.

Consider the equation $x^2 + \alpha x + \beta = 0$ having roots $\alpha, \beta$ such that $\alpha \neq \beta$. Also consider the inequality $||y - \beta| - \alpha| < \alpha$,then:

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