If the quadratic equation formed by eliminati | vedclass

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(A) From $xy+l(x+y)+m=0$,we have $x(y+l) = -(ly+m)$,so $x = -\frac{ly+m}{y+l}$. Substituting this into $x^2+\alpha x+\beta=0$ gives: $\left(-\frac{ly+

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$\{m, \alpha l-m\}$

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(A) From $xy+l(x+y)+m=0$,we have $x(y+l) = -(ly+m)$,so $x = -\frac{ly+m}{y+l}$. Substituting this into $x^2+\alpha x+\beta=0$ gives: $\left(-\frac{ly+m}{y+l}\right)^2 + \alpha\left(-\frac{ly+m}{y+l}\right) + \beta = 0$ $(ly+m)^2 - \alpha(ly+m)(y+l) + \beta(y+l)^2 = 0$ $(l^2y^2 + m^2 + 2lmy) - \alpha(ly^2 + l^2y + my + ml) + \beta(y^2 + 2ly + l^2) = 0$ $(l^2 - \alpha l + \beta)y^2 + (2lm - \alpha l^2 - \alpha m + 2\beta l)y + (m^2 - \alpha ml + \beta l^2) = 0$ Since this equation has the same roots as $x^2 + \alpha x + \beta = 0$,the ratios of coefficients must be equal: $\frac{l^2 - \alpha l + \beta}{1} = \frac{2lm - \alpha l^2 - \alpha m + 2\beta l}{\alpha} = \frac{m^2 - \alpha ml + \beta l^2}{\beta}$ Equating the first and third terms: $\beta(l^2 - \alpha l + \beta) = m^2 - \alpha ml + \beta l^2$ $\beta l^2 - \beta \alpha l + \beta^2 = m^2 - \alpha ml + \beta l^2$ $\beta^2 - \beta \alpha l - m^2 + \alpha ml = 0$ $\beta^2 - \beta m + \beta m - \beta \alpha l - m^2 + \alpha ml = 0$ $\beta(\beta - m) + (m - \alpha l)(\beta - m) = 0$ $(\beta - m)(\beta + m - \alpha l) = 0$ Thus,$\beta = m$ or $\beta = \alpha l - m$.

List-$I$	List-$II$
$(A)$ The minimum value of $2x^2 + 4x + 5$	$(I)$ $-1$
$(B)$ The maximum value of $\frac{x^2 + 4x + 1}{x^2 + x + 1}$	$(II)$ $1$
$(C)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $b =$	$(III)$ $2$
$(D)$ If $1 \leq \frac{3x^2 - 5x + 6}{x^2 + 1} \leq 2$, $\forall x \in [a, b]$ then $a =$	$(IV)$ $3$
	$(V)$ $4$

If the quadratic equation formed by eliminating $x$ from $x^2+\alpha x+\beta=0$ and $xy+l(x+y)+m=0$ has the same roots as that of the given quadratic equation,then the set of values of $\beta$ is

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