Let alpha, beta be the roots of x^2 + (3 - la | vedclass

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(C) Given the quadratic equation $x^2 + (3 - \lambda)x - \lambda = 0$.By Vieta's formulas,the sum of the roots $\alpha + \beta = - (3 - \lambda) = \la

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$2$

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(C) Given the quadratic equation $x^2 + (3 - \lambda)x - \lambda = 0$.By Vieta's formulas,the sum of the roots $\alpha + \beta = - (3 - \lambda) = \lambda - 3$ and the product of the roots $\alpha \beta = - \lambda$.We want to minimize $S = \alpha^2 + \beta^2$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we substitute the values:$S = (\lambda - 3)^2 - 2(-\lambda) = \lambda^2 - 6\lambda + 9 + 2\lambda = \lambda^2 - 4\lambda + 9$.To find the minimum value,we can complete the square:$S = (\lambda^2 - 4\lambda + 4) + 5 = (\lambda - 2)^2 + 5$.Since $(\lambda - 2)^2 \ge 0$,the minimum value of $S$ is $5$,which occurs when $\lambda - 2 = 0$,i.e.,$\lambda = 2$.

Let $\alpha, \beta$ be the roots of $x^2 + (3 - \lambda)x - \lambda = 0$. The value of $\lambda$ for which $\alpha^2 + \beta^2$ is minimum,is

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