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(A) Since $\alpha$ and $\beta$ are the roots of the equation $x^{2}-3 \lambda x+\lambda^{2}=0$,we have:Sum of roots: $\alpha+\beta = 3 \lambda$Product

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$\pm \frac{1}{2}$

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(A) Since $\alpha$ and $\beta$ are the roots of the equation $x^{2}-3 \lambda x+\lambda^{2}=0$,we have:Sum of roots: $\alpha+\beta = 3 \lambda$Product of roots: $\alpha \beta = \lambda^{2}$Given that $\alpha^{2}+\beta^{2} = \frac{7}{4}$.Using the identity $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2 \alpha \beta$,we substitute the values:$(3 \lambda)^{2}-2 \lambda^{2} = \frac{7}{4}$$9 \lambda^{2}-2 \lambda^{2} = \frac{7}{4}$$7 \lambda^{2} = \frac{7}{4}$$\lambda^{2} = \frac{1}{4}$$\lambda = \pm \frac{1}{2}$

If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-3 \lambda x+\lambda^{2}=0,$ find $\lambda$ if $\alpha^{2}+\beta^{2}=\frac{7}{4}$

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