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(D) Given equations are:$x^{2}-4 \lambda x+5=0$ with roots $\alpha, \beta$ $\implies \alpha+\beta=4 \lambda$ and $\alpha \beta=5$.$x^{2}-(3 \sqrt{2}+2

Vedclass · Accepted Answer

$98$

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(D) Given equations are:$x^{2}-4 \lambda x+5=0$ with roots $\alpha, \beta$ $\implies \alpha+\beta=4 \lambda$ and $\alpha \beta=5$.$x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+(7+3 \lambda \sqrt{3})=0$ with roots $\alpha, \gamma$ $\implies \alpha+\gamma=3 \sqrt{2}+2 \sqrt{3}$ and $\alpha \gamma=7+3 \lambda \sqrt{3}$.Subtracting the sum of roots: $(\alpha+\gamma)-(\alpha+\beta) = (3 \sqrt{2}+2 \sqrt{3}) - 4 \lambda \implies \gamma-\beta = 3 \sqrt{2}+2 \sqrt{3}-4 \lambda$.Given $\beta+\gamma=3 \sqrt{2}$.Adding the two equations: $2 \gamma = 6 \sqrt{2}+2 \sqrt{3}-4 \lambda \implies \gamma = 3 \sqrt{2}+\sqrt{3}-2 \lambda$.Subtracting: $2 \beta = 4 \lambda - 2 \sqrt{3} \implies \beta = 2 \lambda - \sqrt{3}$.Since $\alpha+\beta=4 \lambda$,we have $\alpha = 4 \lambda - (2 \lambda - \sqrt{3}) = 2 \lambda + \sqrt{3}$.Using $\alpha \beta = 5$: $(2 \lambda + \sqrt{3})(2 \lambda - \sqrt{3}) = 5 \implies 4 \lambda^{2}-3 = 5 \implies 4 \lambda^{2}=8 \implies \lambda^{2}=2$.Since $\alpha = 2 \lambda + \sqrt{3}$,$\beta = 2 \lambda - \sqrt{3}$,and $\gamma = 3 \sqrt{2}+2 \sqrt{3}-\alpha = 3 \sqrt{2}+2 \sqrt{3}-(2 \lambda+\sqrt{3}) = 3 \sqrt{2}+\sqrt{3}-2 \lambda$.We need $(\alpha+2 \beta+\gamma)^{2} = (\alpha+\beta+\beta+\gamma)^{2} = (4 \lambda + 3 \sqrt{2})^{2}$.Since $\lambda^{2}=2$,let $\lambda = \sqrt{2}$ (assuming positive root for consistency): $(4 \sqrt{2}+3 \sqrt{2})^{2} = (7 \sqrt{2})^{2} = 49 	imes 2 = 98$.

Let $\alpha, \beta$ be the roots of the equation $x^{2}-4 \lambda x+5=0$ and $\alpha, \gamma$ be the roots of the equation $x^{2}-(3 \sqrt{2}+2 \sqrt{3}) x+7+3 \lambda \sqrt{3}=0$. If $\beta+\gamma=3 \sqrt{2}$,then $(\alpha+2 \beta+\gamma)^{2}$ is equal to

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