If alpha is a common root of x^2-5x+lambda=0 | vedclass

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(C) Given that $\alpha$ is a common root of $x^2-5x+\lambda=0$ and $x^2-8x-2\lambda=0$. Substituting $\alpha$ in both equations: $\alpha^2-5\alpha+\la

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$1$

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(C) Given that $\alpha$ is a common root of $x^2-5x+\lambda=0$ and $x^2-8x-2\lambda=0$. Substituting $\alpha$ in both equations: $\alpha^2-5\alpha+\lambda=0$ ... $(i)$ $\alpha^2-8\alpha-2\lambda=0$ ... (ii) Subtracting (ii) from $(i)$: $(x^2-5\alpha+\lambda) - (\alpha^2-8\alpha-2\lambda) = 0$ $3\alpha+3\lambda=0 \Rightarrow \alpha=-\lambda$. Since $\alpha$ is a root of $x^2-5x+\lambda=0$,substitute $\alpha=-\lambda$: $(-\lambda)^2-5(-\lambda)+\lambda=0$ $\lambda^2+5\lambda+\lambda=0 \Rightarrow \lambda^2+6\lambda=0$. Since $\lambda 
eq 0$,we have $\lambda=-6$. Thus,$\alpha = -(-6) = 6$. For $x^2-5x-6=0$,the roots are $\alpha=6$ and $\beta=-1$ (since $\alpha+\beta=5$). For $x^2-8x+12=0$,the roots are $\alpha=6$ and $\gamma=2$ (since $\alpha+\gamma=8$). Finally,$\alpha+\beta+\gamma+\lambda = 6 + (-1) + 2 + (-6) = 1$.

If $\alpha$ is a common root of $x^2-5x+\lambda=0$ and $x^2-8x-2\lambda=0$ $(\lambda \neq 0)$ and $\beta, \gamma$ are the other roots of them respectively,then $\alpha+\beta+\gamma+\lambda=$

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