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+\lamb

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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)$$\alpha^2-8\alpha-2\lambda=0$ ... $(ii)$Subtracting $(ii)$ from $(i)$:$(\alpha^2-5\alpha+\lambda) - (\alpha^2-8\alpha-2\lambda) = 0$$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+6\lambda=0 \Rightarrow \lambda(\lambda+6)=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 the sum of roots is $5$, $6+\beta=5$).For $x^2-8x+12=0$, the roots are $\alpha=6$ and $\gamma=2$ (since the sum of roots is $8$, $6+\gamma=8$).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 $\beta$ and $\gamma$ are the other roots of these equations respectively, then $\alpha+\beta+\gamma+\lambda=$

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