If $\alpha$ is a common root of $x^2-5 x+\lambda=0$ and $x^2-8 x-2 \lambda$ $=0(\lambda \neq 0)$ and $\beta,…
If $\alpha$ is a common root of $x^2-5 x+\lambda=0$ and $x^2-8 x-2 \lambda$ $=0(\lambda \neq 0)$ and $\beta, \gamma$ are the other roots of them, then $\alpha+\beta+\gamma+\lambda=$
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-1
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Solution
$\alpha$ is root of $x^2-5 x+\lambda=0$ and $x^2-8 x-2 \lambda=0$
$\Rightarrow \alpha^2-5 \alpha+\lambda=0$...(i)
and $\alpha^2-8 \alpha-2 \lambda=0$...(ii)
Subtracting (ii) from (i), we get $3 \alpha+3 \lambda=0 \Rightarrow \alpha=-\lambda$
As $\alpha$ is root of $x^2-5 x+\lambda=0$
$\begin{aligned}
& \Rightarrow \lambda^2+5 \lambda+\lambda=0 \Rightarrow \lambda^2+6 \lambda=0 \Rightarrow \lambda=-6[\because \lambda \neq 0] \\
& \Rightarrow a=6
\end{aligned}$ Now, $\alpha, \beta$ are roots of $x^2-5 x+(-6)=0$
$\Rightarrow \alpha+\beta=5 \Rightarrow \beta=-1$ And, $\alpha, \gamma$ are roots of $x^2-8 x+12=0$
$\Rightarrow \alpha+\gamma=8 \Rightarrow \gamma=2$ So, $\alpha+\beta+\gamma+\lambda=6-1+2-6=1$.