If the difference of the roots of $x^2-\lambda x+8=0$ be $2$, the value of $\lambda$ is

Given quadratic equation is $x^2-\lambda x+8=0$ and the difference of roots is $2$.

We have to find the value of $\lambda$.

Let the roots of the above quadratic equation be $\alpha$ and $\beta$. Then,

Sum of roots $=\frac{-b}{a}=\frac{-\left(-\lambda \right)}{1}=\lambda \left[\begin{array}{c}\because b=-\lambda \\ a=1\end{array}\right]$

$\Rightarrow \alpha +\beta =\lambda \cdots \cdots \cdots \cdots \cdots \left(1\right)$

and, product of roots $=\frac{c}{a}=\frac{8}{1}=8 \left[\begin{array}{c}\because c=8\\ a=1\end{array}\right]$

$\Rightarrow \lambda \beta =8 \cdots \cdots \cdots \cdots \cdots \left(2\right)$

and $\alpha -\beta =2$  (given the difference of roots)

Now, using the fact that  ${\left(\alpha +\beta \right)}^2={\left(\alpha -\beta \right)}^2+4\alpha \beta$, we get

$$\begin{gathered} \Rightarrow {\left(\lambda \right)}^2={\left(2\right)}^2+\left(4\times 8\right) \left[\text{Using equation }\left(1\right), \left(2\right), \left(3\right)\right] \\ \Rightarrow {\lambda }^2=\left(4+32\right)=36 \\ \Rightarrow \lambda =\pm \sqrt{36}=\pm 6 \end{gathered}$$

So, $\lambda =\pm 6$

Hence option $\left(3\right)$ is correct.