Answer
There is no extraneous solution.
Work Step by Step
The given equation is
$\Rightarrow \sqrt{6x-5}=x$
Check $x=5$.
$\Rightarrow \sqrt{6x-5}=x$
$\Rightarrow \sqrt{6(5)-5}=5$
$\Rightarrow \sqrt{30-5}=5$
$\Rightarrow \sqrt{25}=5$
$\Rightarrow 5=5$
True.
Check $x=1$.
$\Rightarrow \sqrt{6x-5}=x$
$\Rightarrow \sqrt{6(1)-5}=1$
$\Rightarrow \sqrt{6-5}=1$
$\Rightarrow \sqrt{1}=1$
$\Rightarrow 1=1$
True.
Hence, there is no extraneous solution.
