What is the product of all values of x that satisfies the equation:

First, make sure that \(\sqrt{5x}\) and \(\sqrt{7x-3}\) have real value so \(x \geq 0\) and \(x \geq \frac{3}{7}\). Hence \(x \geq \frac{3}{7}\).

Now, solve that equation
\(\begin{align}
\quad \sqrt{5x}+1 &= \sqrt{7x-3} \\
5x + 2\sqrt{5x}+1 &= 7x-3 \\
2\sqrt{5x} &= 2x-4 \\
\sqrt{5x} &= x-2 \\
x - \sqrt{5x} - 2 &=0
\end{align}\)

Set \(t = \sqrt{x} \geq \sqrt{3/7}\) we have \(t^2 -t\sqrt{5} -2 =0\)

\(\delta = 5 + 4*2 = 13 \implies t_{12}=\frac{\sqrt{5} \pm \sqrt{13}}{2}\)

We have \(t_1 = \frac{\sqrt{5}+ \sqrt{13}}{2} \approx 2.9 > 1 > \sqrt{\frac{3}{7}}\). Choose this root.
\(t_2 = \frac{\sqrt{5}- \sqrt{13}}{2} \approx -0.7 < 0 < \sqrt{\frac{3}{7}}\). Eliminate this root.

Hence \(x=t_1^2=\frac{9+\sqrt{65}}{2}\).

The answer E.