If$$\sqrt{2x -1} - \sqrt{x - 4} = 2$$​ , then find the value of ‘x’.

​$$\begin{aligned}&\textbf{Given:} \\&\sqrt{2x - 1} = 2 + \sqrt{x - 4} \\ &\textbf{Solution:} \\&\text{Square both sides:} \\&(\sqrt{2x - 1})^2 = (2 + \sqrt{x - 4})^2 \\&2x - 1 = 4 + 4\sqrt{x - 4} + (x - 4) \\&2x - 1 = x + 4 + 4\sqrt{x - 4} \\&x - 1 = 4\sqrt{x - 4} \\&\text{Square both sides again:} \\&(x - 1)^2 = (4\sqrt{x - 4})^2 \\&(x - 1)^2 = 16(x - 4) \\&x^2 - 2x + 1 = 16x - 64 \\&x^2 - 18x + 65 = 0 \\&\text{Using the quadratic formula:} \\&x = \frac{18 \pm \sqrt{18^2 - 4(1)(65)}}{2(1)} \\&x = \frac{18 \pm \sqrt{64}}{2} \\&x = \frac{18 \pm 8}{2} \\&\text{The possible solutions are:} \\&x = \frac{18 + 8}{2} = 13 \\&x = \frac{18 - 8}{2} = 5\end{aligned}$$ 

therefore, x = 13, 5 

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