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

$$\sqrt{2x-1}$$ = 2 + $$\sqrt{x-4}$$

Squaring both sides we will get

$$(\sqrt{(2x-1)})^{2}$$ =$$(2+\sqrt{(x-4)})^{2}$$

2x - 1 = 4 + (x-4) + 4 $$\sqrt{x-4}$$

2x -x - 1 = 4 $$\sqrt{x-4}$$

x - 1 = 4 $$\sqrt{x-4}$$ 

Again squaring both sides ,

$$(x-1)^{2}$$ = 4$$(\sqrt{(x-4)})^{2}$$ 

$$x^2$$ + 1 -2x = 16(x-4)

$$x^2$$ + 1 - 2x = 16x - 64

$$x^2$$ -18x + 65 = 0

On solving the above equation ,we get  two rational roots that are 13 and 5