For x 2,the equation sqrt{x+2} - sqrt{x-2} | vedclass

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(D) Given the equation for $x > 2$: $\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$ Squaring both sides,we get: $(x+2) + (x-2) - 2\sqrt{(x+2)(x-2)} = 4x - 2$

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(D) Given the equation for $x > 2$: $\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$ Squaring both sides,we get: $(x+2) + (x-2) - 2\sqrt{(x+2)(x-2)} = 4x - 2$ $2x - 2\sqrt{x^2-4} = 4x - 2$ $-2\sqrt{x^2-4} = 2x - 2$ $-\sqrt{x^2-4} = x - 1$ Squaring again: $x^2 - 4 = (x-1)^2$ $x^2 - 4 = x^2 - 2x + 1$ $-4 = -2x + 1$ $2x = 5$ $x = 2.5$ Now,check the value $x = 2.5$ in the original equation: $LHS$: $\sqrt{2.5+2} - \sqrt{2.5-2} = \sqrt{4.5} - \sqrt{0.5} = \sqrt{9 	imes 0.5} - \sqrt{0.5} = 3\sqrt{0.5} - \sqrt{0.5} = 2\sqrt{0.5} = \sqrt{4 	imes 0.5} = \sqrt{2}$ $RHS$: $\sqrt{4(2.5) - 2} = \sqrt{10 - 2} = \sqrt{8} = 2\sqrt{2}$ Since $LHS$ $
eq$ $RHS$,there is no solution. Thus,option $D$ is correct.

For $x > 2$,the equation $\sqrt{x+2} - \sqrt{x-2} = \sqrt{4x-2}$ has

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