If sqrt{x}=sqrt{3}-sqrt{5}, then the value of | vedclass

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(C) Given: $\sqrt{x} = \sqrt{3} - \sqrt{5}$.Squaring both sides:$x = (\sqrt{3} - \sqrt{5})^2$$x = 3 + 5 - 2\sqrt{15}$$x = 8 - 2\sqrt{15}$Rearranging t

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$2$

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(C) Given: $\sqrt{x} = \sqrt{3} - \sqrt{5}$.Squaring both sides:$x = (\sqrt{3} - \sqrt{5})^2$$x = 3 + 5 - 2\sqrt{15}$$x = 8 - 2\sqrt{15}$Rearranging the terms:$x - 8 = -2\sqrt{15}$Squaring both sides again:$(x - 8)^2 = (-2\sqrt{15})^2$$x^2 - 16x + 64 = 4 	imes 15$$x^2 - 16x + 64 = 60$Subtracting $60$ from both sides:$x^2 - 16x + 4 = 0$We need to find the value of $x^2 - 16x + 6$:$x^2 - 16x + 6 = (x^2 - 16x + 4) + 2$Since $x^2 - 16x + 4 = 0$,we have:$0 + 2 = 2$.

If $\sqrt{x}=\sqrt{3}-\sqrt{5},$ then the value of $x^{2}-16x+6$ is

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