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(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.Rearranging the terms,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides:$(\sqrt{3x + 1})^

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(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.Rearranging the terms,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides:$(\sqrt{3x + 1})^2 = (\sqrt{x} - 1)^2$$3x + 1 = x + 1 - 2\sqrt{x}$.Simplifying the equation:$2x = -2\sqrt{x}$$x = -\sqrt{x}$.Squaring again:$x^2 = x$$x^2 - x = 0$$x(x - 1) = 0$,which gives $x = 0$ or $x = 1$.Checking $x = 0$: $\sqrt{3(0) + 1} + 1 = 1 + 1 = 2$,while $\sqrt{0} = 0$. Since $2 
eq 0$,$x = 0$ is not a solution.Checking $x = 1$: $\sqrt{3(1) + 1} + 1 = \sqrt{4} + 1 = 2 + 1 = 3$,while $\sqrt{1} = 1$. Since $3 
eq 1$,$x = 1$ is not a solution.Therefore,there are no real roots for the given equation.

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

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