The roots of the equation $\sqrt {3x + 1} + 1 = \sqrt x $ are
A) 0
B) 1
C) 0, 1
D) None

Hint: To begin with the roots question, firstly take out all the possible intervals of x where it can belong and secondly try to remove all the roots to treat it as a polynomial function and find x.

Complete step by step solution: As we begin, well find all the x where it can belong. Now we need to remove squares. So try to simplify the equation.

$\sqrt {3x + 1} + 1 = \sqrt x $
Shifting 1 to the RHS and squaring is quite easier, So we going to try this
$\sqrt {3x + 1} = \sqrt x - 1$
squaring both the sides we get.
$3x + 1 = x + 1 - 2\sqrt x $
$2x = - 2\sqrt x $
$x = - \sqrt x $
Still we have a square root on the right hand side. We square it once more. So the resultant will be
$\begin{array}{l}
{x^2} = x\\
{x^2} - x = 0\\
x\left( {x - 1} \right) = 0\\
Therefore, x = 0,1
\end{array}$
Here, we got 2 values of x and it satisfies the condition obtained $x \ge 0$

Now we are going to check whether $x = \left\{ {0,1} \right\}$ Satisfies the initial function or not
So, for $x = 0$ Let’s check it out
$\sqrt {3\left( 0 \right) + 1} + 1 = \sqrt 0 $
$2 = 0,$ which is a false statement now, Let’s check it up for $x = 1,$
$\sqrt {3\left( 1 \right) + 1} + 1 = \sqrt 1 $
3 = 1, and this is too a false statement. So the conclusion is x = (0, 1) doesn’t satisfies our equation
Thus we have none solution.

Option D, is the correct option

Note: While squaring a function, some solution might satisfy the equation. But we know, $3x + 1 > 0$ as the negative value square root gives out complex value. So x should be greater than $ - \dfrac{1}{3}$ i.e, $x \ge - \dfrac{1}{3}$
Secondly, $x \ge 0$ , as in$\sqrt x $ , x cannot be negative.
Using above both the condition, conclusion out is x should be greater than equal to 0. this might be used to eliminate the solutions, which might not satisfy the initial equation.