Answer:
[tex]x = 0[/tex] or [tex]x = 1[/tex].
Step-by-step explanation:
Start by adding [tex]1[/tex] to both sides of this equation:
[tex](x - 1) + 1 = (\sqrt{x} - 1) + 1[/tex].
[tex]x = \sqrt{x}[/tex].
If two numbers are equal, their square should also be equal. Therefore, since[tex]x = \sqrt{x}[/tex], it must be true that [tex]x^{2} = (\sqrt{x})^{2}[/tex]. That is: [tex]x^{2} = x[/tex].
Notice that since [tex]x[/tex] is under a square root, the result must ensure that [tex]x \ge 0[/tex].
Subtract [tex]x[/tex] from both sides of the equation:
[tex]x^{2} - x = x - x[/tex].
[tex]x^{2} - x = 0[/tex].
Factor [tex]x[/tex] out:
[tex]x\, (x - 1) = 0[/tex].
Hence, by the Factor Theorem, [tex]x = 0[/tex] and [tex]x = 1[/tex] would satisfy this rearranged equation. Because of the square root in the original equation, these two value must be non-negative ([tex]x \ge 0[/tex]) to qualify as actual roots of that equation.
In this example, both [tex]x = 0[/tex] and [tex]x = 1[/tex] qualify as roots of that equation.
