Answer
$x=4$
Work Step by Step
$$\sqrt {x} + 2 = x$$
$$\sqrt{x} = x - 2$$
$$(\sqrt{x})^{2} = (x - 2)^{2}$$
$$x = x^{2} - 4x + 4$$
$$0 = x^{2} - 4x - x + 4$$
$$0 = x^{2} -5x + 4$$
Since two factors of 4 that, when added, give 5 are 4 and 1, we can factorize this quadratic function as:
$$0 = (x - 4)(x-1)$$
and now solve for $x$:
$0 = x-4$ or $0 = x - 1$
$4 = x$ or $1 = x$
We now must plug these values into the original equation to make sure that they satisfy the original conditions:
$$\sqrt {x} + 2 = x$$
$$\sqrt {4} + 2 = 4$$
$$ 2 + 2 = 4$$
$$and$$
$$\sqrt {x} + 2 = x$$
$$\sqrt {1} + 2 = 1$$
$$1 + 2 = 1$$
$$3\ne 1$$
Therefore, we conclude that the only solution to the original equation is $x=4$.
