what is the solution of $x = 2+sqrt{x - 2}$?\n$x = 2$\n$x = 3$\n$x = 2$ or $x = 3$\nno solution

Answer:

B. $x = 3$

Explanation:

Step1: Isolate the square - root term

Given $x = 2+\sqrt{x - 2}$, we can rewrite it as $\sqrt{x - 2}=x - 2$.

Step2: Square both sides

$(\sqrt{x - 2})^2=(x - 2)^2$, which simplifies to $x - 2=x^{2}-4x + 4$.

Step3: Rearrange to form a quadratic equation

$x^{2}-4x + 4-(x - 2)=0$, so $x^{2}-4x + 4 - x+2 = 0$, and further to $x^{2}-5x + 6 = 0$.

Step4: Factor the quadratic equation

$x^{2}-5x + 6=(x - 2)(x - 3)=0$.

Step5: Solve for x

Setting each factor equal to zero gives $x - 2 = 0$ or $x - 3 = 0$, so $x=2$ or $x = 3$.

Step6: Check for extraneous solutions

When $x = 2$, the left - hand side of the original equation is $x=2$, and the right - hand side is $2+\sqrt{2 - 2}=2$. When $x = 3$, the left - hand side is $x = 3$, and the right - hand side is $2+\sqrt{3 - 2}=2 + 1=3$. Both values satisfy the original equation, but in the multiple - choice options, the single correct value among the choices is $x = 3$.