Respuesta :
We need to solve the equation:
[tex]x=2+\sqrt[]{x-2}[/tex]First, we can subtract 2 from both sides of the equation and then square both sides:
[tex]\begin{gathered} x-2=2+\surd\mleft(x-2\mright)-2 \\ \\ x-2=\surd(x-2) \\ \\ (x-2)^2=\surd(x-2)^2 \\ \\ x^{2}-4x+4=x-2 \end{gathered}[/tex]Now, we apply some operations on both sides to obtain:
[tex]\begin{gathered} x^{2}-4x+4-x+2=x-2-x+2 \\ \\ x^{2}-5x+6=0 \end{gathered}[/tex]And we can apply the quadratic formula to obtain the solutions. Notice, though, that x-2 cant be negative because this expression was originally inside the square root. So, we have the following condition for the solution:
[tex]\begin{gathered} x-2\ge0 \\ \\ x\ge2 \end{gathered}[/tex]Now, using the quadratic formula, we obtain:
[tex]\begin{gathered} x=\frac{5\pm\sqrt[]{(-5)^{2}-4(1)(6)}}{2(1)} \\ \\ x=\frac{5\pm\sqrt[]{25-24}}{2} \\ \\ x=\frac{5\pm\sqrt[]{1}}{2} \\ \\ x=\frac{5\pm1}{2} \\ \\ x_1=\frac{5-1}{2}=2 \\ \\ x_2=\frac{5+1}{2}=3 \end{gathered}[/tex]Therefore, since both solutions satisfy the condition imposed by the square root, the solutions are
x = 2 and x = 3
