How do you solve \[x=\sqrt{x+6}\] and find any extraneous solutions?

Hint: To solve the above equation and to get the extraneous root we will first simplify the expression. The first step is to remove the square root on the right side of the equation. This can be done by squaring on both sides. After removing the square root, we will rearrange the equation into a quadratic equation form which is \[a{{x}^{2}}+bx+c=0\].

Complete step by step answer:
Now, this question belongs to the concept of radical and geometry construction and the above equation is a radical equation. A radical is basically the root of the equation.
In the above equation, \[x=\sqrt{x+6}\] we will first remove the square root on the right side to get a simpler form of the equation.
Squaring on both sides
\[{{(x)}^{2}}={{\left( \sqrt{x+6} \right)}^{2}}\]
 \[\Rightarrow {{(x)}^{2}}=x+6\]
Thus, we get the equation in \[a{{x}^{2}}+bx+c=0\] form. After rearranging the equation, we get
\[{{x}^{2}}-x-6=0\]
As we can see that the above equation is in the form of \[a{{x}^{2}}+bx+c=0\] therefore, we will further solve this quadratic equation using either factorization method or by completing the square method.
Here we will use the factorization method. It is an easy method to solve quadratic equations.
First, we have to split the middle term \[bx\] in two terms, such that the product of the two terms is equal to the constant \[c\] and the sum of the two terms is equal to the middle term \[bx\]. Then factor the first two terms and the last two terms.
Here we can see the middle term is \[-x\] and the constant is \[-6\] thus the equation can be written as
\[{{x}^{2}}+2x-3x-6=0\]. We can see that the product of the two terms is equal to the constant \[-6\] and the sum of the two terms is equal to the middle term \[-x\]. Therefore, we will further factorize and get the roots.
\[\begin{align}
  & {{x}^{2}}+2x-3x-6=0 \\
 & \Rightarrow x(x+2)-3(x+2)=0 \\
\end{align}\]
Take \[(x+2)\] as a common factor.
\[\Rightarrow (x+2)(x-3)=0\]
\[\Rightarrow x+2=0\] or \[x-3=0\]
\[\Rightarrow x=-2,3\]
Thus \[-2,3\] are the extraneous root of the given equation.

Note:
While finding the roots of the quadratic equation you can use completing the square method. After finding the extraneous roots, substitute it in the original equation which is given in the question so, that you can verify your answer. Do not forget to change the radical equation in the correct form of the quadratic equation.