Given the radical equation:
![q-6=\sqrt[]{27-2q}](https://img.qammunity.org/2023/formulas/mathematics/college/dfnc46pzn7z09qcab0eke4svu7fp9004j6.png)
Squaring both sides to eliminate the root.

Factor the equation to find the roots:

we will check ( q = 1 and q = 9 ) by substitution into the given equation:
When q = 1
![\begin{gathered} q-6=1-6=-5 \\ \sqrt[]{27-2q}=\sqrt[]{27-2}=\sqrt[]{25}=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8tbt0kik5qsc02s9mr08pwya9ha4iukqlk.png)
So, ( q = 1 ) is an extraneous solution.
When q = 9
![\begin{gathered} q-6=9-6=3 \\ \sqrt[]{27-2q}=\sqrt[]{27-18}=\sqrt[]{9}=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/64ujzne1h5dtn8ewxra6x95wf58v20d1ip.png)
So, ( q = 9 ) is the solution of the given equation.
So, the answer will be:
The extraneous solution to the radical equation is 1