.We have the following equation,
[tex]\sqrt[]{7-x}=x+5[/tex]First, we can note that the term into the radical must be zero or positive, that is,
[tex]\begin{gathered} 7-x\ge0 \\ \text{which means} \\ x\le7 \end{gathered}[/tex]Then, by squaring both sides, we have
[tex]\begin{gathered} 7-x=(x+5)^2 \\ or\text{ equivalently,} \\ 7-x=x^2+10x+25 \end{gathered}[/tex]Now, by subtracting 7 to both sides, we have
[tex]-x=x^2+10x+18[/tex]and by adding x to both sides, we get
[tex]\begin{gathered} 0=x^2+11x+18 \\ or\text{ equivalently,} \\ x^2+11x+18=0 \end{gathered}[/tex]Now, we can use the quadratic formula, that is,
[tex]x=\frac{-11\pm\sqrt[]{11^2-4(1)(18)}}{2}[/tex]which gives
[tex]\begin{gathered} x=\frac{-11\pm\sqrt[]{121-72}}{2} \\ x=\frac{-11\pm\sqrt[]{49}}{2} \\ x=\frac{-11\pm7}{2} \end{gathered}[/tex]then, the solutions are
[tex]\begin{gathered} x=\frac{-4}{2}=-2 \\ \text{and} \\ x=\frac{-18}{2}=-9 \end{gathered}[/tex]Now, we know that an extraneous solution is a solution that does not work. Then, by substituting x=-9 into the given equation, we have
[tex]\sqrt[]{7-(-9)}=-9+5[/tex]which gives
[tex]\begin{gathered} \sqrt[]{7+9}=-4 \\ \sqrt[]{16}=-4 \\ 4=-4 \end{gathered}[/tex]which is an absurd result. Then, x=-9 is an extraneous solution.
On the other hand, by substituting x=-2, we obtain
[tex]\begin{gathered} \sqrt[]{7-(-2)}=-2+5 \\ \sqrt[]{7+2}=3 \\ \sqrt[]{9}=3 \\ 3=3 \end{gathered}[/tex]which is correct.
Therefore, the answers are:
[tex]\begin{gathered} \text{Extraneous solution: x=-9} \\ \text{True solution: x=-2} \end{gathered}[/tex]