An extraneous solution is one in which when we plug back the answer into the equation, it does not work.
For option A:
[tex]\begin{gathered} \sqrt[5]{x\text{ }}=\text{ -2} \\ \text{raise both sides to the 5th power} \\ x=(-2)^5 \\ x\text{ = -32} \end{gathered}[/tex]For option B:
[tex]\begin{gathered} \sqrt[3]{x}+\text{ 1 = 10} \\ \text{subtract 1 from both sides} \\ \sqrt[3]{x}\text{ = 10 -1 = 9} \\ Get\text{ the cube of both sides} \\ \text{x =9}^3 \end{gathered}[/tex]For option C:
[tex]\begin{gathered} \sqrt[]{x}\text{ = -5} \\ \text{square both sides} \\ x=(-5)^2 \\ x\text{ = 25} \end{gathered}[/tex]For option D:
[tex]\begin{gathered} \sqrt[4]{x}-\text{ 5 = 3} \\ \text{Add 5 to both sides} \\ \sqrt[4]{x}\text{ = 8} \\ \text{Raise both sides to the 4th power} \\ x=8^4 \end{gathered}[/tex]But out of the 4 options, only option C is extraneous because if we plug back x = 25 into the original equation, you will not get back the original question.
i.e.
[tex]\sqrt[]{25}\text{ = 5 and not -5}[/tex]Thus, option C is the answer
