Question

Which equation has an extraneous solution?A. ^5/x = -2B. ^3/x + 1 = 10C. /x = -5D. ^4/x- 5= 3

Answer 1

An extraneous solution is one in which when we plug back the answer into the equation, it does not work.

For option A:

`\begin{gathered} \sqrt[5]{x\text{ }}=\text{ -2} \\ \text{raise both sides to the 5th power} \\ x=(-2)^5 \\ x\text{ = -32} \end{gathered}`

For option B:

`\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}`

For option C:

`\begin{gathered} \sqrt[]{x}\text{ = -5} \\ \text{square both sides} \\ x=(-5)^2 \\ x\text{ = 25} \end{gathered}`

For option D:

`\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}`

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.

`\sqrt[]{25}\text{ = 5 and not -5}`

Thus, option C is the answer