Question

Show work to justify whether x=2 or x=-2 has an extraneous solution.Hint: Only one of the equations is extraneous

Answer 1

Given the equations:

`\begin{gathered} \sqrt[]{x}=2\ldots(1) \\ \sqrt[]{x}=-2\ldots(2) \end{gathered}`

To find the solutions, we take the square on both sides.

For equation (1):

`\begin{gathered} (\sqrt[]{x})^2=2^2 \\ x=4 \end{gathered}`

To find out if this is an extraneous solution, we evaluate this value on the original equation:

`\begin{gathered} \sqrt[]{4}=2 \\ 2=2\text{ (Correct)} \end{gathered}`

We conclude that this is not an extraneous solution.

Now, for equation (2):

`\begin{gathered} (\sqrt[]{x})^2=(-2)^2 \\ x=4 \end{gathered}`

We evaluate this result on the original function:

`\begin{gathered} \sqrt[]{4}=-2 \\ 2=-2\text{ (Incorrect)} \end{gathered}`

We conclude that this is an extraneous solution.