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

Show work to justify whether x=2 or x=-2 has an extraneous solution.Hint: Only one-example-1

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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.

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