Solve the following equation and check for extraneous solutions. $\sqrt[4]{6 x^{2}-8}=-x$
Enter the solution(s) below. Round your answer to three decimal places.
There are four solutions. They are:
\[
\begin{array}{l}
x_{1}= \\
x_{2}= \\
x_{3}= \\
x_{4}= \\
\end{array}
\]
There are two solutions. They are:
\[
x_{1}=\quad x_{2}=
\]
There is one solution. It is:
\[
x_{1}=
\]
There is no solution.
Answer
Final Answer: There is one solution. It is: \(x_{1}=\boxed{-2}\)
Step 1 ：The given equation is \(\sqrt[4]{6 x^{2}-8}=-x\).
Step 2 ：Square both sides to get rid of the fourth root, resulting in \(6x^2 - 8 = x^2\).
Step 3 ：Solve this quadratic equation to find the possible values of x.
Step 4 ：Check these solutions in the original equation to make sure they are not extraneous solutions.
Step 5 ：After solving, we found two potential solutions: -2 and approximately -1.414.
Step 6 ：However, only -2 was a valid solution. The other solution was extraneous.
Step 7 ：Final Answer: There is one solution. It is: \(x_{1}=\boxed{-2}\)
