Question

what are the solutions of the equation $x^4 - 9x^2 + 8 = 0$? use $u$ substitution to solve.\
\bigcirc $x = 1$ and $x = 2\sqrt{2}$\
\bigcirc $x = \pm 1$ and $x = \pm 2\sqrt{2}$\
\bigcirc $x = \pm i$ and $x = \pm 2i\sqrt{2}$\
\bigcirc $x = \pm i$ and $x = 2\sqrt{2}$

Explanation:

Step1: Let \( u = x^2 \)

Substitute \( u \) into the equation: \( u^2 - 9u + 8 = 0 \)

Step2: Factor the quadratic

\( (u - 1)(u - 8) = 0 \)

Step3: Solve for \( u \)

\( u - 1 = 0 \) gives \( u = 1 \); \( u - 8 = 0 \) gives \( u = 8 \)

Step4: Substitute back \( u = x^2 \)

For \( u = 1 \): \( x^2 = 1 \) ⇒ \( x = \pm1 \)
For \( u = 8 \): \( x^2 = 8 \) ⇒ \( x = \pm2\sqrt{2} \)

Answer:

\( x = \pm1 \) and \( x = \pm2\sqrt{2} \) (corresponding to the option: \( x = \pm1 \) and \( x = \pm2\sqrt{2} \))