Question

what are the solutions of the equation $9x^4 - 2x^2 - 7 = 0$? use $u$ substitution to solve.
\bigcirc $x = \pm \sqrt{\frac{7}{9}}$ and $x = \pm 1$
\bigcirc $x = \pm \sqrt{\frac{7}{9}}$ and $x = \pm i$
\bigcirc $x = \pm i\sqrt{\frac{7}{9}}$ and $x = \pm 1$
\bigcirc $x = \pm i\sqrt{\frac{7}{9}}$ and $x = \pm i$

Explanation:

Step1: Define substitution

Let \( u = x^2 \). Then the equation \( 9x^4 - 2x^2 - 7 = 0 \) becomes \( 9u^2 - 2u - 7 = 0 \) (since \( x^4=(x^2)^2 = u^2 \)).

Step2: Solve quadratic for u

Use quadratic formula \( u=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) for \( 9u^2 - 2u - 7 = 0 \), where \( a = 9 \), \( b=-2 \), \( c = -7 \).
Discriminant: \( (-2)^2 - 4(9)(-7)=4 + 252 = 256 \).
\( u=\frac{2\pm\sqrt{256}}{18}=\frac{2\pm16}{18} \).
Two solutions:

• \( u=\frac{2 + 16}{18}=\frac{18}{18}=1 \)
• \( u=\frac{2-16}{18}=\frac{-14}{18}=-\frac{7}{9} \)

Step3: Substitute back \( u = x^2 \)

• For \( u = 1 \): \( x^2 = 1 \Rightarrow x=\pm1 \).
• For \( u = -\frac{7}{9} \): \( x^2=-\frac{7}{9} \Rightarrow x=\pm\sqrt{-\frac{7}{9}}=\pm i\sqrt{\frac{7}{9}} \) (since \( \sqrt{-a}=i\sqrt{a} \) for real \( a>0 \)).

Answer:

\( x=\pm i\sqrt{\frac{7}{9}} \) and \( x = \pm1 \) (corresponding to the third option: \( x=\pm i\sqrt{\frac{7}{9}} \) and \( x = \pm1 \))