9x^4-10x^2+1=0

To solve the equation \( 9x^4 - 10x^2 + 1 = 0 \), we can use a substitution method to simplify it. Let's set \( y = x^2 \). Then, we can rewrite the equation in terms of \( y \):

\[ 9y^2 - 10y + 1 = 0 \]

Now, we will use the quadratic formula to solve for \( y \):

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = 9 \), \( b = -10 \), and \( c = 1 \). Now, we can calculate the discriminant:

\[ b^2 - 4ac = (-10)^2 - 4 \cdot 9 \cdot 1 = 100 - 36 = 64 \]

Now applying the quadratic formula:

\[ y = \frac{-(-10) \pm \sqrt{64}}{2 \cdot 9} = \frac{10 \pm 8}{18} \]

This gives us two possible solutions for \( y \):

1. \( y = \frac{10 + 8}{18} = \frac{18}{18} = 1 \)
2. \( y = \frac{10 - 8}{18} = \frac{2}{18} = \frac{1}{9} \)

Now we substitute back \( y = x^2 \):

1. For \( y = 1 \): \[ x^2 = 1 \implies x = \pm 1 \]

2. For \( y = \frac{1}{9} \): \[ x^2 = \frac{1}{9} \implies x = \pm \frac{1}{3} \]

Thus, the complete set of solutions for the original equation \( 9x^4 - 10x^2 + 1 = 0 \) is:

\[ x = 1, \; x = -1, \; x = \frac{1}{3}, \; x = -\frac{1}{3} \]