9x4-10x2+1=0

To solve the quadratic equation \( 9x^4 - 10x^2 + 1 = 0 \), let's make a substitution to simplify it. Let \( y = x^2 \). Then, the equation can be rewritten as:

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

Now, we can solve this quadratic equation in \( y \) using the quadratic formula:

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

Where \( a = 9 \), \( b = -10 \), and \( c = 1 \).

Substituting these values into the formula, we have:

\[ y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} \]

Calculating further:

\[ y = \frac{10 \pm \sqrt{100 - 36}}{18} \]

\[ y = \frac{10 \pm \sqrt{64}}{18} \]

\[ y = \frac{10 \pm 8}{18} \]

Now, solving for the two possible values of \( y \):

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

Now we revert back to \( x \) by substituting \( y = x^2 \):

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 solutions to the original equation \( 9x^4 - 10x^2 + 1 = 0 \) are:

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

Final solutions: \[ x = 1, \quad x = -1, \quad x = \frac{1}{3}, \quad x = -\frac{1}{3} \]