X^4-13x^2+36=0

To solve the equation \( x^4 - 13x^2 + 36 = 0 \), we can use a substitution method. Let \( y = x^2 \). Then, the equation becomes:

\[ y^2 - 13y + 36 = 0 \]

Now we can factor this quadratic equation. We need to find two numbers that multiply to \( 36 \) and add up to \( -13 \). The numbers that work are \( -9 \) and \( -4 \). Thus, we can factor the equation as:

\[ (y - 9)(y - 4) = 0 \]

Now, we set each factor equal to zero:

1. \( y - 9 = 0 \) which gives \( y = 9 \)
2. \( y - 4 = 0 \) which gives \( y = 4 \)

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

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

2. For \( y = 4 \): \[ x^2 = 4 \implies x = \pm 2 \]

Thus, the solutions to the original equation \( x^4 - 13x^2 + 36 = 0 \) are:

\[ x = 3, -3, 2, -2 \]

In summary, the complete set of solutions is:

\[ x = 3, -3, 2, -2 \]