Solve by Factoring x^4-12x^2+36=0

$x^{4} - 12 x^{2} + 36 = 0$

Step 1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 12 x^{2} + 36 = 0$

Step 2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} - 12 u + 36 = 0$

Step 3

Factor using the perfect square rule.

Rewrite $36$ as $6^{2}$ .

$u^{2} - 12 u + 6^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$12 u = 2 \cdot u \cdot 6$

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 6 + 6^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 6$ .

${(u - 6)}^{2} = 0$

Step 4

Replace all occurrences of $u$ with $x^{2}$ .

${(x^{2} - 6)}^{2} = 0$

Step 5

Set the $x^{2} - 6$ equal to $0$ .

$x^{2} - 6 = 0$

Step 6

Solve for $x$ .

Add $6$ to both sides of the equation.

$x^{2} = 6$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{6}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt{6}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt{6}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt{6}, - \sqrt{6}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{6}, - \sqrt{6}$

Decimal Form:

$x = 2.44948974 \dots, - 2.44948974 \dots$