Solve by Factoring x^4-18x^2+81=0

$x^{4} - 18 x^{2} + 81 = 0$

Step 1

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

${(x^{2})}^{2} - 18 x^{2} + 81 = 0$

Step 2

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

$u^{2} - 18 u + 81 = 0$

Step 3

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$u^{2} - 18 u + 9^{2} = 0$

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

$18 u = 2 \cdot u \cdot 9$

Rewrite the polynomial.

$u^{2} - 2 \cdot u \cdot 9 + 9^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 9$ .

${(u - 9)}^{2} = 0$

Step 4

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

${(x^{2} - 9)}^{2} = 0$

Step 5

Rewrite $9$ as $3^{2}$ .

${(x^{2} - 3^{2})}^{2} = 0$

Step 6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 3$ .

${((x + 3) (x - 3))}^{2} = 0$

Step 7

Apply the product rule to $(x + 3) (x - 3)$ .

${(x + 3)}^{2} {(x - 3)}^{2} = 0$

Step 8

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

${(x + 3)}^{2} = 0$

${(x - 3)}^{2} = 0$

Step 9

Set ${(x + 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x + 3)}^{2}$ equal to $0$ .

${(x + 3)}^{2} = 0$

Solve ${(x + 3)}^{2} = 0$ for $x$ .

Set the $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 10

Set ${(x - 3)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 3)}^{2}$ equal to $0$ .

${(x - 3)}^{2} = 0$

Solve ${(x - 3)}^{2} = 0$ for $x$ .

Set the $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 11

The final solution is all the values that make ${(x + 3)}^{2} {(x - 3)}^{2} = 0$ true.

$x = - 3, 3$

Solve by Factoring x^4-18x^2+81=0

Related Questions

Support LCMGCF.com by disabling your adblocker.