Solve for x x^2+18x+81=0

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

Step 1

Factor using the perfect square rule.

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

$x^{2} + 18 x + 9^{2} = 0$

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

$18 x = 2 \cdot x \cdot 9$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 9 + 9^{2} = 0$

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

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

Step 2

Set the $x + 9$ equal to $0$ .

$x + 9 = 0$

Step 3

Subtract $9$ from both sides of the equation.

$x = - 9$