Solve for x x^2+22x+121=0

$x^{2} + 22 x + 121 = 0$

Step 1

Factor using the perfect square rule.

Rewrite $121$ as $11^{2}$ .

$x^{2} + 22 x + 11^{2} = 0$

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

$22 x = 2 \cdot x \cdot 11$

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 11 + 11^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 11$ .

${(x + 11)}^{2} = 0$

Step 2

Set the $x + 11$ equal to $0$ .

$x + 11 = 0$

Step 3

Subtract $11$ from both sides of the equation.

$x = - 11$