Question

solve for n.
( n^2 + 22n + 121 = 0 )
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
( n = square )
save answer

Explanation:

Step1: Recognize the quadratic form

The equation \(n^2 + 22n + 121 = 0\) is a quadratic equation. Notice that it might be a perfect square trinomial. Recall that \((a + b)^2=a^2 + 2ab + b^2\). Here, \(a = n\), \(2ab=22n\), and \(b^2 = 121\). Solving for \(b\) from \(b^2=121\), we get \(b = 11\) (since \(2\times n\times11=22n\), which matches the middle term). So the left - hand side can be factored as a perfect square.

Step2: Factor the quadratic

We can factor \(n^2 + 22n + 121\) as \((n + 11)^2\). So the equation becomes \((n + 11)^2=0\).

Step3: Solve for n

To solve \((n + 11)^2=0\), we take the square root of both sides. The square root of a number squared is the number itself (with a consideration of positive and negative, but since we have a square of a number equal to zero, the number must be zero). So \(n+11 = 0\). Subtracting 11 from both sides, we get \(n=- 11\).

Answer:

\(n=-11\)