Question

solve for n.\\(9n^2 + 42n + 49 = 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 =

Explanation:

Step1: Recognize the quadratic form

The equation \(9n^2 + 42n + 49 = 0\) is a quadratic equation. Notice that it might be a perfect square trinomial. Recall that \((a + b)^2=a^2 + 2ab + b^2\). Here, \(a = 3n\) (since \((3n)^2=9n^2\)) and \(b = 7\) (since \(7^2 = 49\)). Then \(2ab=2\times(3n)\times7 = 42n\), which matches the middle term. So we can factor the quadratic as a perfect square.
\(9n^2+42n + 49=(3n + 7)^2\)
So the equation becomes \((3n + 7)^2=0\)

Step2: Solve for n

Set \(3n+7 = 0\) (since a square is zero only when the base is zero).
Subtract 7 from both sides: \(3n=-7\)
Divide both sides by 3: \(n =-\frac{7}{3}\)

Answer:

\(-\frac{7}{3}\)