Question

1. $n^{4}-49n^{2}=0$

Explanation:

Step1: Factor out common term

Let \(x = n^{2}\), then the equation \(n^{4}-49n^{2}=0\) becomes \(x^{2}-49x = 0\). Factor out \(x\): \(x(x - 49)=0\). Substituting back \(x = n^{2}\), we get \(n^{2}(n^{2}-49)=0\).

Step2: Use difference - of - squares

Since \(n^{2}-49=(n + 7)(n - 7)\) (using \(a^{2}-b^{2}=(a + b)(a - b)\) where \(a=n\) and \(b = 7\)), the factored form of the original equation is \(n^{2}(n + 7)(n - 7)=0\).

Step3: Solve for \(n\)

Set each factor equal to zero:
If \(n^{2}=0\), then \(n = 0\).
If \(n+7=0\), then \(n=-7\).
If \(n - 7=0\), then \(n = 7\).

Answer:

\(n = 0\), \(n=-7\), \(n = 7\)