Question

solve for n.
$98n^2 - 28n + 2 = 0$
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: Simplify the equation

Divide the entire equation \(98n^{2}-28n + 2=0\) by 2. We get \(49n^{2}-14n + 1 = 0\).

Step2: Recognize the perfect square trinomial

Notice that \(49n^{2}-14n + 1=(7n - 1)^{2}\). So the equation becomes \((7n - 1)^{2}=0\).

Step3: Solve for n

Take the square root of both sides: \(7n-1 = 0\). Then, add 1 to both sides: \(7n=1\). Divide both sides by 7: \(n=\frac{1}{7}\). Since the square has a multiplicity of 2, but the solution is repeated, so the only solution is \(\frac{1}{7}\).

Answer:

\(\frac{1}{7}\)