Question

station 1 simplifying radicals \
simplify the following radical: \\(\sqrt{98}\\) \
\\(\circ\\ 12\sqrt{5}\\) \
\\(\circ\\ 5\sqrt{2}\\) \
\\(\circ\\ 6\sqrt{2}\\) \
\\(\circ\\ 7\sqrt{2}\\) \
\\(\circ\\ 6\sqrt{5}\\) \
\\(\circ\\ 12\sqrt{2}\\)

Explanation:

Step1: Factor the radicand

We need to factor 98 into a product of a perfect square and another number. We know that \(98 = 49\times2\), and 49 is a perfect square (\(7^2\)). So we can rewrite \(\sqrt{98}\) as \(\sqrt{49\times2}\).

Step2: Use the property of square roots

The property of square roots states that \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (for \(a\geq0,b\geq0\)). Applying this property to \(\sqrt{49\times2}\), we get \(\sqrt{49}\times\sqrt{2}\).

Step3: Simplify \(\sqrt{49}\)

Since \(7^2 = 49\), \(\sqrt{49}=7\). So substituting back, we have \(7\times\sqrt{2}=7\sqrt{2}\).

Answer:

\(7\sqrt{2}\) (corresponding to the option: \(7\sqrt{2}\))