Question

$\sqrt{98} = \dots \sqrt{2}$

Explanation:

Step1: Factorize 98

We know that \(98 = 49\times2\), and \(49 = 7^2\). So we can rewrite \(\sqrt{98}\) as \(\sqrt{49\times2}\).

Step2: Use square - root property

According to the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{49\times2}=\sqrt{49}\times\sqrt{2}\).
Since \(\sqrt{49} = 7\) (because \(7^2=49\)), then \(\sqrt{98}=7\sqrt{2}\). Also, for \(\sqrt{- 98}\), in the complex number system, \(\sqrt{-98}=\sqrt{98}\times\sqrt{- 1}=7\sqrt{2}i\), but if we consider the problem in the context of simplifying the radical form of the positive number 98 first (maybe there is a typo and the negative sign is not supposed to be there), the coefficient of \(\sqrt{2}\) is 7.

Answer:

If we consider the non - negative square root (ignoring the negative sign by mistake), the answer is \(7\) (and if we consider the complex number, \(\sqrt{-98} = 7\sqrt{2}i\), but the coefficient of \(\sqrt{2}\) is \(7\)). So the value in the blank is \(7\) (or \(7i\) for the complex case).