Simplify the expression $\frac{\sqrt{98x^2y^4}}{\sqrt{2x^2}}$. We will assume that $x > 0$ so that $\sqrt{x^2} = x$.

1. Problem Description

Simplify the expression $\frac{\sqrt{98x^2y^4}}{\sqrt{2x^2}}$. We will assume that $x > 0$ so that $\sqrt{x^2} = x$.

2. Solution Steps

First, we can combine the two square roots:

$\frac{\sqrt{98x^2y^4}}{\sqrt{2x^2}} = \sqrt{\frac{98x^2y^4}{2x^2}}$

Simplify the fraction inside the square root:

$\sqrt{\frac{98x^2y^4}{2x^2}} = \sqrt{49y^4}$

Since $49 = 7^2$, we have

$\sqrt{49y^4} = \sqrt{7^2(y^2)^2} = \sqrt{(7y^2)^2}$

Since we are taking the principal square root, we can assume that the result is non-negative. Therefore

$\sqrt{(7y^2)^2} = 7y^2$

3. Final Answer

$7y^2$