Question

simplify \\(\sqrt{x^4y^2}\\)

Explanation:

Step1: Recall square root property

The square root of a product is the product of the square roots, i.e., $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (for $a\geq0, b\geq0$). Also, $\sqrt{x^n}=x^{\frac{n}{2}}$ when $x\geq0$.
So, $\sqrt{x^4y^2}=\sqrt{x^4}\cdot\sqrt{y^2}$.

Step2: Simplify each square root

For $\sqrt{x^4}$, using the exponent rule for square roots, $n = 4$, so $\sqrt{x^4}=x^{\frac{4}{2}}=x^2$.
For $\sqrt{y^2}$, using the exponent rule for square roots, $n = 2$, so $\sqrt{y^2}=y^{\frac{2}{2}}=y$ (assuming $y\geq0$; if we consider absolute value, it would be $|y|$, but often in algebraic simplification with even roots and variables, we assume non - negative for simplicity or consider the principal root).

Combining these two results, we get $\sqrt{x^4y^2}=x^2y$.

Answer:

$x^{2}y$ (or $x^{2}|y|$ if we consider the absolute value of $y$)