Simplify square root of (16y^7)/(x^8)

Solution:

Step 1

Express $\frac{16y^7}{x^8}$ as ${\left(\frac{4y^3}{x^4}\right)}^2\cdot y$.

Step 1.1

Extract the square of the perfect square $4y^3$ from $16y^7$ to get $\sqrt{\frac{(4y^3{)}^2\cdot y}{x^8}}$.

Step 1.2

Extract the square of the perfect square $x^4$ from $x^8$ to form $\sqrt{\frac{(4y^3{)}^2\cdot y}{(x^4{)}^2\cdot 1}}$.

Step 1.3

Reorganize the fraction $\frac{(4y^3{)}^2\cdot y}{(x^4{)}^2\cdot 1}$ to $\sqrt{{\left(\frac{4y^3}{x^4}\right)}^2\cdot y}$.

Step 2

Remove terms from under the square root to obtain $\frac{4y^3}{x^4}\sqrt{y}$.

Step 3

Merge $\frac{4y^3}{x^4}$ with $\sqrt{y}$ to finalize the simplification as $\frac{4y^3\sqrt{y}}{x^4}$.

Knowledge Notes:

To simplify the square root of a fraction, we can use the property that the square root of a product is equal to the product of the square roots of the individual factors. We start by identifying perfect squares within the numerator and denominator that can be factored out. In this case, $16y^7$ can be rewritten as $(4y^3{)}^2\cdot y$ because $16=4^2$ and $y^7=(y^3{)}^2\cdot y$. Similarly, $x^8$ can be rewritten as $(x^4{)}^2$ because $x^8=(x^4{)}^2$.

Once we have factored out the perfect squares, we can take the square root of the entire expression. The square root of a perfect square is simply the base of the square, so $(4y^3{)}^2$ under the square root becomes $4y^3$, and $(x^4{)}^2$ under the square root becomes $x^4$. The remaining term under the square root, which is not a perfect square, stays within the square root.

Finally, we combine the terms that are outside the square root with those that remain inside to complete the simplification. It is important to ensure that the expression is simplified as much as possible, which means that there should be no perfect squares left under the square root and no factors common to the numerator and denominator that can be canceled out.