Simplify square root of 27y^2

Solution:

Simplification Process:

Step 1:

Express $27 y^{2}$ as the product of a perfect square and another term: $(3y)^{2} \cdot 3$.

Step 1.1:

Extract the square root of $9$ from $27$: $\sqrt{9 \cdot 3 y^{2}}$.

Step 1.2:

Represent $9$ as a square of $3$: $\sqrt{3^{2} \cdot 3 y^{2}}$.

Step 1.3:

Rearrange the terms: $\sqrt{3^{2} y^{2} \cdot 3}$.

Step 1.4:

Identify $3^{2} y^{2}$ as a square of $3y$: $\sqrt{(3y)^{2} \cdot 3}$.

Step 2:

Extract the square root of the perfect square: $3y \sqrt{3}$.

Knowledge Notes:

To simplify the square root of an expression involving variables and constants, we can apply the following knowledge points:

1. Prime Factorization: Break down numbers into their prime factors to identify perfect squares.

2. Properties of Square Roots: The square root of a product is equal to the product of the square roots of the individual factors, provided that all the quantities involved are non-negative.

3. Simplifying Perfect Squares: When a term under the square root is a perfect square, it can be taken out of the square root as its base.

4. Algebraic Manipulation: When dealing with variables under a square root, look for powers that are multiples of 2, which indicate perfect squares.

In this problem, $27$ can be factored into $9 \cdot 3$, where $9$ is a perfect square. Since $y^{2}$ is also a perfect square, we can take the square root of both these terms out of the radical. The remaining factor that is not a perfect square stays under the square root.

The expression $\sqrt{27y^{2}}$ simplifies to $3y\sqrt{3}$ because $27$ is factored into $9 \cdot 3$, and $9$ is rewritten as $3^{2}$. Since $3^{2}$ and $y^{2}$ are both perfect squares, their square roots are $3$ and $y$, respectively. The remaining factor $3$ stays under the square root.