Certainly! Let's solve the expression [tex]\(\sqrt{28 y^4}\)[/tex] step-by-step, assuming [tex]\(y\)[/tex] is a positive real number.
1. Simplify the Radicand:
The expression inside the square root is [tex]\(28 y^4\)[/tex].
2. Factor the Radicand into Prime Factors:
We break down 28 into its prime factors and recognize the powers of [tex]\(y\)[/tex].
[tex]\[
28 = 4 \times 7 = (2^2) \times 7
\][/tex]
Therefore, we can rewrite [tex]\(28 y^4\)[/tex] as:
[tex]\[
28 y^4 = (2^2 \times 7) \times y^4
\][/tex]
3. Express the Entire Radicand as a Product:
[tex]\[
28 y^4 = (2^2 y^4) \times 7
\][/tex]
4. Distribute the Square Root:
Since the square root of a product is the product of the square roots, we can write:
[tex]\[
\sqrt{28 y^4} = \sqrt{2^2 y^4 \times 7} = \sqrt{2^2 y^4} \times \sqrt{7}
\][/tex]
5. Simplify the Square Roots:
Consider [tex]\(\sqrt{2^2 y^4}\)[/tex]:
[tex]\[
\sqrt{2^2 y^4} = 2 y^2
\][/tex]
This is because [tex]\(\sqrt{2^2} = 2\)[/tex] and [tex]\(\sqrt{y^4} = y^2\)[/tex] for [tex]\(y > 0\)[/tex].
6. Combine the Simplified Results:
Now combine the simplified parts:
[tex]\[
\sqrt{28 y^4} = 2 y^2 \cdot \sqrt{7}
\][/tex]
7. Final Simplified Form:
[tex]\[
\sqrt{28 y^4} = 2 \sqrt{7} y^2
\][/tex]
So, the expression [tex]\(\sqrt{28 y^4}\)[/tex] simplifies to [tex]\(2 \sqrt{7} y^2\)[/tex], assuming [tex]\(y\)[/tex] is a positive real number.
