Question

1. \\(\sqrt{75x^5}\\) 18) \\(\sqrt{32n^7}\\)\
2. \\(\sqrt{98n^{12}}\\) 20) \\(\sqrt{28k^9}\\)\
3. \\(\sqrt{18x^4y}\\) 22) \\(\sqrt{24mn^2}\\)\
4. \\(\sqrt{50x^3y^4}\\) 24) \\(\sqrt{320mn^4}\\)\
5. \\(\sqrt{192a^3b}\\) 26) \\(\sqrt{80x^2y^4}\\)\
6. \\(\sqrt{112x^3y^4}\\) 28) \\(\sqrt{256x^4y^4}\\)\
7. \\(\sqrt{216x^3y^3}\\) 30) \\(\sqrt{192u^4v^4}\\)

Explanation:

Let's solve problem 18: $\boldsymbol{\sqrt{32n^7}}$

Step1: Factor the radicand

Factor 32 and $n^7$ into perfect squares and remaining factors.
$32 = 16 \times 2$ (16 is a perfect square), and $n^7 = n^6 \times n$ ( $n^6=(n^3)^2$ is a perfect square).
So, $\sqrt{32n^7} = \sqrt{16 \times 2 \times n^6 \times n}$.

Step2: Apply square - root property

Use the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a\geq0,b\geq0$) and $\sqrt{a^2}=a$ ($a\geq0$).
$\sqrt{16 \times 2 \times n^6 \times n}=\sqrt{16}\cdot\sqrt{n^6}\cdot\sqrt{2n}$
Since $\sqrt{16} = 4$ and $\sqrt{n^6}=n^3$ (because $(n^3)^2=n^6$), we get:
$4\times n^3\times\sqrt{2n}=4n^3\sqrt{2n}$

Answer:

$4n^3\sqrt{2n}$