Question

1. \\(\frac{4\sqrt{14}}{8\sqrt{2}}\\)\
2. \\(\frac{2\sqrt{20}}{6\sqrt{4}}\\)\

mixed practice:\

1. \\(3\sqrt{12} \cdot 6\\)\
2. \\(\frac{3\sqrt{20}}{2\sqrt{4}}\\)\
3. \\(\sqrt{6}(\sqrt{6})\\)\
4. \\(\sqrt{5} \cdot -4\sqrt{20}\\)\
5. is the product of \\(\sqrt{3}\\) and \\(\sqrt{9}\\) rational or irrational? explain your answer.\
6. \\(\frac{\sqrt{50}}{\sqrt{2}}\\)\
7. \\(\sqrt{10} \cdot \sqrt{5}\\)\
8. \\(\frac{8\sqrt{40}}{4\sqrt{5}}\\)

Explanation:

Step1: Simplify $3\sqrt{12} \cdot 6$

First simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$.
$3 \cdot 2\sqrt{3} \cdot 6 = 6\sqrt{3} \cdot 6 = 36\sqrt{3}$

Step2: Simplify $\frac{3\sqrt{20}}{2\sqrt{4}}$

Simplify radicals: $\sqrt{20}=2\sqrt{5}$, $\sqrt{4}=2$.
$\frac{3 \cdot 2\sqrt{5}}{2 \cdot 2} = \frac{6\sqrt{5}}{4} = \frac{3\sqrt{5}}{2}$

Step3: Simplify $\sqrt{6}(\sqrt{6})$

Use property $\sqrt{a} \cdot \sqrt{a} = a$.
$\sqrt{6} \cdot \sqrt{6} = 6$

Step4: Simplify $\sqrt{5} - 4\sqrt{20}$

Simplify $\sqrt{20}=2\sqrt{5}$.
$\sqrt{5} - 4 \cdot 2\sqrt{5} = \sqrt{5} - 8\sqrt{5} = -7\sqrt{5}$

Step5: Classify product of $\sqrt{3}$ and $\sqrt{9}$

First calculate the product: $\sqrt{3} \cdot \sqrt{9} = \sqrt{3} \cdot 3 = 3\sqrt{3}$.
A rational number can be written as $\frac{a}{b}$ where $a,b$ are integers, $b
eq0$. $3\sqrt{3}$ cannot be expressed this way, so it is irrational.

Step6: Simplify $\frac{\sqrt{50}}{\sqrt{2}}$

Use property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$.
$\sqrt{\frac{50}{2}} = \sqrt{25} = 5$

Step7: Simplify $\sqrt{10} \cdot \sqrt{5}$

Use property $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$.
$\sqrt{10 \cdot 5} = \sqrt{50} = 5\sqrt{2}$

Step8: Simplify $\frac{8\sqrt{40}}{4\sqrt{5}}$

Simplify radicals and coefficients: $\frac{8}{4} = 2$, $\frac{\sqrt{40}}{\sqrt{5}} = \sqrt{8} = 2\sqrt{2}$.
$2 \cdot 2\sqrt{2} = 4\sqrt{2}$

Answer:

1. $36\sqrt{3}$
2. $\frac{3\sqrt{5}}{2}$
3. $6$
4. $-7\sqrt{5}$
5. Irrational; the product is $3\sqrt{3}$, which cannot be written as a ratio of two integers.
6. $5$
7. $5\sqrt{2}$
8. $4\sqrt{2}$