Let's go ahead and simplify the given square root expressions step by step.

(a) $4\sqrt{48} + 3\sqrt{147} - 5\sqrt{12}$

We first simplify each square root:

• $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
• $\sqrt{147} = \sqrt{49 \times 3} = 7\sqrt{3}$
• $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$

Now substitute these into the expression: $4(4\sqrt{3}) + 3(7\sqrt{3}) - 5(2\sqrt{3}) = 16\sqrt{3} + 21\sqrt{3} - 10\sqrt{3}$ Simplifying the terms: $16\sqrt{3} + 21\sqrt{3} - 10\sqrt{3} = 27\sqrt{3}$

(b) $\sqrt{2} - 8\sqrt{2}$

Combine the like terms: $\sqrt{2} - 8\sqrt{2} = -7\sqrt{2}$

(c) $\sqrt{8} - 3\sqrt{2}$

First, simplify $\sqrt{8}$: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$ Now substitute and combine the terms: $2\sqrt{2} - 3\sqrt{2} = -\sqrt{2}$

(d) $\sqrt{50} - \sqrt{32}$

Simplify each square root:

• $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
• $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$

Now subtract: $5\sqrt{2} - 4\sqrt{2} = \sqrt{2}$

(e) $\sqrt{90} - 5\sqrt{40} - 2\sqrt{10}$

Simplify each square root:

• $\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$
• $\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$

Substitute into the expression: $3\sqrt{10} - 5(2\sqrt{10}) - 2\sqrt{10} = 3\sqrt{10} - 10\sqrt{10} - 2\sqrt{10}$ Combine like terms: $3\sqrt{10} - 10\sqrt{10} - 2\sqrt{10} = -9\sqrt{10}$

(f) $\sqrt{2} - 2\sqrt{2} - 3\sqrt{8}$

First simplify $\sqrt{8}$: $\sqrt{8} = 2\sqrt{2}$ Now substitute and combine: $\sqrt{2} - 2\sqrt{2} - 3(2\sqrt{2}) = \sqrt{2} - 2\sqrt{2} - 6\sqrt{2}$ $\sqrt{2} - 2\sqrt{2} - 6\sqrt{2} = -7\sqrt{2}$

(g) $2\sqrt{5} + \sqrt{45}$

Simplify $\sqrt{45}$: $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ Now substitute and combine the terms: $2\sqrt{5} + 3\sqrt{5} = 5\sqrt{5}$

(h) $\sqrt{5} + 4\sqrt{5} + 3\sqrt{125}$

Simplify $\sqrt{125}$: $\sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}$ Now substitute and combine: $\sqrt{5} + 4\sqrt{5} + 3(5\sqrt{5}) = \sqrt{5} + 4\sqrt{5} + 15\sqrt{5}$ $\sqrt{5} + 4\sqrt{5} + 15\sqrt{5} = 20\sqrt{5}$

(i) $\sqrt{12} - \sqrt{27}$

Simplify both square roots:

• $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
• $\sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$

Now subtract: $2\sqrt{3} - 3\sqrt{3} = -\sqrt{3}$

(j) $2\sqrt{3} - \sqrt{48}$

Simplify $\sqrt{48}$: $\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$ Now substitute and subtract: $2\sqrt{3} - 4\sqrt{3} = -2\sqrt{3}$

Do you have any questions or would you like further clarification?

Here are 5 related questions you might explore:

1. How do you simplify square roots with non-perfect square factors?
2. What is the process for combining like terms with square roots?
3. How can you simplify expressions with cube roots or higher-order roots?
4. What are the key properties of radicals that help in simplification?
5. How do you rationalize the denominator when dealing with square roots?

Tip: Always try to break square roots into their prime factorization to simplify them easily!