To solve the expression $4\sqrt{63} + 3\sqrt{28}$, let's start by simplifying each square root term.

Step 1: Simplify $\sqrt{63}$

$\sqrt{63} = \sqrt{9 \times 7} = \sqrt{9} \times \sqrt{7} = 3\sqrt{7}$

Step 2: Simplify $\sqrt{28}$

$\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}$

Step 3: Substitute back into the expression

Now, substitute these simplified forms back into the original expression: $4\sqrt{63} + 3\sqrt{28} = 4(3\sqrt{7}) + 3(2\sqrt{7})$

Step 4: Multiply the coefficients

$4 \times 3\sqrt{7} + 3 \times 2\sqrt{7} = 12\sqrt{7} + 6\sqrt{7}$

Step 5: Combine like terms

Since both terms have $\sqrt{7}$ as a common factor, you can add the coefficients: $12\sqrt{7} + 6\sqrt{7} = (12 + 6)\sqrt{7} = 18\sqrt{7}$

So, the simplified form of the expression is $\boxed{18\sqrt{7}}$.

Would you like further details or have any questions?

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Related Questions:

1. How do you simplify square roots involving prime factorization?
2. What are the properties of square roots that help in simplifying expressions?
3. Can you show the steps to rationalize a denominator involving square roots?
4. How would this process differ if we used cube roots instead of square roots?
5. How do you combine like terms in expressions involving square roots?

Tip: When simplifying square roots, always look for the largest perfect square factor to break it down easily.