Multiply.
$\sqrt{3}(\sqrt{14}-6 \sqrt{3})$
Simplify your answer as much as possible.
The problem asks to calculate the product of \(\sqrt{3}\) and the binomial \((\sqrt{14}-6\sqrt{3})\). This requires applying the distributive property of multiplication over subtraction. The distributive property states that for any real numbers \(a\), \(b\), and \(c\), the expression \(a(b - c)\) is equivalent to \(ab - ac\). Additionally, we will need to use properties of square roots, specifically \(\sqrt{m} \cdot \sqrt{n} = \sqrt{m \cdot n}\) and \(\sqrt{k} \cdot \sqrt{k} = k\) for any positive real number \(k\).
To find the product \(\sqrt{3}(\sqrt{14}-6\sqrt{3})\), we distribute \(\sqrt{3}\) into each term inside the parentheses:
\[\sqrt{3}(\sqrt{14}-6\sqrt{3}) = (\sqrt{3} \cdot \sqrt{14}) - (\sqrt{3} \cdot 6\sqrt{3})\]
Now, we simplify each of the two terms.
For the first term, \(\sqrt{3} \cdot \sqrt{14} = \sqrt{3 \cdot 14} = \sqrt{42}\).
For the second term, we have \(\sqrt{3} \cdot 6\sqrt{3} = 6 \cdot (\sqrt{3} \cdot \sqrt{3}) = 6 \cdot 3 = 18\).
Now, combine the simplified results from the previous step. We have the expression \(\sqrt{42} - 18\). This expression cannot be simplified further because \(\sqrt{42}\) contains no perfect square factors other than 1 (its prime factorization is \(2 \times 3 \times 7\)), and it is not a rational number that can be combined with the integer \(-18\). Thus, the final simplified form is \(\sqrt{42} - 18\).
Calculating with radical expressions is useful in geometry and architecture. For example, if you need to calculate the dimensions of a room that involve irrational lengths, such as finding the area or diagonal of a space with side lengths expressed as radicals, you would perform operations like these to find exact measurements before approximating them for construction. This ensures precision in engineering designs.