Answered

Simplify $\sqrt{6}(\sqrt{3}+2)$


A$3 \sqrt{2}+2 \sqrt{6}$
B$3+2 \sqrt{6}$
C$3 \sqrt{8}+2 \sqrt{6}$

Verified Answer :

  • Use the distributive property to expand \(\sqrt{6}(\sqrt{3} + 2)\) into \(\sqrt{6} \cdot \sqrt{3} + \sqrt{6} \cdot 2\).
  • Simplify the first part \(\sqrt{6 \cdot 3} = \sqrt{18} = 3\sqrt{2}\) and the second part \(2\sqrt{6}\).
  • Combine the results to get \(3\sqrt{2} + 2\sqrt{6}\), which matches option (a).
  • The correct option is \(\boxed{(a)}\).

Explanation

  1. Applying the Distributive Property

To simplify the expression \(\sqrt{6}(\sqrt{3} + 2)\), we apply the distributive property of multiplication over addition, which states that \(a(b + c) = ab + ac\). Applying this property, we get:
\[\sqrt{6}(\sqrt{3} + 2) = \sqrt{6} \cdot \sqrt{3} + \sqrt{6} \cdot 2\]

  1. Simplifying the Terms

Next, we simplify each term individually.
For the first term, we use the property of radicals that \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\).
\[\sqrt{6} \cdot \sqrt{3} = \sqrt{6 \cdot 3} = \sqrt{18}\]
We can further simplify \(\sqrt{18}\) by factoring out the largest perfect square:
\[\sqrt{18} = \sqrt{9 \cdot 2} = \sqrt{9} \cdot \sqrt{2} = 3\sqrt{2}\]
For the second term, we simply write \(\sqrt{6} \cdot 2\) as \(2\sqrt{6}\).

  1. Final Simplification and Selection

Combining the simplified terms, we obtain:
\[3\sqrt{2} + 2\sqrt{6}\]
Comparing this result with the given options:

A\(3\sqrt{2} + 2\sqrt{6}\)
B\(3 + 2\sqrt{6}\)
C\(3\sqrt{8} + 2\sqrt{6}\)

The expression matches option (a).

Examples

Simplifying radical expressions is fundamental in various fields, such as engineering and physics, where calculations often involve vectors and complex numbers. For instance, if you are designing a structural brace that forms a right-angled triangle, calculating the length of the hypotenuse using the Pythagorean theorem often leads to expressions involving roots. Being able to simplify these expressions, like converting \(\sqrt{18}\) to \(3\sqrt{2}\), makes it much easier to estimate values, perform further algebraic manipulations, or compare your results with standardized engineering formulas to ensure precision.