Question

ding and subtracting numeric radical expressions
find the sum: $4\sqrt{3} + 11\sqrt{12}$
$15\sqrt{15}$ $48\sqrt{3}$ $26\sqrt{3}$
$15\sqrt{3}$

Explanation:

Step1: Simplify \(\sqrt{12}\)

We know that \(12 = 4\times3\), so \(\sqrt{12}=\sqrt{4\times3}\). By the property of square roots \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (where \(a = 4\), \(b=3\) and \(a\geq0,b\geq0\)), we have \(\sqrt{4\times3}=\sqrt{4}\times\sqrt{3}\). Since \(\sqrt{4} = 2\), then \(\sqrt{12}=2\sqrt{3}\).

Step2: Substitute and simplify the expression

The original expression is \(4\sqrt{3}+11\sqrt{12}\). Substitute \(\sqrt{12}=2\sqrt{3}\) into the expression, we get \(4\sqrt{3}+11\times(2\sqrt{3})\). First, calculate \(11\times2 = 22\), so the expression becomes \(4\sqrt{3}+22\sqrt{3}\). Now, since the terms have the same radical part (\(\sqrt{3}\)), we can combine like terms. Using the distributive property \(a\sqrt{c}+b\sqrt{c}=(a + b)\sqrt{c}\) (where \(a = 4\), \(b = 22\), \(c=3\)), we have \((4 + 22)\sqrt{3}=26\sqrt{3}\).

Answer:

\(26\sqrt{3}\)