Adding radicals is much like adding variables. Note that x=1x; the 1 is understood to be there, but never written. Similarly, \u221a3=1\u221a3. Therefore, 2\u221a5 + \u221a5 = 3\u221a5<\/p>\n
You can’t simplify x plus y; x+y is just x+y. Similarly, \u221a6+\u221a7 doesn’t simplify. A calculator will give you a decimal for it, but that’s not an exact value.<\/p>\n
So what about the following:<\/p>\n
Example 1: Simplify \u221a(12)+\u221a(75)+\u221a3<\/p>\n
You may want to read up on simplifying radicals in my earlier post<\/a> to understand what follows.<\/p>\n Solution:<\/strong> On the face of it, we can’t add the radicals because they’re not the same kind. However, we can simplify some of them as follows:<\/p>\n \u221a(12) = 2\u221a3<\/p>\n \u221a(75) = 5\u221a3<\/p>\n Simplifying reveals them to be the same kind. We indeed can<\/em> add them.<\/p>\n \u221a(12)+\u221a(75)+\u221a3 becomes<\/p>\n 2\u221a3 + 5\u221a3 + \u221a3 = 8\u221a3<\/p>\n By a similar principle, \u221a(50)-\u221a8=5\u221a2-2\u221a2=3\u221a2<\/p>\n I’ll be covering some different examples in an upcoming post:)<\/p>\n