Use regex.test() if all you want is a boolean result:

console.log(/^([a-z0-9]{5,})$/.test('abc1')); // false

console.log(/^([a-z0-9]{5,})$/.test('abc12')); // true

console.log(/^([a-z0-9]{5,})$/.test('abc123')); // true

...and you could remove the () from your regexp since you've no need for a capture.

Although the match function doesn't accept string literals as regex patterns, you can use the constructor of the RegExp object and pass that to the String.match function:

var re = new RegExp(yyy, 'g');
xxx.match(re);

Any flags you need (such as /g) can go into the second parameter.

You can use a string as a regular expression using the RegExp Object:

var myPattern = new RegExp('.'+myWord,'g'); 

Fiddle Example

Doing a single match in your case, is simply changed the second parameter in RegExp from g to m (which means to make one match per line for multi lines, but in this case a strings is simply all one line). For finding the word "wood" from "ood","od","d", or other cases. You can do this:

var myPattern = new RegExp("\\b\\w+"+myWord+"\\b",'m'); 

Note I have a solution in the comments below, but this one is better.

The items \\b ... \\b mean word boundaries. Basically ensuring that it matches a single word. Then the \\w means any valid "word character". So overall the regexp means (using myWord = "od"):

|word boundary| + (1 or more word characters) + "od" + |word boundary|

This will ensure that it matches any words in the string than end with the characters "od". Or in a more general case you can do:

var myPattern = new RegExp("\\b\\w*"+myWord+"\\w*\\b",'m');