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This module covers techniques for the simplification of radicals.

Simplifying radicals

The property ab size 12{ sqrt { ital "ab"} } {} = a size 12{ sqrt {a} } {} b size 12{ sqrt {b} } {} can be used to simplify radicals. The key is to break the number inside the root into two factors, one of which is a perfect square .

Simplifying a radical

75 size 12{ sqrt {"75"} } {}
= 25 3 because 25•3 is 75, and 25 is a perfect square
= 25 3 because ab size 12{ sqrt { ital "ab"} } {} = a size 12{ sqrt {a} } {} b size 12{ sqrt {b} } {}
= 5 3 size 12{ sqrt {3} } {} because 25 =5

So we conclude that 75 size 12{ sqrt {"75"} } {} =5 3 size 12{ sqrt {3} } {} . You can confirm this on your calculator (both are approximately 8.66).

We rewrote 75 as 25 3 because 25 is a perfect square. We could, of course, also rewrite 75 as 5 15 , but—although correct—that would not help us simplify, because neither number is a perfect square.

Simplifying a radical in two steps

180
= 9•20 because 9 20 is 180, and 9 is a perfect square
= 9 20 because ab size 12{ sqrt { ital "ab"} } {} = a size 12{ sqrt {a} } {} b size 12{ sqrt {b} } {}
= 3 20 So far, so good. But wait! We’re not done!
= 3 4•5 There’s another perfect square to pull out!
= 3 4 5
= 3 ( 2 ) 5
= 6 5 Now we’re done.

The moral of this second example is that after you simplify, you should always look to see if you can simplify again .

A secondary moral is, try to pull out the biggest perfect square you can. We could have jumped straight to the answer if we had begun by rewriting 180 as 36 5 .

This sort of simplification can sometimes allow you to combine radical terms, as in this example:

Combining radicals

75 size 12{ sqrt {"75"} } {} 12 size 12{ sqrt {"12"} } {}
= 5 3 size 12{ sqrt {3} } {} –2 3 size 12{ sqrt {3} } {} We found earlier that 75 size 12{ sqrt {"75"} } {} = 5 3 size 12{ sqrt {3} } {} . Use the same method to confirm that 12 size 12{ sqrt {"12"} } {} = 2 3 size 12{ sqrt {3} } {} .
= 3 3 size 12{ sqrt {3} } {} 5 of anything minus 2 of that same thing is 3 of it, right?

That last step may take a bit of thought. It can only be used when the radical is the same. Hence, 2 size 12{ sqrt {2} } {} + 3 size 12{ sqrt {3} } {} cannot be simplified at all. We were able to simplify 75 size 12{ sqrt {"75"} } {} 12 size 12{ sqrt {"12"} } {} only by making the radical in both cases the same .

So why does 5 3 size 12{ sqrt {3} } {} –2 3 size 12{ sqrt {3} } {} = 3 3 size 12{ sqrt {3} } {} ? It may be simplest to think about verbally: 5 of these things, minus 2 of the same things, is 3 of them. But you can look at it more formally as a factoring problem, if you see a common factor of 3 size 12{ sqrt {3} } {} .

5 3 size 12{ sqrt {3} } {} –2 3 size 12{ sqrt {3} } {} = 3 size 12{ sqrt {3} } {} ( 5 2 ) = 3 size 12{ sqrt {3} } {} ( 3 ) .

Of course, the process is exactly the same if variable are involved instead of just numbers!

Combining radicals with variables

x 3 2 + x 5 2
= x 3 + x 5 Remember the definition of fractional exponents!
= x 2 * x + x 4 * x As always, we simplify radicals by factoring them inside the root...
x 2 * x + x 4 * x and then breaking them up...
= x x + x 2 x and then taking square roots outside!
= ( x 2 + x ) x Now that the radical is the same, we can combine.

Rationalizing the denominator

It is always possible to express a fraction with no square roots in the denominator.

Is it always desirable? Some texts are religious about this point: “You should never have a square root in the denominator.” I have absolutely no idea why. To me, 1 2 size 12{ { {1} over { sqrt {2} } } } {} looks simpler than 2 2 size 12{ { { sqrt {2} } over {2} } } {} ; I see no overwhelming reason for forbidding the first or preferring the second.

However, there are times when it is useful to remove the radicals from the denominator: for instance, when adding fractions. The trick for doing this is based on the basic rule of fractions: if you multiply the top and bottom of a fraction by the same number, the fraction is unchanged. This rule enables us to say, for instance, that 2 3 size 12{ { {2} over {3} } } {} is exactly the same number as 2 3 3 3 size 12{ { {2 cdot 3} over {3 cdot 3} } } {} = 6 9 size 12{ { {6} over {9} } } {} .

In a case like 1 2 size 12{ { {1} over { sqrt {2} } } } {} , therefore, you can multiply the top and bottom by 2 size 12{ sqrt {2} } {} .

1 2 size 12{ { {1} over { sqrt {2} } } } {} = 1 * 2 2 * 2 = 2 2 size 12{ { { sqrt {2} } over {2} } } {}

What about a more complicated case, such as 12 1 + 3 size 12{ { { sqrt {"12"} } over {1+ sqrt {3} } } } {} ? You might think we could simplify this by multiplying the top and bottom by ( 1 + 3 size 12{ sqrt {3} } {} ), but that doesn’t work: the bottom turns into ( 1 + 3 ) 2 = 1 + 2 3 size 12{ sqrt {3} } {} + 3 , which is at least as ugly as what we had before.

The correct trick for getting rid of ( 1 + 3 size 12{ sqrt {3} } {} ) is to multiply it by ( 1 3 size 12{ sqrt {3} } {} ). These two expressions, identical except for the replacement of a+ by a- , are known as conjugates . What happens when we multiply them? We don’t need to use FOIL if we remember that

( x + y ) ( x - y ) = x 2 - y 2

Using this formula, we see that

( 1 + 3 ) ( 1 - 3 ) = 1 2 - ( 3 ) 2 = 1 - 3 = - 2

So the square root does indeed go away. We can use this to simplify the original expression as follows.

Rationalizing using the conjugate of the denominator

12 1 + 3 size 12{ { { sqrt {"12"} } over {1+ sqrt {3} } } } {} = 12 ( 1 - 3 ) ( 1 + 3 ) ( 1 - 3 ) = 12 - 36 1 - 3 = 2 3 - 6 -2 = - 3 + 3

As always, you may want to check this on your calculator. Both the original and the simplified expression are approximately 1.268.

Of course, the process is the same when variables are involved.

Rationalizing with variables

1 x x size 12{ { {1} over {x - sqrt {x} } } } {} = 1 x + x x x x + x size 12{ { {1 left (x+ sqrt {x} right )} over { left (x - sqrt {x} right ) left (x+ sqrt {x} right )} } } {} = x + x x 2 x size 12{ { {x+ sqrt {x} } over {x rSup { size 8{2} } - x} } } {}

Once again, we multiplied the top and the bottom by the conjugate of the denominator : that is, we replaced a- with a+ . The formula ( x + a ) ( x - a ) = x 2 - a 2 enabled us to quickly multiply the terms on the bottom, and eliminated the square roots in the denominator.

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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