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This module covers some properties of radicals.

What is x 2 + 9 ? Many students will answer quickly that the answer is ( x + 3 ) and have a very difficult time believing this answer is wrong. But it is wrong.

x 2 is x *

I’m fudging a bit here: x 2 is x only if you ignore negative numbers. For instance, if x = –3 , then x 2 = 9 , and x 2 is 3; so in that case, x 2 is not x . In general, x 2 = | x | . However, this subtlety is not relevant to the overall point, which is that you cannot break up two terms that are added under a radical.
and 9 is 3, but x 2 + 9 is not ( x + 3 ) .

Why not? Remember that x 2 + 9 is asking a question: “what squared gives the answer x 2 + 9 ?” So ( x + 3 ) is not an answer, because ( x + 3 ) 2 = x 2 + 6x+9 , not x 2 + 9 .

As an example, suppose x = 4 . So x 2 + 9 = 4 2 + 9 = 25 = 5 . But ( x + 3 ) = 7 .

If two numbers are added or subtracted under a square root, you cannot split them up. In symbols: a + b a + b or, to put it another way, x 2 + y 2 a + b

x 2 + 9 cannot, in fact, be simplified at all. It is a perfectly valid function, but cannot be rewritten in a simpler form.

How about 9 x 2 ? By analogy to the previous discussion, you might expect that this cannot be simplified either. But in fact, it can be simplified:

9 x 2 = 3 x

Why? Again, 9 x 2 is asking “what squared gives the answer 9 x 2 ?” The answer is 3 x because ( 3 x ) 2 = 9 x 2 .

Similarly, 9 x 2 = 3 x , because 3 x 2 = 9 x 2 size 12{ left ( { {3} over {x} } right ) rSup { size 8{2} } = { {9} over {x rSup { size 8{2} } } } } {} .

If two numbers are multiplied or divided under a square root, you can split them up. In symbols: ab = a b , a b = a b

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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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