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Radical Concepts -- Properties of Radicals

Module by: Kenny M. Felder. E-mail the author

Summary: This module covers some properties of radicals.

What is x2+9x2+9? Many students will answer quickly that the answer is (x+3)(x+3) and have a very difficult time believing this answer is wrong. But it is wrong.

x2x2 is x*x* 1 and 99 is 3, but x2+9x2+9 is not (x+3)(x+3).

Why not? Remember that x2+9x2+9 is asking a question: “what squared gives the answer x2+9x2+9 ?” So (x+3)(x+3) is not an answer, because (x+3)2 =x2+6x+9(x+3)2=x2+6x+9, not x2+9x2+9 .

As an example, suppose x=4x=4. So x2+9 = 42+9 = 25 =5x2+9=42+9=25=5. But (x+3)=7(x+3)=7.

Common Error:

If two numbers are added or subtracted under a square root, you cannot split them up. In symbols: a+ba+ba+ba+b or, to put it another way, x2 + y2 a+b x2 + y2 a+b

x2+9x2+9 cannot, in fact, be simplified at all. It is a perfectly valid function, but cannot be rewritten in a simpler form.

How about 9x29x2 ? By analogy to the previous discussion, you might expect that this cannot be simplified either. But in fact, it can be simplified:

9x2 =3x9x2=3x

Why? Again, 9x29x2 is asking “what squared gives the answer 9x2 9x2 ?” The answer is 3x3x because (3x)2=9x2 (3x)2=9x2.

Similarly, 9x2 =3x9x2=3x, because 3x2=9x23x2=9x2 size 12{ left ( { {3} over {x} } right ) rSup { size 8{2} } = { {9} over {x rSup { size 8{2} } } } } {}.

On the other hand...:

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

Footnotes

  1. I’m fudging a bit here: x2x2 is xx only if you ignore negative numbers. For instance, if x=–3x=–3, then x2=9x2=9, and x2x2 is 3; so in that case, x2x2 is not xx. In general, x2 =|x|x2=|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.

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