Algebra: Simplifying Radical Expressions
Simplifying Radical Expressions
Think of a radical symbol like a prison, and the pieces of the radicand as inmates. Not all the prisoners are doomed to a life sentence, trapped inside the dank (and foul-smelling) radical big house—there is a chance of parole. However, in order to be released from the radical sign, you must meet its parole requirements.
Specifically, a radical will only release things raised to a power that matches its index. So, square roots will only release pieces of its radicand that are raised to the second power, and a radical with an index of 5 will only release things raised to the fifth power. When asked to simplify radicals, what you're actually doing is paroling the factors within that meet the requirements, and leaving the rest inside to rot.
Example 1: Simplify the radical expressions.
- Solution: Start by factoring the radicand's coefficient; in other words, write it as a product of smaller numbers. However, since the index of the radical is 3, you want the factors to be powers of 3 if possible. Therefore, instead of writing 16 as 16 · 1 or 4 · 4, write it as 8 · 2, or 23 · 2, since 8 is a perfect cube and can be written as 23.
- Now turn your attention to the variables. You can rewrite x4 as x3 · x (since x3 · x = x3 + 1 = x4 so it contains an exponent of 3. Luckily, y6 is a perfect cube, since y2 = y6, so write it as with that all-important power of 3 as well: (y2)3.
- Of all the pieces in the radicand, only 23, x3, and (y2)3 contain powers of 3. Yank them out in front of the radical, stripping away the third power as they exit the prison, which leaves only 2 and x inside.
To make simplifying radicals easier, memorize the first 15 perfect squares (1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225) and the first 5 perfect cubes (1, 8, 27, 64, 125).
Why do you rewrite y6 as (y2)3 in Example 1(a)? Basically, you're trying to make groups of three things, so that they can be released from the radical. It might help to think of (y2)3 as a group of three y2's, and (y2)3 = y6 thanks to exponential Rule 3 from Encountering Expressions.
- Solution: Since this is a square root, you want as much of the radicand as possible to be raised to the second power. The coefficient 18 only has one factor that's a perfect square: 9; so, rewrite 18 as the product 2 · 9 (or 2 · 32). The x2 term already has an exponent of 2, but you should rewrite the y3 term as y2 · y, to identify the y2 as a candidate for release.
- Pull everything with an exponent of 2 outside the radical (and take away the power as you do), leaving what's left as the new radicand.
You've got to be wondering where the heck those absolute value signs came from! I kind of sprung them on you, and I apologize. However, there's a rule in algebra that says if you ever have the expression (the power of the exponent matches the index of the radicand), and n is even, then the simplified result is |x|. (This is because you always want an even-powered root to have a positive answer, and those absolute values make sure that no matter what the variable's value is, the answer will be positive.)
Don't forget, if you have a variable in a radicand raised to the same, even power as its index, you should surround it with absolute value symbols once it is released.
Since you're releasing both x2 and y2 from this square root, and those exponents match the index of the radical, you've got to toss them inside absolute value signs once they're paroled.
You've Got Problems
Problem 1: Simplify the radical √300x6y3.
Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.