Algebra (Period 1)

SIMPLIFYING RADICALS 11-3

SIMPLIFYING NONPERFECT NUMBERS UNDER THE RADICAL:
A simplified radical expression is one where there is no perfect squares left under the radical sign
You can factor the expression under the radical to find any perfect squares in the number:
EXAMPLE: square root of 50 = SQRT(25 * 2)
Next, simplify the SQRT of the perfect square and leave the nonperfect factor under the radical:
SQRT(25 * 2) = SQRT(25) *SQRT(2) = 5 SQRT( 2 )
***Imagine the square root sign top extends over all numbers****
√(25 * 2) = √25 * √2 = 5√2

HELPFUL HINTS:
When you are factoring the radicand, you're looking for the LARGEST PERFECT SQUARE that is a FACTOR of the radicand.
So start with:
Does 4 go into it?
Does 9 go into it?
Does 16 go into it?
Does 25 go into it?
You can even use all the perfect squares that you have memorized... it really helps// just remember when you pull out a number... you are pulling out the number which multiples by itself to get that perfect square... NOT the perfect square itself...


Another method: Inverted Division or Factor Trees
Factor the radicand completely into its prime factors
(remember this from Pre-Algebra?)
Find the prime factorization either way in order from least to greatest.
Circle factors in PAIRS
Every time you have a pair, you have a factor that is squared!
Then, you can take that factor out of the radical sign.
Remember that you are just taking one of those factors out!
Example: SQRT 250
Prime factorization = 2 * 5 * 5 * 5
Circle the first two 5's
5 * 5 is 25 and so you can take the square root of 25 = 5 out of the radicand
Everything else is not in a pair (squared) so it must remain under the radical
Final answer: SQRT 250 = 5 SQRT 10
√250 = 5√10


VARIABLES UNDER THE SQUARE ROOT SIGN:
An even power of a variable just needs to be divided by two to find its square root
EXAMPLE: SQRT (x¹⁰ ) = x⁵ or
√x¹⁰ = x⁵
We saw this already in factoring!!!


SIMPLIFYING NONPERFECT VARIABLES:
If the variable has an even power:
EXAMPLE: The square root of 75x¹⁰ = SQRT [(25)(3)(x10 )] =
SQRT [ (5)(5)](3)(x⁵x⁵)

***Imagine the square root sign top extends over all numbers****
√75x¹⁰ = √[(25)(3)(x¹⁰ )] = √[ (5)(5)](3)(x⁵x⁵)


Simplified, you can pull out a factor of 5 and x⁵
Final simplified radical = 5x⁵ SQRT(3)
or
5x⁵√3



If the variable has an odd power:
If you have an odd power variable, simply express it as the even power one below that odd power times that variable to the 1 power:
Example: x⁵ = x⁴ x
so if you have the SQRT( x⁵ ) = SQRT (x⁴ x) = x²SQRT(x)

√x⁵ =√x⁴ x = x²√x



FACTORING A GCF FIRST, THEN FINDING A BINOMIAL SQUARED:
Sometimes you will need to factor what's under the radical before you start to simplify
Example: SQRT(3x² + 12x + 12) or

***Imagine the square root sign top extends over all numbers****
√3x² + 12x + 12

First factor out a 3:

SQRT [3 (x² + 4x + 4) ] or
***Keep imagining that line across all the numbers under the radical √ ****
√3(x² + 4x + 4)

Now factor the trinomial: SQRT [3 (x + 2) (x + 2) ] or

√3(x + 2)²

The (x + 2)² is a perfect square so SQRT [3 (x + 2) (x + 2) ] = (x + 2) SQRT( 3 )

or (x+2)√3

you need to include the HUGS around the (x +2) because you want that binomial to be multiplied by the √3 .. not just the 2.