Algebra (Period 1)

Radical Expressions 11-2


If an expression under the square root sign is NEGATIVE, it does not exist in the REAL numbers!


There is no number that you can square and get a NEGATIVE PRODUCT



VARIABLES UNDER THE SQUARE ROOT SIGN: 
If you have a variable under the square root sign,
you need to determine what values of the variable will keep the radicand greater than or equal zero


The square root of x then is only real when x is greater than or equal to zero


The SQRT (x + 2) is only real when x + 2 is greater than or equal to 0

Set x + 2 greater than or equal to 0 and solve as an inequality!

You will find that x must be greater than or equal to -2



SPECIAL CASE!!!! a variable squared plus a positive integer under radical: 
If you're trying to find the principal square root of x² (or any variable squared) plus a positive integer, then all numbers will work because a squared number will always end up either positive or zero!


Example: √(x² + 3) under the radical, any number positive or negative will keep the radicand positive (real), because once you square it, it is positive.


Then you're just adding another positive number.



If there is a variable squared and then a negative number (subtraction), the square will need to be equal or greater than that negative number to stay zero or positive under the radical.

EXAMPLE: √(x² - 10)

x² must be equal or greater than 10, so x must be at least the square root of 10

(the square root of 10 squared is 10)



ANOTHER SPECIAL CASE!!!!!!!!!!

ANY RADICAL EXPRESSION THAT HAS A VARIABLE SQUARED IS SIMPLIFIED TO THE ABSOLUTE VALUE OF THE VARIABLE.


Example: The square root of x² is the absolute value (positive) of x SHown here as: ( I x I )


Why? 
Because it is assumed that you're finding the PRINCIPAL (positive) square root.


EXAMPLE:
x = -3
 √x² = √(-3)² = √9 = 3 (not -3)

so you have to put absolute value signs around the answer


IF THERE IS A VARIABLE SQUARED
(see p. 489 #17-30)


TRINOMIALS UNDER THE RADICAL:

What do you think you would do if you saw x² + 10x + 25 under the radical sign????


FACTOR IT! 


IT MAY BE A PERFECT SQUARE (a binomial squared!)



EXAMPLE:
√( x² + 10x + 25) factors to
√(x + 5)² = I x + 5 I

Determine the values for the variable that will make each expression a real number
√m(m+3)

you know that m(m+3) ≥ 0 so m ≥0 OR m ≤ -3

√x²(x-3)

again set x²(x-3) ≥ 0 and you discover x = 0 or x ≥3

Given a and c, what must be true of b to make

√b² -4ac
a real number?

a = -3 and c = 2
substitute in and we have

√b² -4(-3)(2)

√b² +24
b can be any real number!!

But what if we have
a = 2 and c = 8
√b² -4ac
substitute in
√b² -4(2)(8)
√b² -64
b is either
b≤ -8 or b ≥ 8


Determine whether each of the follow statements is sometimes, always or never true:

√a² + b ² is a real number ---> ALWAYS

√3 - t is a real number for t ≥ 3 ---> SOMETIMES

√a² - b ² is a real number ---> SOMETIMES

√a² + 2ab + b ² is a real number ---> ALWAYS


For a polynomial in the form ax² + bx + c = 0 to have real solutions,
√b² -4ac must be a real number. Which of the following polynomials have real solutions?

x² - 12x + 3 = 0 is real because √b² -4ac = √12² -4(1)(3) =
√144 -12 √132 and that's real

x² + 5x + 7 = 0 is NOT REAL because √5² -4(1)(7) = √25 -28 = √-3 which is NOT REAL