Transcript Dividing Radicals, Solving Radical Equations

Chapter 9 Continued – Radical Expressions and Equations
Multiplication & Division of Radical Expressions
When multiplying radical expressions, you can just put
everything that’s under a radical sign together under one big
radical sign.
Multiply :
3 x 4 2 x 2 y 6 xy2
 3 x 4  2 x 2 y  6 xy2
 36 x 7 y 3
 36 x 6  x  y 2  y
 6 x 3 y xy

Distributive Property:
2x x  2x

 x 2 x  ( 2 x )( 2 x )
 x 2x  2x
BE CAREFUL!
Also,
2 x does not equal 2 x
2 x should be written as x 2
to avoid confusion.
2 x does not equal 2 x
since the 2 on the outside of
is not 2
x
Example 3 Multiply
2

x  y 5 x 2 y

Use FOIL
2 x 5 x  2 x  2 y   y 5 x   y  2 y 
 10 x
 4 xy
 5 xy
 2y
 10 x  9 xy  2 y
You try th is one :
Multiply
3
Example
Multiply
2


x  y 5 x 2 y
x 7 2 x 7
  

 

 
 2 x 2 x   7  2 x  7  2 x  (7)( 7)
 4 x  49
Notice that this was the special factoring case of the difference of two squares :
(a  b)(a-b)  a 2 -b 2
(a  b) and (a-b) are called conjugates of each other.
Notice that when ra adical expression has two terms, all
radicals disappear when you multiply the expression by its
conjugate.
Try this one:
x 1 x 1



Radical Expressions in Simplest Form
A radical expression is in simplest form if:
1. The radicand contains no factor greater than 1 that is a perfect
square.
2. There is no fraction under the radical sign.
3. There is no radical in the denominator of a fraction.
Quotient Property of Square Roots
a
a

b
b
is not in simplest form because there is a fraction under the
radical sign. This can be simplified by taking the square root
of the numerator and the denominator.
4x2
4x2 2x

 3
6
6
z
z
z
Simplify
4x2 y
xy

4x2 y
xy
 4x
2 x
2
3
Is not in simplest form because there is a radical expression in
the denominator;
The way to simplify is to multiply both numerator and
denominator by 3
2
3 2 3


3
3 3
This doesn’t always work when there is a two-term expression with
at least one radical term added to another term.
2y
y 3

y 3
y 3

2y


y 3
 y 6
2
UGH!
y 9
The trick for these types is to multiply the numerator and
denominator by the conjugate.
2y
y 3

y 3
y 3




2y


y 3

y 3

y 3
2 y2  3 2 y
 y 3
2
2
y 2 3 2y
y 9
SIMPLIFIED!
Solving Equations Containing Radical Expressions
Property of Squaring Both Sides of an Equation
If a and b are real numbers and a=b, then a2=b2
3x  2  5
Solve :
3x  3
Square both sides to get x out from under the radical sign.
 3x 
2
 32
3x  9
x3
Check :
3(3)  2  5
925
3  2  5 TRUE
It’s very important to check your solution because some
“solutions” actually make the original equation untrue.
Example:
x  12  3
x  9
Notice that when you get the constants on one side, your equation
says that the radical expression must equal a negative number.
This is impossible! Therefore there is NO SOLUTION to an
equation like this.
Example :
30  x  x
30-x  x 2
0  x 2  x-30
0  ( x  6)( x  5)
x  6 or x  5
Check x  6
30  (6)  6
square both sides
This is now a degree 2 equation so put it in
standard form, factor it, then use zero-product
rule.
Impossible because the principal square root
of a number can never be negative.
Therefore -6 is not a possible solution.
Check x  5
30  (5)  5
25  5
55
OK
Therefore, only solution is {5}
You try!
Solve:
a=
Solve equation and exclude any extraneous solutions:
m=
Solve:
x  x 5 1
In this case put the radical expression s on opposites of each other
before squaring both sides.
x  1 x  5
 x   1 
2
x 5

2
x  1  2 x  5   x  5
This got rid of the radical on the left but we
still have a radical on the right. Combine like term
then start the process again.
x  1  2 x  5   x  5
x  x  1 2 x  5  x  x  5
0  4  2 x  5
42  2
x5
16  4( x  5)
16  4 x  20
36  4 x
9x
CHECK :

9  95 1
3 4 1
3 2 1
2