Calculus and Vectors – How to get an A+
1.1 Radical Expressions: Rationalizing Denominators
Ex 1. Simplify:
A Radicals
a a =a
n
a)
( a) = a
n
m
( a )=
m
an
3 3 =3
b) ( 5 ) 3 = ( 5 ) 2 5 = 5 5
n
c) (3 7 ) 5 = (3 7 ) 3 (3 7 ) 2 = 7(3 7 ) 2 = 7 3 7 2 = 7 3 49
= (n a ) m
Note: If n is even, then a ≥ 0 for
n
a.
B Rationalizing Denominators (I)
a
a
c a c
=
=
bc
b c b c c
Ex 2. Rationalize:
2
2
5 2 5 2 5
=
=
=
3 5 3 5 5 3 × 5 15
C Conjugate Radicals
a+ b ⇔a− b
Ex 3. For each expression, find the conjugate radical.
a) 2 + 3
⇒ 2− 3
a+ b⇔ a− b
b)
2− 3
a +b c ⇔ a −b c
c)
3+2 5
a b +c d ⇔a b −c d
⇒
2+ 3
⇒
d) 2 5 + 3 7
3−2 5
⇒ 2 5 −3 7
Ex 4. Use the difference of squares identity to simplify:
D Difference of squares identity
(a + b)(a − b) = a 2 − b 2
a) (a + b )(a − b ) = a 2 − ( b ) 2 = a 2 − b
b) ( a + b )( a − b ) = ( a ) 2 − ( b ) 2 = a − b
c) ( a + b c )( a − b c ) = ( a ) 2 − (b c ) 2 = a − b 2 c
E Rationalizing Denominators (II)
Hint: Multiply and divide by the
conjugate radical e of the denominator.
Ex 5. Rationalize the denominator:
3(1 + 2 )
3(1 + 2 )
3
3 1+ 2
=
= 2
=
= −3(1 + 2 )
a)
2
1− 2
1 − 2 1− 2 1 + 2 1 − ( 2)
b)
4
2+3 5
=
4
2−3 5
2+3 5 2−3 5
=
4(2 − 3 5 )
2
2 − (3 5 )
2
=
4(2 − 3 5 )
4(2 − 3 5 )
=−
4 − 9×5
41
c)
2
3− 6
=
3+ 6
2
3− 6
3+ 6
=
2( 3 + 6 )
2
( 3) − ( 6 )
2
=
2( 3 + 6 )
2( 3 + 6 )
=−
3−6
3
F Rationalizing Numerators
Hint: Multiply and divide by the
conjugate radical of the numerator.
Ex 6. Rationalize the numerator:
( 5)2 − ( 3)2
5− 3
5− 3 5+ 3
2
=
=
=
2 −1
2 − 1 5 + 3 ( 2 − 1)( 5 + 3 ) ( 2 − 1)( 5 + 3 )
G Equivalent Expressions
Hint: You may get equivalent
expressions by rationalizing the
numerator or denominator.
Note: State restrictions.
Ex 7. Find equivalent expressions by rationalizing. State restrictions.
x −1
x − 1 x + 1 ( x − 1)( x + 1)
a)
=
=
= x + 1, x ≥ 0, x ≠ 1
x −1
x −1
x −1 x +1
b)
x+9 −3
=
x
x+9 −3
x
x+9 +3
x+9 +3
=
x ≥ −9, x ≠ 0
1.1 Radical Expressions: Rationalizing Denominators
© 2010 Iulia & Teodoru Gugoiu - Page 1 of 2
x
x( x + 9 + 3)
=
1
x+9 +3
,
Calculus and Vectors – How to get an A+
1
1
1
1
−
x+h
x
x+h x
=
h
⎛ 1
c)
1
h⎜⎜
+
x
⎝ x+h
x + h > 0, x > 0, h ≠ 0
H More algebraic identities
a 3 − b 3 = (a − b)(a 2 + ab + b 2 )
a 3 + b 3 = (a + b)(a 2 − ab + b 2 )
a 4 − b 4 = (a − b)(a + b)(a 2 + b 2 )
−
⎞
⎟
⎟
⎠
=−
1
⎛ 1
1 ⎞
⎟
x( x + h)⎜⎜
+
x ⎟⎠
⎝ x+h
,
Ex 8. For each case, the numerator and denominator have a common
zero. Use algebraic identities to eliminate the common zero. State
restrictions.
x − 1 (3 x ) 3 − 1 (3 x − 1)((3 x ) 2 + 3 x + 1) 3 2 3
a)
=
=
= x + x + 1, x ≠ 1
3
3
3
x −1
x −1
x −1
b)
x4 −1
x3 − 1
=
( x − 1)( x + 1)( x 2 + 1)
( x − 1)( x 2 + x + 1)
=
( x + 1)( x 2 + 1)
x2 + x +1
Reading: Nelson Textbook, Pages 6-8
Homework: Nelson Textbook: Page 9, #1a, 2a, 3a, 4a, 5, 6a, 7ac
1.1 Radical Expressions: Rationalizing Denominators
© 2010 Iulia & Teodoru Gugoiu - Page 2 of 2
,
x ≠1